|
|
A215084
|
|
a(n) = sum of the sums of the k first n-th powers.
|
|
4
|
|
|
0, 1, 6, 46, 470, 6035, 93436, 1695036, 35277012, 828707925, 21693441550, 626254969978, 19766667410282, 677231901484775, 25031756512858200, 992872579254244088, 42066929594261568840, 1896157095455962952169, 90601933352843530354170, 4574495282686422755339734, 243359175218492577008763726
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
First term a(0) may be computed as 1 by starting the inner sum at j=0 and taking the convention 0^0 = 1.
|
|
LINKS
|
|
|
FORMULA
|
a(n) = Sum_{k=1..n} Sum_{j=1..k} j^n.
a(n) = Sum_{k=1..n} H_k^{-n} where H_k^{-n} is the k-th harmonic number of order -n.
a(n) = Sum_{k=1..n} (B(n+1, k+1) - B(n+1, 1))/(n+1), where B(n, x) are the Bernoulli polynomials. - Daniel Suteu, Jun 25 2018
a(n) ~ c * n^n, where c = 1/(1 - 2*exp(-1) + exp(-2)) = 2.50265030107711874333... - Vaclav Kotesovec, Nov 06 2021
|
|
EXAMPLE
|
a(3) = (1^3) + (1^3 + 2^3) + (1^3 + 2^3 + 3^3) = (1^3 + 1^3 + 1^3) + (2^3 + 2^3) + (3^3) = 3 * 1^3 + 2 * 2^3 + 1 * 3^3 = 46. - David A. Corneth, Jun 27 2018
|
|
MATHEMATICA
|
Table[Sum[Sum[j^n, {j, 1, k}], {k, 0, n}], {n, 0, 20}]
a[n_] := (n+1)*HarmonicNumber[-1, -n] - HarmonicNumber[n, -n-1] + (n+1)*HarmonicNumber[n, -n]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Mar 05 2013 *)
Table[Total[Accumulate[Range[n]^n]], {n, 0, 20}] (* Harvey P. Dale, Mar 29 2020 *)
|
|
PROG
|
(PARI) a(n) = sum(k=1, n, sum(j=1, k, j^n)); \\ Michel Marcus, Jun 25 2018
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|