Inductors and lamps puzzle
Ioannis
pressure sodium lamps and metal halide lamps (all these are relatively high
power lamps that are used to illuminate streets and various other spaces) are
what are called "negative resistance" devices:
When they are connected to the AC outlet directly, they cause more and more
current to be drawn, until they either blow the fuse or destroy themselves.
To prevent this from happening, they are connected in series with what's
called a "ballast". A ballast is an inductive coil which limits the circuit
current to i_b Amperes. The value i_b is characteristic of the ballast used.
In other words, there is a one-to-one correspondance between ballast type and
the amperage it provides when its connected in series with a specific lamp.
The ballast is (almost) pure inductive resistance R_b, therefore the ballast
current and resistance are related via Ohm's Law for AC: i_b = 220V (or 110V)
/R_b. Let's assume that we are working with 220V AC.
Since ballasts are AC resistors, they can be wired in series or in parallel.
Example: A HID lamp that requires 1.3A to operate, can use a ballast with R_b
~= 169.23 Ohm in AC.
220V/169.23Ohm ~= 1.3A
But it can also use two "larger"[*] ballasts with R_{b1} = R_{b2} ~= 84.6 Ohm
wired in series:
220V/(84.6+84.6)Ohm ~= 1.3A
The circuit can also use two "smaller" ballasts with R_{b1} = R_{b2}~=338.4Ohm
wired in parallel:
In this case the total resistance R_T is given by:
1/R_T = 1/R_{b1} + 1/R_{b2}, =>
R_T = R_{b1}*R_{b2}/(R_{b1} + R_{b2}) =
=338.4^2/(2*338.4)
=338.4/2
=169.23 Ohm.
Therefore the circuit's total amperage i_T will again be:
i_T = 220/R_T ~= 1.3A.
Additionally, the circuit can also use two /different/ ballasts, with R_{b1} +
R_{b2} = 169.23Ohm, wired in series.
And it can also use two different ballasts in parallel, provided the total
resistance R_T from the parallel connection expression, comes out to be 169.23
Ohm. I am too lazy to cook up an example.
The puzzle: You are given n different ballasts, which provide for
corresponding circuit amperages {i_1, i_2,...i_n}.
1) Determine how many different circuit amperages i_m you can get with these n
ballasts. (Without using brute force?)
2) Specifically[**]: I have a lamp that requires amperage i_L. Can I find a
combined connection with some of my n ballasts (some of them in parallel,
others in series, etc) that will provide a "close" amperage i_B for my HID
lamp? Let's say close is |i_L - i_B| <= epsilon, for some specified epsilon.
3) Same questions as 1) and 2) but allow some of the n ballasts to be
identical (i.e. some of the ballasts in the collection have i_k = i_m)
Many thanks,
[*] By convention ballasts with smaller R_b are termed "larger" because they
drive lamps of larger ratings, in Watts.
[**] This is an actual problem I encountered with by collection of lamps and
ballasts, so if anyone can come up with a Maple solution to the above puzzle,
I'd appreciate it.
--
Ioannis
-------
The best way to predict reality, is to know exactly what you DON'T want.
Ioannis
news:1164583793.491748@athnrd02...
[snip]
> The puzzle: You are given n different ballasts, which provide for
> corresponding circuit amperages {i_1, i_2,...i_n}.
>
> 1) Determine how many different circuit amperages i_m you can get with these
n
> ballasts. (Without using brute force?)
Just to clarify: 1) asks how many different amperages i_m one can get using
the n ballasts, but in different wiring combinations. So, for example, when
there are only two ballasts {b_1, b_2} of amperages {i_1, i_2}, there are
obviously four solutions: the current of ballast b_1 alone in the circuit, the
current of ballast b_2 alone in the circuit, the current of ballasts b_1 and
b_2 in series and the current of ballasts b_1 and b_2 in parallel. So the
solution for n=2, {i_1, i_2} is:
i_m \in {i_1, i_2, i_1*i_2/(i_1 + i_2), i_1 + i_2}
> 2) Specifically[**]: I have a lamp that requires amperage i_L. Can I find a
> combined connection with some of my n ballasts (some of them in parallel,
> others in series, etc) that will provide a "close" amperage i_B for my HID
> lamp? Let's say close is |i_L - i_B| <= epsilon, for some specified epsilon.
Obviously in 2) I am interested for i_L =/= i_n for all n. Otherwise choose
the ballast that gives i_n.
James Waldby
> "Ioannis" ... wrote ...
> [snip]
> > The puzzle: You are given n different ballasts, which provide for
> > corresponding circuit amperages {i_1, i_2,...i_n}.
> >
> > 1) Determine how many different circuit amperages i_m you can get
> > with these n ballasts. (Without using brute force?)
>
> Just to clarify: 1) asks how many different amperages i_m one can get using
> the n ballasts, but in different wiring combinations. So, for example, when
> there are only two ballasts {b_1, b_2} of amperages {i_1, i_2}, there are
(Insert left-out-_'s at obvious places in following.)
The following recursive approach almost works:
Let L = {L1, L2 ...} be the given values {i_1, i_2, ...}.
Let Nj be the set of possible circuits with up to j items from
among j items. Let Mj,k be the set of possible circuits with
exactly k items from among j items. (That is, if C is in Mj,k,
then C is a properly connected circuit made up of k items.)
Thus, Nj = union over k=1 to j of Mj,k.
We suppose that item values L1, L2... are completely general, such
that if C is in Mj,k and Lq is in C and Lp is not, and p < j+1,
then substituting Lp for Lq gives C' in Mj,k, and C != C'.
Now consider augmenting Nj with Lr, r = j+1. For each circuit
C in Mj,k, we can form a member of Mr,(k+1) with Lr in series with C,
and another with it in parallel, and can form k members of Mr,k by
substituting Lr in place of each of the k different Li in C. Then
|Mr,k| = 2*|Mj,(k-1)| + k*|Mj,k| if 1 < k < r. (When k = r, this doesn't
hold; the approach gives an easy and rigorous upper bound, however.)
-jiw
Ioannis
news:456A9A5A...@pat7.com...
[snip]
> The following recursive approach almost works:
>
> Let L = {L1, L2 ...} be the given values {i_1, i_2, ...}.
> Let Nj be the set of possible circuits with up to j items from
> among j items. Let Mj,k be the set of possible circuits with
> exactly k items from among j items. (That is, if C is in Mj,k,
> then C is a properly connected circuit made up of k items.)
> Thus, Nj = union over k=1 to j of Mj,k.
>
> We suppose that item values L1, L2... are completely general, such
> that if C is in Mj,k and Lq is in C and Lp is not, and p < j+1,
> then substituting Lp for Lq gives C' in Mj,k, and C != C'.
>
> Now consider augmenting Nj with Lr, r = j+1. For each circuit
> C in Mj,k, we can form a member of Mr,(k+1) with Lr in series with C,
> and another with it in parallel, and can form k members of Mr,k by
> substituting Lr in place of each of the k different Li in C. Then
> |Mr,k| = 2*|Mj,(k-1)| + k*|Mj,k| if 1 < k < r. (When k = r, this doesn't
> hold; the approach gives an easy and rigorous upper bound, however.)
Thanks. I think you almost got it. I say "almost", because just as soon as I
went to bed, I also came up with a complete recursive solution. I think that
the step n = 2 also solves the recursive step.
Let I_n = {i_1, i_2, ..., i_n} be the set of amperages from connections
obtained with n ballasts.
Now let i_k be in I_n. Adding one more ballast with amperage i_{n+1}, and
treating i_k as a single connection/amperage, we have the following recurence:
I_{n+1} = {i_k, i_{n+1}, i_k*i_{n+1}/(i_k + i_{n+1}), i_k + i_{n+1}: i_k \in
I_n}
I think the above gives a complete recursive enumeration of all amperages.
For question 2) then, we have to enumerate I_n and calculate the set S =
min{|i_L - i_B|: i_B \in I_n}
If the set S contains members s_i for which s_i <= epsilon holds, we are done.
We pick one of them. Otherwise we say there is no solution.
> -jiw
Ioannis
[snip]
> Thanks. I think you almost got it. I say "almost", because just as soon as I
> went to bed, I also came up with a complete recursive solution. I think that
> the step n = 2 also solves the recursive step.
>
> Let I_n = {i_1, i_2, ..., i_n} be the set of amperages from connections
> obtained with n ballasts.
>
> Now let i_k be in I_n. Adding one more ballast with amperage i_{n+1}, and
> treating i_k as a single connection/amperage, we have the following
recurence:
>
> I_{n+1} = {i_k, i_{n+1}, i_k*i_{n+1}/(i_k + i_{n+1}), i_k + i_{n+1}: i_k \in
> I_n}
>
> I think the above gives a complete recursive enumeration of all amperages.
To clarify, let us look at the enumeration sets I_n for n \in {1,2,3}:
I_1 = {i_1}
I_2 = {i_1, i_2, i_1*i_2/(i_1 + i_2), i_1 + i_2}
I_3 = {i_k, i_3, i_k*i_3/(i_k + i_3), i_k + i_3: i_k \in I_2} =
= {i_1, i_2, i_1*i_2/(i_1 + i_2), i_1 + i_2,
i_3,
i_1*i_3/(i_1 + i_3), i_2*i_3/(i_2 + i_3),
i_1*i_2/(i_1 + i_2)*i_3/(i_1*i_2/(i_1 + i_2) + i_3),
(i_1 + i_2)*i_3/(i_1 + i_2 + i_3),
i_1 + i_3, i_2 +i_3, i_1*i_2/(i_1 + i_2) + i_3, i_1 + i_2 + i_3}
if I've got all the substitutions right on the recursion.
I have no clue how to program such an enumeration with Maple or how to get
|I_n|, though.
So far it looks like |I_1| = 1, |I_2| = 4 and |I_3| = 13.
Any takers?
Thanks!
The Last Danish Pastry
I do not think that your recursion is correct.
Let: i_1=A, i_2=B, i_3=C
You miss out, for example, ((A in parallel with C) in series with B).
--
Clive Tooth
http://www.shutterstock.com/cat.mhtml?gallery_id=61771
Joe Riel
> So far it looks like |I_1| = 1, |I_2| = 4 and |I_3| = 13.
|I_3| = 17:
1. 1
2. 2
3. 3
4. 1 + 2
5. 1 + 3
6. 2 + 3
7. 1 | 2
8. 1 | 3
9. 2 | 3
10. 1 + 2 | 3
11. 2 + 1 | 3
12. 3 + 1 | 2
13. (1 + 2) | 3
14. (1 + 3) | 2
15. (2 + 3) | 1
16. 1 + 2 + 3
17. 1 | 2 | 3
.|. is the parallel operator. 1 = L1, 2 = L2, 3 = L3.
--
Joe Riel
Ioannis
news:4t0o2uF...@mid.individual.net...
[snip]
> I do not think that your recursion is correct.
>
> Let: i_1=A, i_2=B, i_3=C
>
> You miss out, for example, ((A in parallel with C) in series with B).
Hmmm. I see. You and Joe are right. Switching to Joe's notation, it's missing
at least the 1 | 3 + 2 connection. I think the reason the recursion misses
those is because it considers i_k \in I_n as a single block. Obviously this is
not the case with your example above.
Perhaps there is no recursive relationship that describes all the connections.
From Joe's response I smell combinatorics here and that scares me.
Thanks to both.
> --
> Clive Tooth
Ioannis
> The following recursive approach almost works:
>
> Let L = {L1, L2 ...} be the given values {i_1, i_2, ...}.
> Let Nj be the set of possible circuits with up to j items from
> among j items. Let Mj,k be the set of possible circuits with
> exactly k items from among j items. (That is, if C is in Mj,k,
> then C is a properly connected circuit made up of k items.)
> Thus, Nj = union over k=1 to j of Mj,k.
>
> We suppose that item values L1, L2... are completely general, such
> that if C is in Mj,k and Lq is in C and Lp is not, and p < j+1,
> then substituting Lp for Lq gives C' in Mj,k, and C != C'.
>
> Now consider augmenting Nj with Lr, r = j+1. For each circuit
> C in Mj,k, we can form a member of Mr,(k+1) with Lr in series with C,
> and another with it in parallel, and can form k members of Mr,k by
> substituting Lr in place of each of the k different Li in C. Then
> |Mr,k| = 2*|Mj,(k-1)| + k*|Mj,k| if 1 < k < r. (When k = r, this doesn't
> hold; the approach gives an easy and rigorous upper bound, however.)
Back to your solution. I /think/ I am starting to understand it now.
The only sad thing to me seems that you are using a counting argument which
allows only for the calculation of |Mr,k|.
Looks like an explicit enumeration of Mr,k may be harder...
But thanks again, anyway. Your solution (and any others) will eventually go to
my webpages.
> -jiw
The Last Danish Pastry
> "The Last Danish Pastry" <cli...@gmail.com> wrote in message
> news:4t0o2uF...@mid.individual.net...
> [snip]
>
>> I do not think that your recursion is correct.
>>
>> Let: i_1=A, i_2=B, i_3=C
>>
>> You miss out, for example, ((A in parallel with C) in series with
>> B).
>
> Hmmm. I see. You and Joe are right. Switching to Joe's notation,
> it's missing
> at least the 1 | 3 + 2 connection. I think the reason the recursion
> misses
> those is because it considers i_k \in I_n as a single block.
> Obviously this is
> not the case with your example above.
>
> Perhaps there is no recursive relationship that describes all the
> connections.
> From Joe's response I smell combinatorics here and that scares me.
You might want to look at
http://www.research.att.com/~njas/sequences/A000084
A000084[n] is the number of ways of arranging n resistors in a circuit
consisting entirely of series and parallel elements.
Gerry Myerson
wrote:
Unfortunately, the online encyclopedia of integer sequences gives
110 sequences with 1, 4, 17. Anyone have I_4 ?
Just looking at the first page, this one looks a possibility:
A056542
a(n) = n*a(n-1) + 1, a(1) = 0.
0, 1, 4, 17, 86, 517, 3620, 28961, 260650, 2606501, 28671512, 344058145,
4472755886, 62618582405, 939278736076, 15028459777217, 255483816212690,
4598708691828421, 87375465144740000, 1747509302894800001
COMMENT
For n>=2 also operation count to create all permutations of n distinct
elements using Algorithm L (lexicographic permutation generation) from
Knuth's The Art of Computer Programming, Vol. 4, chapter 7.2.1.2.
Sequence gives number of loop repetitions of the j search loop in step
L2. - Hugo Pfoertner (hugo(AT)pfoertner.org), Feb 06 2003
This sequence shares divisibility properties with A000522; each of the
primes in A072456 divide only a finite number of terms of this sequence.
- T. D. Noe (noe(AT)sspectra.com), Jul 07 2005
REFERENCES
Donald E. Knuth: The Art of Computer Programming, Volume 4,
Combinatorial Algorithms, Volume 4A, Enumeration and Backtracking.
Pre-fascicle 2B, A draft of section 7.2.1.2: Generating all
permutations. Available online; see link.
LINKS
Donald E. Knuth, TAOCP Vol. 4, Pre-fascicle 2b (generating all
permutations).
Hugo Pfoertner, FORTRAN implementation of Knuth's Algorithm L for
lexicographic permutation generation.
Tom Muller, Prime and Composite Terms in Sloane's Sequence A056542,
Journal of Integer Sequences, Vol. 8 (2005), Article 05.3.3.
T. D. Noe, Table of n, a(n) for n=1..100
FORMULA
a(n) = floor[(e-2)*n! ] =A002627(n)-n! =A000522(n)-2*n! =n!-A056543(n).
a:=n->sum((n-j)!*binomial(n,j),j=2..n). - Zerinvary Lajos
(zerinvarylajos(AT)yahoo.com), Jul 31 2006
EXAMPLE
a(4)=4*a(3)+1=4*4+1=17
MAPLE
a:=n->sum((n-j)!*binomial(n, j), j=2..n): seq(a(n), n=1..20); -
Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 31 2006
MATHEMATICA
tmp=0; Join[{tmp}, Table[tmp=n*tmp+1, {n, 2, 100}]] (T. D. Noe
(noe(AT)sspectra.com), Jul 12 2005)
FoldList[ #1*#2 + 1 &, 0, Range[2, 21]] (from Robert G. Wilson v
(rgwv(at)rgwv.com), Oct 11 2005)
AUTHOR
Henry Bottomley (se16(AT)btinternet.com), Jun 20 2000
EXTENSIONS
More terms from James A. Sellers (sellersj(AT)math.psu.edu), Jul 04 2000
If you search for resistors at that site, you get a number of hits,
none of which seems to be exactly right. The closest is
A005840
Expansion of (1-x)*e^x/(2-e^x).
1, 1, 2, 8, 46, 332, 2874, 29024, 334982, 4349492, 62749906, 995818760,
17239953438, 323335939292, 6530652186218, 141326092842416,
3262247252671414, 80009274870905732, 2077721713464798210,
56952857434896699992, 1643312099715631960910 (list; graph; listen)
COMMENT
Also number of distinct resistances possible for n arbitrary resistors
each connected in series or parallel with previous ones (cf. A051045).
The n-th term of A051045 uses the n different resistances 1, ..., n
ohms, whereas the problem corresponding to A005840 allows arbitrary
general resistances a1, a2, ..., an, chosen so as to give the maximum
possible number of distinct equivalent resistances - E. W. Weisstein..
Stanley's Problem 5.4(a) involves threshold graphs; Problem 5.4(c)
involves hyperplane arrangements.
REFERENCES
R. P. Stanley, ``A zonotope associated with graphical degree
sequences,'' in Applied Geometry and Discrete Combinatorics. DIMACS
Series in Discrete Math., Amer. Math. Soc., Vol. 4, pp. 555-570, 1991.
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see
Problem 5.4(a).
LINKS
E. W. Weisstein, Link to a section of The World of Mathematics.
EXAMPLE
exp(x)*(1-x)/(2-exp(x)) = 1 + x + x^2 + 4/3*x^3 + 23/12*x^4 + 83/30*x^5
+ 479/120*x^6 + 1814/315*x^7 + O(x^8); then the coefficients are
multiplied by n! to get 1, 1, 2, 8, 46, 332, 2874, 29024, ...
AUTHOR
Simon Plouffe (plouffe(AT)math.uqam.ca)
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Ioannis
news:gerry-532D45....@sunb.ocs.mq.edu.au...
[snip for brevity]
> COMMENT
> Also number of distinct resistances possible for n arbitrary resistors
> each connected in series or parallel with previous ones (cf. A051045).
Hmmm. That would be it, except there is a minor point here: You wrote
"previous ones", above, but the page on A051045 says (quote) "previous ONE". I
think this makes some difference.
Thinking about the original problem, yes, it sounds like it's equivalent to
finding all distinct resistances of n resistors 1, 2, 3, ..., n, connected
arbitrarilly. So A051045 does not seem to be it (because of the "ONE").
A005840 looks like it, but why wouldn't they mention the resistor combo there?
> The n-th term of A051045 uses the n different resistances 1, ..., n
> ohms, whereas the problem corresponding to A005840 allows arbitrary
> general resistances a1, a2, ..., an, chosen so as to give the maximum
> possible number of distinct equivalent resistances - E. W. Weisstein..
Ok, A005840 says "ONES", so it should be it, but the numbers don't match.
> Stanley's Problem 5.4(a) involves threshold graphs; Problem 5.4(c)
> involves hyperplane arrangements.
[snip]
> --
> Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
bill
Ioannis wrote:
> High Intensity Discharge lamps, such as high pressure mercury lamps, high
> pressure sodium lamps and metal halide lamps (all these are relatively high
> power lamps that are used to illuminate streets and various other spaces) are
> what are called "negative resistance" devices:
>
> When they are connected to the AC outlet directly, they cause more and more
> current to be drawn, until they either blow the fuse or destroy themselves.
>
> To prevent this from happening, they are connected in series with what's
> called a "ballast". A ballast is an inductive coil which limits the circuit
> current to i_b Amperes. The value i_b is characteristic of the ballast used.
> In other words, there is a one-to-one correspondance between ballast type and
> the amperage it provides when its connected in series with a specific lamp.
>
> The ballast is (almost) pure inductive resistance R_b, therefore the ballast
> current and resistance are related via Ohm's Law for AC: i_b = 220V (or 110V)
> /R_b. Let's assume that we are working with 220V AC.
I thought that the voltage drop across an inductor was V_L =
L*(di/dt) ?
Gerry Myerson
But that sequence starts 1, 2, 4, 10, 24, ...,
so it doesn't seem to be counting the same thing.
With 2 resistors, we're taking that to mean at most 2 resistors,
while A000084 takes it to mean exactly 2 resistors. That's going
to lead to a very different sequence.
Gerry Myerson
"Ioannis" <morp...@olympus.mons> wrote:
> "Gerry Myerson" <ge...@maths.mq.edi.ai.i2u4email> wrote in message
> news:gerry-532D45....@sunb.ocs.mq.edu.au...
> [snip for brevity]
>
> > COMMENT
> > Also number of distinct resistances possible for n arbitrary resistors
> > each connected in series or parallel with previous ones (cf. A051045).
>
> Hmmm. That would be it, except there is a minor point here: You wrote
> "previous ones", above, but the page on A051045 says (quote) "previous ONE". I
> think this makes some difference.
>
> Thinking about the original problem, yes, it sounds like it's equivalent to
> finding all distinct resistances of n resistors 1, 2, 3, ..., n, connected
> arbitrarilly. So A051045 does not seem to be it (because of the "ONE").
> A005840 looks like it, but why wouldn't they mention the resistor combo there?
>
> > The n-th term of A051045 uses the n different resistances 1, ..., n
> > ohms, whereas the problem corresponding to A005840 allows arbitrary
> > general resistances a1, a2, ..., an, chosen so as to give the maximum
> > possible number of distinct equivalent resistances - E. W. Weisstein..
>
> Ok, A005840 says "ONES", so it should be it, but the numbers don't match.
Maybe the reason the numbers don't match is that in this thread we've
been counting circuits with at most n resistors while A005840 is
counting circuits with exactly n resistors. I think that makes the next
number in our sequence 94.
Joe Riel
>> The ballast is (almost) pure inductive resistance R_b, therefore the ballast
>> current and resistance are related via Ohm's Law for AC: i_b = 220V (or 110V)
>> /R_b. Let's assume that we are working with 220V AC.
>
> I thought that the voltage drop across an inductor was V_L =
> L*(di/dt) ?
It is. The R in the previous paragraph should really be X, the
impedance of the inductor. For a specific frequency, X = 2*Pi*f*L.
That is, if the current is a sine wave I = Ipk*sin(2*Pi*f*t), then
dI/dt = Ipk*2*Pi*f*cos(2*Pi*f*t), so the voltage is
Ipk*2*Pi*f*L*cos(2*Pi*f*t) = Ipk*X*cos(2*Pi*f*t). This is more
readily expressed by V = I*X, where V and I correspond to the
rms values.
Joe
Alec Mihailovs
"Gerry Myerson" <ge...@maths.mq.edi.ai.i2u4email> wrote
>> Ok, A005840 says "ONES", so it should be it, but the numbers don't match.
>
> Maybe the reason the numbers don't match is that in this thread we've
> been counting circuits with at most n resistors while A005840 is
> counting circuits with exactly n resistors. I think that makes the next
> number in our sequence 94.
Yes, in this case the sequence would look like
0, 1, 4, 17, 94, 667, 5752, 58053, 669970, 8698991, 125499820,
1991637529, 34479906886, 646671878595, 13061304372448,
282652185684845, 6524494505342842, 160018549741811479,
4155443426929596436, 113905714869793400001
The exponential generating function would be
exp(x) (-2 exp(x) + exp(x) x + 2)
---------------------------------
-2 + exp(x)
and one of the possible formulas for it would be a(n)=2*A005840(n) + n - 2.
Alec Mihailovs
http://mihailovs.com/Alec/
The Last Danish Pastry
Yes. I did not mean that A000084 was a complete solution to his
problem, but rather that it addressed an interesting related topic:
the number of ways of arranging n resistors in a circuit consisting
entirely of series and parallel elements.
--
Clive Tooth
http://www.shutterstock.com/cat.mhtml?gallery_id=61771
Ioannis
news:vsSah.6450$ya1....@news02.roc.ny...
Doesn't seem to appear in the Database. Anybody cares to submit it or should
I?
Cheerio!
> Alec Mihailovs
Alec Mihailovs
>
> Doesn't seem to appear in the Database. Anybody cares to submit it or
> should
> I?
Please do.
Alec
Ioannis
Ok, it's in. Just out of curiosity, Alec, can you offer some thoughts on how
you came up with the complete sequence, its generating function and the
expression 2*A005840(n) + n - 2?
Thanks,
> Alec
Alec Mihailovs
>
> Ok, it's in. Just out of curiosity, Alec, can you offer some thoughts on
> how
> you came up with the complete sequence, its generating function and the
> expression 2*A005840(n) + n - 2?
The main credit should go to Gerry Meyerson.
He wrote: "the reason the numbers don't match is that in this thread we've
been counting circuits with at most n resistors while A005840 is
counting circuits with exactly n resistors. I think that makes the next
number in our sequence 94."
Let b(n)=A005840(n) if n>0 and b(0)=0, and let a(n) be the sequence
discussed in this thread.
How to calculate a(4), for example?
a(4)=b(4) + 4 b(3) + 6 b(2) + 4 b(1) + b(0),
because there is one choice of 4 resistors out of 4, 4 choices of 3
resistors out of 4, 6 choices of 2 resistors out of 4, 4 choices of 1
resistor out of 4, and for every such choice the number of circuits
constracted from them is b(..).
In general, we get formula
a(n) = sum C(n,k)*b(n) for k from 0 to n.
In other words, sequence a(0), a(1),... is the inverse binomial transform of
the sequence b(0), b(1), ...
Now, if g_b(x) = sum b(n)/n!*x^n for n from 0 to infinity, it is called an
exponential generating function for b, and
g_a(x) = sum a(n)/n!*x^n is the exponential generating function for a, and
a(0),a(1),... is the inverse binomial transform of b(0),b(1),... then it is
well known and it is easy to check that
g_a(x)=exp(x)*g_b(x).
g_b(x) can be obtain by subtracting 1 from the exponential generating
function for A005840.
That's how I get g_a(x).
It can be rewritten in the form
2*(x-1)*exp(x)/(exp(x)-2) + (x-2)*exp(x)
for n>0. The first term gives 2*b(n), and the second term gives (n-2), so
a(n)=2*b(n)+n-2
It also true for n=0.
Alec
Gerry Myerson
"Alec Mihailovs" <al...@mihailovs.com> wrote:
> From: "Ioannis" <morp...@olympus.mons
> >
> > Ok, it's in. Just out of curiosity, Alec, can you offer some thoughts on
> > how
> > you came up with the complete sequence, its generating function and the
> > expression 2*A005840(n) + n - 2?
>
> The main credit should go to Gerry Meyerson.
If you're referring to me, then
1. the spelling is Myerson, and
2. you're far too kind. While you may have found it routine
to get from my remark to your formulas, I'd say you did
the lion's share of the work. Well, you and whoever did
the original work on A005840. Well done.
Alec Mihailovs
> "Alec Mihailovs" <al...@mihailovs.com> wrote:
>>
>> The main credit should go to Gerry Meyerson.
>
> If you're referring to me, then
>
> 1. the spelling is Myerson, and
Sorry about that. I don't understand how that could happen.
Certainly I meant to write Gerry Myerson,
Alec