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%I #10 Mar 31 2012 10:28:37
%S 1,98,1647,12304,58625,210006,617743,1572992,3587409,7500250,14615711,
%T 26874288,47061937,79060814,128145375,201327616,307755233,459166482,
%U 670405519,960002000,1350818721,1870771078,2553622127,3439857024,4577640625,6023862026,7845269823
%N a(n)=(n^3+3*n^7)/4.
%C Each term is the difference of two cubes because ((n^3+n)/2)^3-((n^3-n)/2)^3=(n^3+3*n^7)/4. More generally, ((n^s+n)/2)^3-((n^s-n)/2)^3 = (n^3+3*n^(2s+1))/4 for any s; in this case, s=3.
%H Nathaniel Johnston, <a href="/A190636/b190636.txt">Table of n, a(n) for n = 1..1000</a>
%H Rafael Parra Machío, <a href="http://hojamat.es/parra/diofantica.pdf">Ecuaciones Diofanticas</a>, p. 24
%H Rafael Parra Machío, <a href="http://hojamat.es/parra/cuarticas.pdf">Ecuaciones Cuarticas</a>, p. 11
%F a(n) = ((n^3+n)/2)^3 - ((n^3-n)/2)^3.
%F G.f.: (z^7+90*z^6+891*z^5+1816*z^4+891*z^3+90*z^2+z)/(z-1)^8.
%e 58625=65^3-60^3=(5^3+3*5^7)/4; 47061937=1105^3-1092^3=(13^3 + 3*13^7)/4
%t Table[((n^3+n)/2)^3 - ((n^3-n)/2)^3, {n, 1, 20}]
%t Table[1/4*(n^3+3 n^7),{n,1,20}]
%K nonn,easy
%O 1,2
%A _Rafael Parra Machio_, May 15 2011
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