A060976 -id:A060976 - cp's OEIS frontend

A121707 Numbers n > 1 such that n^3 divides Sum_{k=1..n-1} k^n = A121706(n).

35, 55, 77, 95, 115, 119, 143, 155, 161, 187, 203, 209, 215, 221, 235, 247, 253, 275, 287, 295, 299, 319, 323, 329, 335, 355, 371, 377, 391, 395, 403, 407, 413, 415, 437, 455, 473, 475, 493, 497, 515, 517, 527, 533, 535, 539, 551, 559, 575, 581, 583, 589, 611

Offset: 1

Alexander Adamchuk, Aug 16 2006

nonn

All terms belong to A038509 (Composite numbers with smallest prime factor >= 5). Many but not all terms belong to A060976 (Odd nonprimes, c, which divide Bernoulli(2*c)).
Many terms are semiprimes:
- the non-semiprimes are {275, 455, 475, 539, 575, 715, 775, 875, 935, ...}: see A321487;
- semiprime terms that are multiples of 5 have indices {7, 11, 19, 23, 31, 43, 47, 59, 67, 71, 79, 83, ...} = A002145 (Primes of form 4*k + 3, except 3, or k > 0; or Primes which are also Gaussian primes);
- semiprime terms that are multiples of 7 have indices {5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, ...} = A003627 (Primes of form 3*k - 1, except 2, or k > 1);
- semiprime terms that are multiples of 11 have indices {5, 7, 13, 17, 19, 23, 37, 43, 47, 53, 59, 67, 73, 79, 83, ...} = Primes of the form 4*k + 1 and 4*k - 1. [Edited by Michel Marcus, Jul 21 2018, M. F. Hasler, Nov 09 2018]
Conjecture: odd numbers n > 1 such that n divides Sum_{k=1..n-1} k^(n-1). - Thomas Ordowski and Robert Israel, Oct 09 2015. Professor Andrzej Schinzel (in a letter to me, dated Dec 29 2015) proved this conjecture. - Thomas Ordowski, Jul 20 2018
Note that n^2 divides Sum_{k=1..n-1} k^n for every odd number n > 1. - Thomas Ordowski, Oct 30 2015
Conjecture: these are "anti-Carmichael numbers" defined; n > 1 such that p - 1 does not divide n - 1 for every prime p dividing n. Equivalently, odd numbers n > 1 such that n is coprime to A027642(n-1). A number n > 1 is an "anti-Carmichael" if and only if gcd(n, b^n - b) = 1 for some integer b. - Thomas Ordowski, Jul 20 2018
It seems that these numbers are all composite terms of A317358. - Thomas Ordowski, Jul 30 2018
a(62) = 697 is the first term not in A267999: see A306097 for all these terms. - M. F. Hasler, Nov 09 2018
If the conjecture from Thomas Ordowski is true, then no term is a multiple of 2 or 3. - Jianing Song, Jan 27 2019
Conjecture: an odd number n > 1 is a term iff gcd(n, A027642(n-1)) = 1. - Thomas Ordowski, Jul 19 2019
Conjecture: Sequence consists of numbers n > 1 such that r = b^n + n - b will produce a prime for one or more integers b > 1. Only when n is in this sequence do one or more prime factors of n fail to divide r for all b.  Also, n and b must be coprime for r to be prime. The above also applies to r = b^n - n - b, ignoring n=3, b=2. - Richard R. Forberg, Jul 18 2020
Odd numbers n > 1 such that Sum_{k(even)=2..n-1}2*k^(n-1) == 0 (mod n). - Davide Rotondo, Oct 28 2020
What is the asymptotic density of these numbers? The numbers A267999 have a slightly lower density. The difference between the densities is equal to the density of the numbers A306097. - Thomas Ordowski, Feb 15 2021
The asymptotic density of this sequence is in the interval (0.253, 0.265) (Ordowski, 2021). - Amiram Eldar, Feb 26 2021

Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 1371 terms from Robert Israel)
T. Ordowski, Density of anti-Carmichael numbers, SeqFan Mailing List, Feb 17 2021.
Don Reble, Comments on A121707

Cf. A000312, A002145, A002997, A027642, A031971, A038509, A060976, A121706, A267999 (probably a subsequence).
Cf. A306097 for terms of this sequence A121707 not in sequence A267999, A321487 for terms which are not semiprimes.
Cf. A191677 (n divides Sum_{kCf. A326478 for a conjectured connection with the Bernoulli numbers.

filter:= n -> add(k &^ n mod n^3, k=1..n-1) mod n^3 = 0:
select(filter, [$2..1000]); # Robert Israel, Oct 08 2015

fQ[n_] := Mod[Sum[PowerMod[k, n, n^3], {k, n - 1}], n^3] == 0; Select[
Range[2, 611], fQ] (* Robert G. Wilson v, Apr 04 2011 and slightly modified Aug 02 2018 *)

is(n)=my(n3=n^3);sum(k=1,n-1,Mod(k,n3)^n)==0 \\ Charles R Greathouse IV, May 09 2013

for(n=2, 1000, if(sum(k=1, n-1, k^n) % n^3 == 0, print1(n", "))) \\ Altug Alkan, Oct 15 2015

# after Andrzej Schinzel
def isA121707(n):
    if n == 1 or is_even(n): return False
    return n.divides(sum(k^(n-1) for k in (1..n-1)))
[n for n in (1..611) if isA121707(n)] # Peter Luschny, Jul 18 2019

Sequence corrected by Robert G. Wilson v, Apr 04 2011

A281648 (Numerator of Bernoulli(2*n)) read mod n.

Original entry on oeis.org

0, 1, 1, 3, 0, 5, 0, 7, 1, 9, 0, 5, 0, 7, 5, 15, 0, 11, 0, 9, 1, 11, 0, 13, 0, 13, 19, 7, 0, 19, 0, 31, 11, 17, 0, 11, 0, 19, 13, 13, 0, 37, 0, 33, 35, 23, 0, 37, 0, 39, 34, 39, 0, 11, 5, 35, 19, 29, 0, 29, 0, 31, 61, 63, 0, 55, 0, 51, 23, 21, 0, 43, 0, 37, 50, 19

Offset: 1

Seiichi Manyama, Jan 26 2017

nonn

Conjecture: a(n) == n-1 (mod n) if only if n = 6, 10 or n = 2^k for k >= 0. This is true for n <= 1024. - Seiichi Manyama, Jan 27 2017

Seiichi Manyama, Table of n, a(n) for n = 1..1000

Cf. A000367, A060976, A069040, A070192, A070193, A281662.

f[n_] := Mod[Numerator[BernoulliB[2 n]], n]; Array[f, 77] (* Robert G. Wilson v, Jan 26 2017 *)

a(n)=numerator(bernfrac(2*n))%n \\ Charles R Greathouse IV, Jan 27 2017

def bernoulli(n)
  ary = []
  a = []
  (0..n).each{|i|
    a << 1r / (i + 1)
    i.downto(1){|j| a[j - 1] = j * (a[j - 1] - a[j])}
    ary << a[0]
  }
  ary
end
def A281648(n)
  a = bernoulli(2 * n)
  (1..n).map{|i| a[2 * i].numerator % i}
end

a(n) = A000367(n) mod n.

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A121707 Numbers n > 1 such that n^3 divides Sum_{k=1..n-1} k^n = A121706(n).

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A281648 (Numerator of Bernoulli(2*n)) read mod n.

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