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A288487 Cuboids that fit in square rings from A288486 obtaining a fifth power.

%I A288487 #23 Nov 23 2024 03:35:04
%S A288487 1,8,75,400,1445,4056,9583,20000,38025,67240,112211,178608,273325,
%T A288487 404600,582135,817216,1122833,1513800,2006875,2620880,3376821,4298008,
%U A288487 5410175,6741600,8323225,10188776,12374883,14921200,17870525,21268920,25165831,29614208
%N A288487 Cuboids that fit in square rings from A288486 obtaining a fifth power.
%C A288487 If we add a(n) and A288487(n) graphically we obtain a bigger cuboid which is a square of cubes (a cuboid with dimensions n^2 * n^2 * n).
%C A288487 a(10^n) is a palindrome in base 10.
%H A288487 Daniel Poveda Parrilla, <a href="/A288487/b288487.txt">Table of n, a(n) for n = 0..10000</a>
%H A288487 Daniel Poveda Parrilla, <a href="/A288487/a288487_1.png">Illustration of initial terms</a>
%H A288487 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F A288487 G.f.: (1 + 2*x + 42*x^2 + 50*x^3 + 25*x^4)/(1 - x)^6.
%F A288487 a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
%F A288487 a(n) = (n + 1)*(n^2 + 1)^2 = (n + 1)*(A002522(n))^2 = (n + 1)*A082044(n).
%F A288487 a(n) = n^5 + A002061(A002061(n+1)).
%F A288487 a(n) = A000584(n+1) - A288486(n)
%F A288487 a(n) = (n + 1)*A059100(n-1) + 4*(n^2 -1)*A014206(n-1) for n > 1.
%t A288487 Table[(1 + n)*(1 + n^2)^2, {n, 0, 28}] (* or *) CoefficientList[Series[(1 + 2 x + 42 x^2 + 50 x^3 + 25 x^4)/(1 - x)^6, {x, 0, 28}], x] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 8, 75, 400, 1445, 4056}, 29]
%o A288487 (PARI) Vec((1 + 2*x + 42*x^2 + 50*x^3 + 25*x^4)/(1 - x)^6 + O(x^28))
%Y A288487 Cf. A000584, A002061, A002522, A014206, A059100, A082044, A288486.
%K A288487 nonn,easy
%O A288487 0,2
%A A288487 _Daniel Poveda Parrilla_, Jun 11 2017