A233045 -id:A233045 - cp's OEIS frontend

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

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

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

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.

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.

A diagonal of array A103438.
Cf. A002144, A014117, A089072, A229303, A256016, A261490.
For a(n) mod n see A182398.

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

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

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

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 *)

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

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

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.

A054377 Primary pseudoperfect numbers: numbers n > 1 such that 1/n + sum 1/p = 1, where the sum is over the primes p | n.

Original entry on oeis.org

2, 6, 42, 1806, 47058, 2214502422, 52495396602, 8490421583559688410706771261086

Offset: 1

Eric W. Weisstein

Primary pseudoperfect numbers are the solutions of the "differential equation" n' = n-1, where n' is the arithmetic derivative of n. - Paolo P. Lava, Nov 16 2009
Same as n > 1 such that 1 + sum n/p = n (and the only known numbers n > 1 satisfying the weaker condition that 1 + sum n/p is divisible by n). Hence a(n) is squarefree, and is pseudoperfect if n > 1. Remarkably, a(n) has exactly n (distinct) prime factors for n < 9. - Jonathan Sondow, Apr 21 2013
From the Wikipedia article: it is unknown whether there are infinitely many primary pseudoperfect numbers, or whether there are any odd primary pseudoperfect numbers. - Daniel Forgues, May 27 2013
Since the arithmetic derivative of a prime p is p' = 1, 2 is obviously the only prime in the sequence. - Daniel Forgues, May 29 2013
Just as 1 is not a prime number, 1 is also not a primary pseudoperfect number, according to the original definition by Butske, Jaje, and Mayernik, as well as Wikipedia and MathWorld. - Jonathan Sondow, Dec 01 2013
Is it always true that if a primary pseudoperfect number N > 2 is adjacent to a prime N-1 or N+1, then in fact N lies between twin primes N-1, N+1? See A235139. - Jonathan Sondow, Jan 05 2014
Also, integers n > 1 such that A069359(n) = n - 1. - Jonathan Sondow, Apr 16 2014

			From _Daniel Forgues_, May 24 2013: (Start)
With a(1) = 2, we have 1/2 + 1/2 = (1 + 1)/2 = 1;
with a(2) = 6 = 2 * 3, we have
  1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = (1*3 + 3)/(2*3) = (1 + 1)/2 = 1;
with a(3) = 42 = 6 * 7, we have
  1/2 + 1/3 + 1/7 + 1/42 = (21 + 14 + 6 + 1)/42 =
  (3*7 + 2*7 + 7)/(6*7) = (3 + 2 + 1)/6 = 1;
with a(4) = 1806 = 42 * 43, we have
  1/2 + 1/3 + 1/7 + 1/43 + 1/1806 = (903 + 602 + 258 + 42 + 1)/1806 =
  (21*43 + 14*43 + 6*43 + 43)/(42*43) = (21 + 14 + 6 + 1)/42 = 1;
with a(5) = 47058 (not oblong number), we have
  1/2 + 1/3 + 1/11 + 1/23 + 1/31 + 1/47058 =
  (23529 + 15686 + 4278 + 2046 + 1518 + 1)/47058 = 1.
For n = 1 to 8, a(n) has n prime factors:
  a(1) = 2
  a(2) = 2 * 3
  a(3) = 2 * 3 *  7
  a(4) = 2 * 3 *  7 * 43
  a(5) = 2 * 3 * 11 * 23 *  31
  a(6) = 2 * 3 * 11 * 23 *  31 * 47059
  a(7) = 2 * 3 * 11 * 17 * 101 *   149 *       3109
  a(8) = 2 * 3 * 11 * 23 *  31 * 47059 * 2217342227 * 1729101023519
If a(n)+1 is prime, then a(n)*[a(n)+1] is also primary pseudoperfect. We have the chains: a(1) -> a(2) -> a(3) -> a(4); a(5) -> a(6). (End)
A primary pseudoperfect number (greater than 2) is oblong if and only if it is not the initial member of a chain. - _Daniel Forgues_, May 29 2013
If a(n)-1 is prime, then a(n)*(a(n)-1) is a Giuga number (A007850). This occurs for a(2), a(3), and a(5). See A235139 and the link "The p-adic order . . .", Theorem 8 and Example 1. - _Jonathan Sondow_, Jan 06 2014

M. A. Alekseyev, J. M. Grau, A. M. Oller-Marcen. Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n). Discrete Applied Mathematics, 2018. doi:10.1016/j.dam.2018.05.022 arXiv:1602.02407 [math.NT]
W. Butske, L. M. Jaje, and D. R. Mayernik, On the Equation Sum_{p|N} 1/p + 1/N = 1, Pseudoperfect numbers and partially weighted graphs, Math. Comput., 69 (1999), 407-420. [Title corrected by _Jonathan Sondow_, Apr 11 2012]
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.
J. M. Grau, A. M. Oller-Marcén, and D. Sadornil, On µ-Sondow Numbers, arXiv:2111.14211 [math.NT], 2021.
John Machacek, Egyptian Fractions and Prime Power Divisors, arXiv:1706.01008 [math.NT], 2017.
J. Sondow and K. MacMillan, Reducing the Erdos-Moser equation 1^n + 2^n + . . . + k^n = (k+1)^n modulo k and k^2, Integers 11 (2011), #A34.
J. Sondow and K. MacMillan, Primary pseudoperfect numbers, arithmetic progressions, and the Erdos-Moser equation, Amer. Math. Monthly, 124 (2017) 232-240; arXiv:math/1812.06566 [math.NT], 2018.
J. Sondow and E. Tsukerman, The p-adic order of power sums, the Erdos-Moser equation, and Bernoulli numbers, arXiv:1401.0322 [math.NT], 2014; see section 4.
Eric Weisstein's World of Mathematics, Primary pseudoperfect number.
Wikipedia, Primary pseudoperfect number.
OEIS Wiki, Primary pseudoperfect numbers.

Cf. A005835, A007850, A069359, A168036, A190272, A191975, A203618, A216825, A216826, A230311, A235137, A235138, A235139, A236433.

pQ[n_] := (f = FactorInteger[n]; 1/n + Sum[1/f[[i]][[1]], {i, Length[f]}] == 1)
Select[Range[2, 10^6], pQ[#] &] (* Robert Price, Mar 14 2020 *)

isok(n) = if (n > 1, my(f=factor(n)[,1]); 1/n + sum(k=1, #f, 1/f[k]) == 1); \\ Michel Marcus, Oct 05 2017

from sympy import primefactors
A054377 = [n for n in range(2,10**5) if sum([n/p for p in primefactors(n)]) +1 == n] # Chai Wah Wu, Aug 20 2014

A031971(a(n)) (mod a(n)) = A233045(n). - Jonathan Sondow, Dec 11 2013
A069359(a(n)) = a(n) - 1. - Jonathan Sondow, Apr 16 2014
a(n) == 36*(n-2) + 6 (mod 288) for n = 2,3,..,8. - Kieren MacMillan and Jonathan Sondow, Sep 20 2017

cp's OEIS Frontend

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

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