A294287 Sum of the cubes of the parts in the partitions of n into two distinct parts.

0, 0, 9, 28, 100, 198, 441, 720, 1296, 1900, 3025, 4140, 6084, 7938, 11025, 13888, 18496, 22680, 29241, 35100, 44100, 52030, 64009, 74448, 90000, 103428, 123201, 140140, 164836, 185850, 216225, 241920, 278784, 309808, 354025, 391068, 443556, 487350, 549081

Offset: 1

Author

Wesley Ivan Hurt, Oct 26 2017

Keywords

• nonn
• easy

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for sequences related to partitions
• Index entries for linear recurrences with constant coefficients, signature (1,4,-4,-6,6,4,-4,-1,1).

Crossrefs

Cf. A294270.

Programs

Mathematica

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

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

PARI

a(n) = sum(i=1, (n-1)\2, i^3 + (n-i)^3); \\ Michel Marcus, Nov 19 2017

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

Formula

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

From David A. Corneth, Oct 27 2017: (Start)

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

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

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

From Colin Barker, Nov 21 2017: (Start)

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

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)