cp's OEIS Frontend

Also, numbers n such that B(n)*n == 6 (mod n), where B(n) is the n-th Bernoulli number. Equivalently, SUM[prime p, (p-1) divides n] n/p == -6 (mod n). There are no other terms below 10^15. - Max Alekseyev, Aug 26 2013

Also, integers n such that B(n)*n == 7 (mod n), where B(n) is the n-th Bernoulli number, or SUM[prime p, (p-1) divides n] n/p == -7 (mod n). It is easy to see that for n>1, every prime divisor p of n, except p=7, must appear in first power, while p=7 may appear in first or second power. Moreover, the multiset P of prime divisors of all such n satisfies the property: if p is in P, then p-1 is the product of distinct elements of P. This multiset is P = {2, 3, 7, 7, 43}, implying that the sequence is finite and complete. - Max Alekseyev, Aug 25 2013

Also, numbers n such that B(n)*n == 8 (mod n), where B(n) is the n-th Bernoulli number. Equivalently, SUM[prime p, (p-1) divides n] n/p == -8 (mod n). There are no other terms below 10^15. - Max Alekseyev, Aug 26 2013

Also, integers n such that B(n)*n == 79 (mod n), where B(n) is the n-th Bernoulli number.

Also, integers n such that Sum_{prime p, (p-1) divides n} n/p == -79 (mod n).

Although this sequence is finite, the prime 79 does not belong to A302345.

Also, integers n such that B(n)*n == 193 (mod n), where B(n) is the n-th Bernoulli number.

Also, integers n such that Sum_{prime p, (p-1) divides n} n/p == -193 (mod n).

Although this sequence is finite, the prime 193 does not belong to A302345.

Numbers n such that the congruence k^(2n+1) == k (mod n) is true for 1<=k<=n. - Michel Lagneau, May 02 2012

In 2005, B. C. Kellner proved E. W. Weisstein's conjecture that denom(B_n) = n only if n = 1806. - Jonathan Sondow, Oct 14 2013.

For n>1, a(n) is even. Alternatively, the even terms of this sequence can be characterized in any of the following ways: (i) even integers n such that n*B(n) == n/2 (mod n), where B(n) is the n-th Bernulli number; OR (ii) integers n such that gcd(n,A027642(n)) = 2; OR (iii) even integers n such that (p-1) does not divide n for every odd prime p dividing n (cf. A124240). - Max Alekseyev, Sep 05 2013

For all k we have k divides psi(k^2). This sequence gives those k such that the quotient is 1.

Apart from 5 exceptional terms, every term is the product of 4 and distinct odd primes. The exceptional terms are precisely the 5 terms in A014117.

Except for k = 1, 2, 6, 42, 1806, k is a term if and only if k = 4*(p_1)*(p_2)*...*(p_m), where p_1 < p_2 < ... < p_m are odd primes, (p_i)-1 | 4*(p_1)*(p_2)*...*(p_(i-1)) for all 1 <= i <= m.

The LCM of two terms is again in this sequence.

Is this sequence infinite? If this sequence is finite, it means that there exists a term of the form k = 4*(p_1)*(p_2)*...*(p_s), where p_1 < p_2 < ... < p_s are odd primes such that: for every (e_0, e_1, ..., e_s) in {0, 1}^(s+1), 2^((e_0)+1)*(p_1)^(e_1)*(p_2)^(e_2)*...*(p_s)^(e_s)+1 is either composite or equal to some p_i. That term must be divisible by all other terms, since there are no more odd primes q other than p_1, p_2, ..., p_s such that q-1 | k.

Numbers k such that b^k == 1 (mod k^2) for every b coprime to k. Proof: these are numbers k such that psi(k^2) divides k, which holds if and only if psi(k^2) = k. Subsequence of A124240 (see my comment there). If k is a term of the sequence and k+1 is prime, then k*(k+1) is also a term. - Thomas Ordowski, Jul 26 2024