A031971 a(n) = Sum_{k=1..n} k^n.

1, 5, 36, 354, 4425, 67171, 1200304, 24684612, 574304985, 14914341925, 427675990236, 13421957361110, 457593884876401, 16841089312342855, 665478473553144000, 28101527071305611528, 1262899292504270591313, 60182438244917445266889, 3031284048960901518840700

Offset: 1

Author

Chris du Feu (chris(AT)beckingham0.demon.co.uk)

Keywords

• nonn
• nice
• easy

Comments

From Alexander Adamchuk, Jul 21 2006: (Start)

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).
• Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 21.

Links

• T. D. Noe, Table of n, a(n) for n = 1..100
• J. M. Grau, A. M. Oller-Marcen, and J. Sondow, On the congruence 1^m + 2^m + ... + m^m == n (mod m) with n|m, arXiv 1309.7941 [math.NT], 2013-2014.

Crossrefs

A diagonal of array A103438.

Cf. A002144, A014117, A089072, A229303, A256016, A261490.

For a(n) mod n see A182398.

Programs

Haskell

a031971 = sum . a089072_row  -- Reinhard Zumkeller, Mar 18 2013

Magma

[&+[(k)^n: k in [0..n]]: n in [1..30]]; // Vincenzo Librandi, Apr 18 2011

Maple

a := n->sum('i'^n,'i'=1..n);
# alternative
A031971 := proc(n)
    (bernoulli(n+1,n+1)-bernoulli(n+1))/(n+1) ;
end proc: # R. J. Mathar, May 10 2013

Mathematica

Table[Zeta[-n] - Zeta[-n, n + 1], {n, 25}] (* Alexander Adamchuk, Jul 21 2006 *)
Table[Total[Range[n]^n], {n,25}] (* T. D. Noe, Apr 19 2011 *)
Table[HarmonicNumber[n, -n], {n, 1, 25}] (* Jean-François Alcover, Apr 09 2015 *)

PARI

a(n)=sum(k=1,n,k^n) \\ Charles R Greathouse IV, Jun 05 2015

Python

from sympy import harmonic
def A031971(n):
    return harmonic(n,-n) # Chai Wah Wu, Feb 15 2020

Formula

a(n) is asymptotic to (e/(e - 1))*n^n. - Benoit Cloitre, Dec 17 2003

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(A054377(n)) = A233045(n). - Jonathan Sondow, Dec 11 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.