With regards to OEIS A166748 (http://research.att.com/~njas/sequences/A166748)
Alexander R.Povolotsky
A166748)
I am using WolphramAlpha (due to lack Mathematica or Maple or similar
capability packages in my possession)
By simply typing in first 19 terms of A166748 (WolphramAlpha wouldn't
take more ...)
I am getting
Possible closed form for all terms given:
a_n = (2^(n-2)*(3*(-1)^n*
Gamma(-3*i)*Gamma(3*i) - (-1)^n*Gamma(1/2-3*i)*Gamma (1/2+3*i) +
3*Gamma(-3*i)*Gamma(3*i) + Gamma(1/2-3*i)*Gamma(1/2+3*i))*Gamma(n/2-
(1/2+3*i))*Gamma(n/2-(1/2-3*i)))/(Gamma(-3*i)*Gamma(3*i)*Gamma(1/2-3*i)
*Gamma(1/2+3*i))
Then I go to quickmath website (another trick of "poor" man ) and use
simplify (fully) option
For above I get:
a(n)=
(3*2^(-3 + n)*gamma(((-1 - 6*i) + n)/2)*gamma(((-1 + 6*i) + n)/2)*((1
+ (-1)^n)*csch(3*pi) - (-1 + (-1)^n)*sech(3*pi))*sinh(6*pi))/pi
Then I go back to WolphramAlpha and calculate the sum (Maple
notations)
sum(1/((3*2^(-3 + n)*gamma(((-1 - 6*i) + n)/2)*gamma(((-1 + 6*i) + n)/
2)*((1 + (-1)^n)*csch(3Pi) - (-1 + (-1)^n)*sech(3*Pi))*sinh(6*Pi))/
Pi),n=0...infinity)
It gives me the real value (can not copy and paste it for some reason
from WolphramAlpha)
7.33...
If I use Pi^2 in the sum's numerator (instead of Pi shown above)
sum(1/((3*2^(-3 + n)*gamma(((-1 - 6*i) + n)/2)*gamma(((-1 + 6*i) + n)/
2)*((1 + (-1)^n)*csch(3Pi) - (-1 + (-1)^n)*sech(3*Pi))*sinh(6*Pi))/
Pi*2),n=0...infinity)
it gives me real value 23.14... - very close to exp(Pi) but NOT
exactly !
Now if I assume that the generating function for OEIS A166748 (http://
research.att.com/~njas/sequences/A166748) is
a(n+2)=(n^2+36)*a(n), a(0)=1, a(1)=6
WolframAlpha resolves above recurrency into
a(n) = -(2^(n-1)*(3 *(-1)^n*Gamma(-3*i)*Gamma(3* i)-(-1)^n* Gamma
(1/2-3*i)*Gamma(1/2+3*i)-3*Gamma(-3*i) Gamma(3*i)-Gamma(1/2-3*i)*Gamma
(1/2+3*i)) Gamma(n/2-3*i)*Gamma(n/2+3*i))/(Gamma(-3*i)*Gamma(3*i)*Gamma
(1/2-3*i)*Gamma(1/2+3*i))
Then I go to quickmath website and use simplify (fully) option
For above I get:
a(n)=
(3*2^n*gamma(-3*i + n/2)*gamma(3*i + n/2)*(cos((n*pi)/2) + i*sin
((n*pi)/2))*sinh(((6 - i*n)*pi)/2))/pi
which is different from the originally obtained
a(n)=
(3*2^(-3 + n)*gamma(((-1 - 6*i) + n)/2)*gamma(((-1 + 6*i) + n)/2)*((1
+ (-1)^n)*csch(3*pi) - (-1 + (-1)^n)*sech(3*pi))*sinh(6*pi))/pi
Tried to do sums for new one in WolframAlpha also but it chokes on it
(the server is busy ....)
Are those two results principally different ?
Which one is correct for OEIS A166748 (http://research.att.com/~njas/
sequences/A166748)
?
ARP
Raymond Manzoni
(snip)
> For above I get:
> a(n)=
> (3*2^n*gamma(-3*i + n/2)*gamma(3*i + n/2)*(cos((n*pi)/2) + i*sin
> ((n*pi)/2))*sinh(((6 - i*n)*pi)/2))/pi
> which is different from the originally obtained
> a(n)=
> (3*2^(-3 + n)*gamma(((-1 - 6*i) + n)/2)*gamma(((-1 + 6*i) + n)/2)*((1
> + (-1)^n)*csch(3*pi) - (-1 + (-1)^n)*sech(3*pi))*sinh(6*pi))/pi
>
> Tried to do sums for new one in WolframAlpha also but it chokes on it
> (the server is busy ....)
>
> Are those two results principally different ?
> Which one is correct for OEIS A166748 (http://research.att.com/~njas/
> sequences/A166748)
> ?
>
> ARP
>
Let's start with your expression :
a(n)= (3*2^n*gamma(-3*i + n/2)*gamma(3*i + n/2)*(cos((n*pi)/2) +
i*sin((n*pi)/2))*sinh(((6 - i*n)*pi)/2))/pi
and simplify further :
for n= 2m this becomes :
a(2m)= (3 4^m gamma(-3i + m) gamma(3 i + m) (-1)^m sinh((3 - i m) pi))/pi
we may simplify g(m)= gamma(-3i + m) gamma(3i + m)
(using gamma(iy) gamma(-iy)= pi/(y sinh(pi y))
and gamma(x+1)= x gamma(x) )
to get (x=0..m-1 and y=3) :
g(m)= (pi/(3 sinh(3 pi))) prod_{k=0}^{m-1} (-3i + k)(3i + k)
g(m)= (pi/(3 sinh(3 pi))) prod_{k=0}^{m-1} (k^2+9)
s(m)= sinh((3 - i m) pi) is
s(m)= sinh(3 pi) cosh(im pi) - cosh(3 pi) sinh(im pi)
(using sinh(i m pi)= i sin(m pi)= 0
and cosh(i m pi)= cos(m pi)= (-1)^m )
s(m)= (-1)^m sinh(3 pi)
so that
a(2m)= (3 4^m (pi/(3 sinh(3 pi))) (-1)^(2m) sinh( 3 pi)/pi
prod_{k=0}^{m-1} (k^2 + 9)
with the rather simple result :
a(2m)= 4^m prod_{k=0}^{m-1} (k^2 + 9)
The same way me may get for n= 2m+1 :
a(2m+1)= 6 4^m prod_{k=0}^{m-1} ((k+1/2)^2+9)
(using gamma(-3*I+m+1/2)*gamma(3*I+m+1/2)= (pi/cosh(3 pi))
prod_{k=0}^{m-1} ((k+1/2)^2+9) and so on...)
Of course a direct derivation of these results could be nicer!
Hoping it helped anyway,
Raymond
Martin Rubey
> with the rather simple result :
>
> a(2m)= 4^m prod_{k=0}^{m-1} (k^2 + 9)
I guess you could read this off the recurrence. The computer can guess
the result, in any case:
(7) -> guess([l.(2*i+1) for i in 0..8], maxLevel==2)
n - 1
++-++ 2
(7) | | 4p + 36
| | 10
p = 0
10
(sorry about the missing parenthesis)
Martin
Raymond Manzoni
Reading more carefully the initial post I notice that it contained a
simple recurrence too :
a(n+2)= (n^2+36)*a(n), a(0)=1, a(1)=6
Nice guess anyway!
Raymond
Alexander R.Povolotsky
1) Are those two results principally different ?
2)Which one is correct for OEIS A166748 (http://research.att.com/
~njas/
sequences/A166748)
Cheers,
ARP
Raymond Manzoni
Shortly :
1) No
2) Both
I didn't answer to your original question but the method I used to
simplify your variant :
a(n)= (3*2^n*gamma(-3*i + n/2)*gamma(3*i + n/2)*(cos((n*pi)/2) +
i*sin((n*pi)/2))*sinh(((6 - i*n)*pi)/2))/pi
may be used too for your initial :
a_i(n)= (3*2^(-3 + n) * gamma(((-1 - 6*i) + n)/2) * gamma(((-1 + 6*i) +
n)/2) * ((1 + (-1)^n)*csch(3*pi) - (-1 + (-1)^n)*sech(3*pi))*sinh(6*pi))/pi
if you notice that the corresponding gamma part :
g_i(n)= gamma(((-1 - 6*i) + n)/2)*gamma(((-1 + 6*i) + n)/2)
g_i(n)= gamma(-3i + (n-1)/2) gamma(3i + (n-1)/2)
is the same than in your variant (the a(n) I studied) :
g(n)= gamma(-3*i + n/2)*gamma(3*i + n/2)
but with a shift of 1 on the n parameter that is :
g(n)= g_i(n+1)
Let's try to simplify a_i(n+1) :
a_i(n+1)= (3*2^(-2 + n) * gamma(-3i + n/2)*gamma(3i + n/2) *
((1-(-1)^n)/sinh(3*pi) - (-1-(-1)^n)/cosh(3*pi))*sinh(6*pi))/pi
(using csch(x)= 1/sinh(x)) and sech(x)=1/cosh(x) )
Let's note :
f_i(n)= ((1-(-1)^n)/sinh(3*pi) - (-1-(-1)^n)/cosh(3*pi))*sinh(6*pi))
then f_i(2m)= 2*sinh(6*pi)/cosh(3*pi)= 4*sinh(3*pi)
and f_i(2m+1)= 2*sinh(6*pi)/sinh(3*pi)= 4*cosh(3*pi)
so that :
a_i(n+1)= (3*2^n * g(n) * f_i(n))/pi
Now notice that if n=2m then we get
(3 4^m g(m) sinh(3 pi))/pi exactly like in my first answer
(since 2^(-2 + 2m)*4= 4^m)
and the same should be true for the case n=2m+1.
So that both answers are identical (up to a shift of 1 on n) :
a(n) = a_i(n+1)
Cheers,
Raymond
Alexander R.Povolotsky
BTW, in http://list.seqfan.eu/pipermail/seqfan/2009-October/002649.html
the following programmatic/calculation Pari (I presume) short syntax
was given
for the same issue:
a(n)=round(2^n*norm(gamma(n/2+3*I))*if(n%2,sinh(3*Pi),cosh(3*Pi))/
Pi*3)
Another (hidden ;-) ) question, which I had in my original posting:
>If I use Pi^2 in the sum's numerator (instead of Pi shown above)
>sum(1/((3*2^(-3 + n)*gamma(((-1 - 6*i) + n)/2)*gamma(((-1 + 6*i) + n)/2)*((1 + (-1)^n)*csch(3Pi) >- (-1 + (-1)^n)*sech(3*Pi))*sinh(6*Pi))/Pi*2),n=0...infinity)
>it gives me real value 23.14... - very close to exp(Pi) but NOT exactly !
Could anyone shad light on above result ?
Cheers,
ARP
M. Hasler
wrote:
> Thanks Raymond,
>
> BTW, inhttp://list.seqfan.eu/pipermail/seqfan/2009-October/002649.html
> the following programmatic/calculation Pari (I presume) short syntax
> was given
> for the same issue:
> a(n)=round(2^n*norm(gamma(n/2+3*I))*if(n%2,sinh(3*Pi),cosh(3*Pi))/
> Pi*3)
For the records:
it should read ...if(n%2,cosh(3*Pi),sinh(3*Pi))...
(my fault...) which a mathematician would rather write
...(exp(3*Pi)-(-1)^n*exp(-3*Pi))...
(with 2^(n-1) instead of 2^n)
OTOH for numerical computations the following is *slightly* more
efficient
(but the n%2 and 2^n are probably negligible w.r.t. the call to gamma
() and cosh/sinh)
A166748(n)=round(norm(gamma(n/2+3*I))/Pi*if(bittest(n,0),cosh
(3*Pi),sinh(3*Pi))*3<<n)
Maximilian