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A206808
Sum_{0<j<n} n^3-j^3.
3
7, 45, 156, 400, 855, 1617, 2800, 4536, 6975, 10285, 14652, 20280, 27391, 36225, 47040, 60112, 75735, 94221, 115900, 141120, 170247, 203665, 241776, 285000, 333775, 388557, 449820, 518056, 593775, 677505, 769792, 871200, 982311
OFFSET
2,1
COMMENTS
a(n) = n^4-p(n), where p(n) is the n-th partial sum of (j^3).
a(n) = t(n)-t(n-1), where t = A206809.
For a guide to related sequences, see A206817.
FORMULA
a(n) = (3*n^4-2*n^3-n^2)/4. G.f.: -x^2*(x^2+10*x+7) / (x-1)^5. - Colin Barker, Jul 11 2014
EXAMPLE
a(2) = 2^3-1^3 = 7.
a(3) = (3^3-1^3) + (3^3-2^3) = 45.
MATHEMATICA
s[k_] := k^3; t[1] = 0;
p[n_] := Sum[s[k], {k, 1, n}];
c[n_] := n*s[n] - p[n];
t[n_] := t[n - 1] + (n - 1) s[n] - p[n - 1]
Table[c[n], {n, 2, 50}] (* A206808 *)
Flatten[Table[t[n], {n, 2, 35}]] (* A206809 *)
PROG
(PARI) vector(100, n, (3*n^4+10*n^3+11*n^2+4*n)/4) \\ Colin Barker, Jul 11 2014
(PARI) Vec(-x^2*(x^2+10*x+7)/(x-1)^5 + O(x^100)) \\ Colin Barker, Jul 11 2014
(Sage) [sum([n^3-j^3 for j in range(1, n)]) for n in range(2, 35)] # Danny Rorabaugh, Apr 18 2015
CROSSREFS
Sequence in context: A107125 A212105 A208826 * A197369 A059937 A278682
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Feb 15 2012
STATUS
approved