|
|
A130863
|
|
Ratio of quadruple Sum of k^2-1 to quadruple sum of k made into an integer sequence: (1/6)*(-1 + n)(2 + n)(3 + n)(7 + n).
|
|
0
|
|
|
0, 30, 100, 231, 448, 780, 1260, 1925, 2816, 3978, 5460, 7315, 9600, 12376, 15708, 19665, 24320, 29750, 36036, 43263, 51520, 60900, 71500, 83421, 96768, 111650, 128180, 146475, 166656, 188848
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
|
|
LINKS
|
|
|
FORMULA
|
a(n) =1/2)*(n + 2)*(n + 3)*(n + 4)*Sum[Sum[Sum[Sum[k^2 - 1, {k, 1, m}], {m, 1, j}], {j, 1, l}], {l, 1, n}]/Sum[Sum[Sum[Sum[k, {k, 1, m}], {m, 1, j}], { j, 1, l}], {l, 1, n}]=(1/6)*(-1 + n)(2 + n)(3 + n)(7 + n)
G.f.: x^2*(-30+50*x-31*x^2+7*x^3)/(-1+x)^5. - R. J. Mathar, Nov 14 2007
|
|
MATHEMATICA
|
h[n_] = (1/2)*(n + 2)*(n + 3)*(n + 4)*Sum[Sum[Sum[Sum[k^2 - 1, {k, 1, m}], {m, 1, j}], {j, 1, l}], {l, 1, n}]/Sum[Sum[Sum[Sum[k, {k, 1, m}], {m, 1, j}], {j, 1, l}], {l, 1, n}]; Table[h[n], {n, 1, 30}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|