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%I #18 Nov 05 2019 07:00:02
%S 12,150,738,2316,5640,11682,21630,36888,59076,90030,131802,186660,
%T 257088,345786,455670,589872,751740,944838,1172946,1440060,1750392,
%U 2108370,2518638,2986056,3515700,4112862,4783050,5531988,6365616,7290090,8311782,9437280,10673388
%N Subsequence of 'Fermat near misses' which is generated by a simple formula based on the cubic binomial expansion along with formulas for the corresponding terms in the expression, x^3 + y^3 = z^3 + 1.
%C From Lewis Mammel (l_mammel(AT)att.net), Aug 21 2008: (Start)
%C In Ramanujan's parametric equation: (ax+y)^3 + (b+x^2y)^3 = (bx+y)^3 + (a+x^2y)^3
%C where a^2 + ab + b^2 = 3xy^2.
%C This sequence is obtained by setting a=0, y=1 and finding the solution to b^2=3x:
%C b=3n, x=3n^2. (End)
%H Colin Barker, <a href="/A141326/b141326.txt">Table of n, a(n) for n = 1..1000</a>
%H Tito Piezas III and Eric Weisstein, <a href="http://mathworld.wolfram.com/DiophantineEquation3rdPowers.html">Diophantine Equation--3rd Powers</a> [Lewis Mammel (l_mammel(AT)att.net), Aug 21 2008]
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = 9*n^4 + 3*n, with b(n) = 9*n^4 and c(n) = 9*n^3 + 1 we have 1 + a(n)^3 = b(n)^3 + c(n)^3.
%F From _Colin Barker_, Oct 25 2019: (Start)
%F G.f.: 6*x*(2 + 15*x + 18*x^2 + x^3) / (1 - x)^5.
%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5.
%F a(n) = 3*(n + 3*n^4).
%F (End)
%e For a(1)=12: 1 + 12^3 = 9^3 + 10^3 = 1729.
%o (PARI) Vec(6*x*(2 + 15*x + 18*x^2 + x^3) / (1 - x)^5 + O(x^40)) \\ _Colin Barker_, Oct 26 2019
%Y Cf. A050791, A050792, A050793, A050794.
%K easy,nonn
%O 1,1
%A Lewis Mammel (l_mammel(AT)att.net), Aug 03 2008
%E Edited by _Joerg Arndt_, Oct 26 2019
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