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

Author

Alexander Adamchuk, Aug 16 2006

Keywords

• nonn

Comments

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

Links

• 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

Crossrefs

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_{k

Cf. A326478 for a conjectured connection with the Bernoulli numbers.

Programs

Maple

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

Mathematica

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 *)

PARI

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

PARI

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

Sage

# 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

Extensions

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

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