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A294270
Sum of the cubes of the parts in the partitions of n into two parts.
10
0, 2, 9, 44, 100, 252, 441, 848, 1296, 2150, 3025, 4572, 6084, 8624, 11025, 14912, 18496, 24138, 29241, 37100, 44100, 54692, 64009, 77904, 90000, 107822, 123201, 145628, 164836, 192600, 216225, 250112, 278784, 319634, 354025, 402732, 443556, 501068, 549081
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
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)
E.g.f.: x*(x*(7 + 8*x + 2*x^2)*cosh(x) + (1 + 4*x + 9*x^2 + 2*x^3)*sinh(x))/8. - Stefano Spezia, Feb 03 2026
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
Sum of k-th powers of the parts in the partitions of n into two parts for k=0..10: A052928 (k=0), A093353 (k=1), A226141 (k=2), this sequence (k=3), A294271 (k=4), A294272 (k=5), A294273 (k=6), A294274 (k=7), A294275 (k=8), A294276 (k=9), A294279 (k=10).
Sequence in context: A345414 A272199 A260074 * A163650 A259777 A365038
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Oct 26 2017
STATUS
approved