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

1,3

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).

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)

Mathematica

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

Prog

(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

Crossrefs

Cf. A294270.

Sequence in context: A296601 A294567 A053819 * A349547 A085292 A198059

Adjacent sequences: A294284 A294285 A294286 * A294288 A294289 A294290

Keyword

nonn,easy

Author

Wesley Ivan Hurt, Oct 26 2017

Status

approved