1^3+3^3+5^3+...+337^3 = (239*169)^2

TPiezas

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Jul 13, 2010, 7:16:41 AM7/13/10

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Hello all,

1^3+3^3+5^3+...+9^3 = (7*5)^2
1^3+3^3+5^3+...+57^3 = (41*29)^2
1^3+3^3+5^3+...+337^3 = (239*169)^2

I'm sure some can recognize in what other context the pairs {7,5},
{41,29}, and {239,169} appear...

http://sites.google.com/site/tpiezas/updates07

- Titus

Danny73

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Jul 13, 2010, 10:08:59 AM7/13/10

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Also interesting is the difference between
succeeding terms that converge on sqrt(8)+3.
Based on that alone, a new set and succeeding
sets of your terms can be found.

29/5 = 5.8 < sqrt(8)+3
41/7 = 5.857142... > sqrt(8)+3
169/29 = 5.827586... < sqrt(8)+3
239/41 = 5.8292682926... > sqrt(8)+3
985/169 = 5.8284023668... < sqrt(8)+3
1393/239 = 5.8284518828... > sqrt(8)+3
5741/985 = 5.8284263959... < sqrt(8)+3
8119/1393 = 5.8284278535... > sqrt(8)+3
33461/5741 = 5.8284271032... < sqrt(8)+3
47321/8119 = 5.8284271462... > sqrt(8)+3
195025/33461 = 5.8284271241... < sqrt(8)+3
275807/47321 = 5.828427125377... > sqrt(8)+3
etc.

Now pairing the second column with every
other 2 in that column --

{7,5},{41,29},{239,169},{1393,985},{8119,5741},{47321,33461},
{275807,195025},.. etc. Your updated list.

Converge on the sqrt(2) but are all < sqrt(2)

Basically the second column is repeated in the first
column except for {7,5}.

Dan

TPiezas

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Jul 13, 2010, 12:08:29 PM7/13/10

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Well, to give the answer in advance, let positive integers {x,y} be a
soln to the Pell eqn x^2-2y^2 = -1. Then,

a) 1^3+3^3+5^3 + ... + (2y-1)^3 = (xy)^2
b) 1^3+2^3+3^3 + ... + (x^2)^3 = (xy)^4

http://sites.google.com/site/tpiezas/updates07

- Tito

OwlHoot

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Jul 13, 2010, 12:09:47 PM7/13/10

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Well there is the well-known sum:

1^3 + 2^3 + ... + n^3 = (n (n + 1) / 2)^2

so you have:

1^3 + 3^3 + .. + (2 n - 1)^3 =

(2 n (2 n + 1) / 2)^2 - 2^3 (n (n + 1) / 2)^2

So, dividing out n^2, your results correspond to integer solutions of:

(2 n + 1)^2 - 2 (n + 1)^2 = m^2

<=> 2 n^2 - 1 = m^2

which is a Pell equation with general integer solution as follows
for t = .. -1, 0, 1, ..

m + n SQRT(2) = (1 + SQRT(2))^(2 t + 1)

Cheers

John Ramsden

Henry

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Jul 14, 2010, 4:22:57 PM7/14/10

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So 7, 41, 239, ... are http://www.research.att.com/~njas/sequences/A002315
alternate terms of http://www.research.att.com/~njas/sequences/A001333
and 5, 29, 169,... are http://www.research.att.com/~njas/sequences/A001653
alternate terms of http://www.research.att.com/~njas/sequences/A000129

The original series gives 1, 1225, 1413721, 1631432881, ...
which are http://www.research.att.com/~njas/sequences/A046177
i.e. all numbers which are (a) odd, (b) square and (c) triangular

Prai Jei

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Jul 14, 2010, 5:50:11 PM7/14/10

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TPiezas set the following eddies spiralling through the space-time
continuum:

> I'm sure some can recognize in what other context the pairs {7,5},
> {41,29}, and {239,169} appear...

Replace the larger number in each pair with the pair of adjacent numbers
which add to it, and you get the series of "almost isosceles" Pythagorean
triples where the legs differ by 1:

{7,5} -> {3, 4, 5}
{41,29} -> {20, 21, 29}
{239,169} -> {119, 120, 169}

--
ξ:) Proud to be curly

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