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A294287
Sum of the cubes of the parts in the partitions of n into two distinct parts.
7
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
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
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Oct 26 2017
STATUS
approved