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Sum of the cubes of the parts in the partitions of n into two distinct parts.
7

%I #27 Jan 04 2018 06:35:34

%S 0,0,9,28,100,198,441,720,1296,1900,3025,4140,6084,7938,11025,13888,

%T 18496,22680,29241,35100,44100,52030,64009,74448,90000,103428,123201,

%U 140140,164836,185850,216225,241920,278784,309808,354025,391068,443556,487350,549081

%N Sum of the cubes of the parts in the partitions of n into two distinct parts.

%H Colin Barker, <a href="/A294287/b294287.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,4,-4,-6,6,4,-4,-1,1).

%F a(n) = Sum_{i=1..floor((n-1)/2)} i^3 + (n-i)^3.

%F From _David A. Corneth_, Oct 27 2017: (Start)

%F For odd n, a(n) = binomial(n, 2)^2 = n^4/4 - n^3/2 + x^2/4.

%F For even n, a(n) = binomial(n, 2)^2 - n^3/8 = n^4/4 - 5*n^3/8 + x^2/4. (End)

%F G.f.: -x^3*(9 + 19*x + 36*x^2 + 22*x^3 + 9*x^4 + x^5) /(1+x)^4 /(x-1)^5. - _R. J. Mathar_, Nov 07 2017

%F From _Colin Barker_, Nov 21 2017: (Start)

%F a(n) = (1/16)*(n^2*(4 - (9 + (-1)^n)*n + 4*n^2)).

%F 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.

%F (End)

%t Table[Sum[i^3 + (n - i)^3, {i, Floor[(n-1)/2]}], {n, 40}]

%o (PARI) first(n) = my(res = vector(n, i, binomial(i,2)^2)); forstep(i=2, n, 2, res[i] -= i^3/8); res \\ _David A. Corneth_, Oct 27 2017

%o (PARI) a(n) = sum(i=1, (n-1)\2, i^3 + (n-i)^3); \\ _Michel Marcus_, Nov 19 2017

%o (PARI) concat(vector(2), Vec(x^3*(9 + 19*x + 36*x^2 + 22*x^3 + 9*x^4 + x^5) / ((1 - x)^5*(1 + x)^4) + O(x^40))) \\ _Colin Barker_, Nov 21 2017

%Y Cf. A294270.

%K nonn,easy

%O 1,3

%A _Wesley Ivan Hurt_, Oct 26 2017