|
|
A031971
|
|
a(n) = Sum_{k=1..n} k^n.
|
|
84
|
|
|
1, 5, 36, 354, 4425, 67171, 1200304, 24684612, 574304985, 14914341925, 427675990236, 13421957361110, 457593884876401, 16841089312342855, 665478473553144000, 28101527071305611528, 1262899292504270591313, 60182438244917445266889, 3031284048960901518840700
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
p^(3k - 1) divides a(p^k) for prime p > 2 and k = 1, 2, 3, 4, ... or p^2 divides a(p) for prime p > 2. p^5 divides a(p^2) for prime p > 2. p^8 divides a(p^3) for prime p > 2. p^11 divides a(p^4) for prime p > 2.
p^2 divides a(2p) for prime p > 3. p^3 divides a(3p) for prime p > 2. p^2 divides a(4p) for prime p > 5. p^3 divides a(5p) for prime p > 3. p^2 divides a(6p) for prime p > 7.
p divides a(2p - 1) for all prime p. p^3 divides a(2p^2 - 1) for all prime p. p^5 divides a(2p^3 - 1) for all prime p.
p divides a((p - 1)/2) for p = 5, 13, 17, 29, 37, 41, 53, 61, ... = A002144 Pythagorean primes: primes of form 4n + 1.
(End)
If p prime then a(p-1) == -1 (mod p) [see De Koninck & Mercier reference]. Example: for p = 7, a(6) = 67171 = 7 * 9596 - 1. - Bernard Schott, Mar 06 2020
|
|
REFERENCES
|
J.-M. De Koninck et A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 327 pp. 48-200, Ellipses, Paris (2004).
|
|
LINKS
|
|
|
FORMULA
|
a(n) = zeta(-n) - zeta(-n, n + 1), where zeta(s) is the Riemann zeta function and zeta(s, a) is the Hurwitz zeta function, a generalization of the Riemann zeta function. - Alexander Adamchuk, Jul 21 2006
a(n) == 1 (mod n) <==> n is in A014117 = 1, 2, 6, 42, 1806 (see the link "On the congruence ..."). - Jonathan Sondow, Oct 18 2013
a(n) = n! * [x^n] exp(x)*(exp(n*x) - 1)/(exp(x) - 1). - Ilya Gutkovskiy, Apr 07 2018
a(n) ~ ((e*n+1)/((e-1)*(n+1))) * n^n. - N. J. A. Sloane, Oct 13 2018, based on email from Claude F. Leibovici who claims this is slightly better than Cloitre's version when n is small.
|
|
MAPLE
|
a := n->sum('i'^n, 'i'=1..n);
# alternative
(bernoulli(n+1, n+1)-bernoulli(n+1))/(n+1) ;
|
|
MATHEMATICA
|
Table[Total[Range[n]^n], {n, 25}] (* T. D. Noe, Apr 19 2011 *)
|
|
PROG
|
(Haskell)
(Python)
from sympy import harmonic
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,nice,easy
|
|
AUTHOR
|
Chris du Feu (chris(AT)beckingham0.demon.co.uk)
|
|
STATUS
|
approved
|
|
|
|