OFFSET
1,2
COMMENTS
a(n) is a square when n is odd. In fact: a(2*k+1) = (2*k^2 + k)^2; a(2*k) = k^2*(4*k^2 - 3*k + 1), where (2*k)^2 > 4*k^2 - 3*k + 1 > (2*k - 1)^2 for k>0. - Bruno Berselli, Nov 20 2017
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,4,-4,-6,6,4,-4,-1,1).
FORMULA
a(n) = Sum_{i=1..floor(n/2)} i^3 + (n-i)^3.
From Colin Barker, Nov 20 2017: (Start)
G.f.: x^2*(2 + 7*x + 27*x^2 + 28*x^3 + 24*x^4 + 7*x^5 + x^6) / ((1 - x)^5*(1 + x)^4).
a(n) = n^2*(4*n^2 - 7*n + 4 + n*(-1)^n)/16.
a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3) - 6*a(n-4) + 6*a(n-5) + 4*a(n-6) - 4*a(n-7) - a(n-8) + a(n-9) for n>9. (End)
MATHEMATICA
Table[Sum[i^3 + (n - i)^3, {i, Floor[n/2]}], {n, 80}]
PROG
(PARI) concat(0, Vec(x^2*(2 + 7*x + 27*x^2 + 28*x^3 + 24*x^4 + 7*x^5 + x^6) / ((1 - x)^5*(1 + x)^4) + O(x^40))) \\ Colin Barker, Nov 20 2017
(Magma) [0] cat &cat[[k^2*(4*k^2-3*k+1), k^2*(2*k+1)^2]: k in [1..20]]; // Bruno Berselli, Nov 22 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Oct 26 2017
STATUS
approved