# [seqfan] Re: A025597 computation

Robert G. Wilson v rgwv at rgwv.com
Sun Jun 24 20:09:08 CEST 2012

```I changed the formula into Mathematica: with \$MaxExtraPrecision=1024;
f[n_] := f[n] = IntegerPart[ Chop[ N[ Simplify[(4/81)*Sum[(-1)^(j +
k)*(Sin[j*Pi/9]*Sin[k*Pi/9])^2*((1 + 2 Cos[j*Pi/9])*(1 + 2 Cos[k*Pi/9]) -
1)^n, {j, 8}, {k, 8}]], 24]]]; Array[f, 33 , 0]
{0, 0, 0, 0, 0, 0, 1, 56, 1309, 20369, 255366, 2782296, 27630317, 256617790,
2269878170, 19345170656, 160223380546, 1297456951652, 10319966008680,
80906898257760, 626886465395595, 4810654849509082, 36623649326935517,
276978367797824968, 2083200939963715125, 15595683813101279420,
116301107197615295905, 864435471139653592500, 6407217503910893277444,
47378067330093217182599, 349630722268949591264588,
2575700394882062294904447}

Bob.

-----Original Message-----
From: seqfan-bounces at list.seqfan.eu [mailto:seqfan-bounces at list.seqfan.eu]
On Behalf Of David Wilson
Sent: Sunday, June 24, 2012 10:20 AM
To: Sequence Fanatics
Subject: [seqfan] A025597 computation

I have been sent the following ostensible formula for A025597.
*
f(n) = (4/81) * sum(j=1...8, k=1...8) [(-1)^(j+k)] *
[sin(j*pi/9)*sin(k*pi/9)]^2 * [(1+2cos(j*pi/9))*(1+2cos(k*pi/9))-1]^n

*Could someone with power tools kindly verify a few terms?
*
*

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```