A226962 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 4 (mod n).

1, 8, 24, 168, 7224

Offset

1,2

Comments

Also, numbers n such that B(n)*n == 4 (mod n), where B(n) is the n-th Bernoulli number. Equivalently, SUM[prime p, (p-1) divides n] n/p == -4 (mod n). There are no other terms below 10^15. - Max Alekseyev, Aug 26 2013

Links

Table of n, a(n) for n=1..5.

Mathematica

Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 4 &]

Prog

(PARI) is(n)=Mod(sumdiv(n, d, if(isprime(d+1), n/(d+1))), n)==-4 \\ Charles R Greathouse IV, Nov 13 2013

Crossrefs

Cf. A031971.

Solutions to 1^n+2^n+...+n^n == m (mod n): A005408 (m=0), A014117 (m=1), A226960 (m=2), A226961 (m=3), this sequence (m=4), A226963 (m=5), A226964 (m=6), A226965 (m=7), A226966 (m=8), A226967 (m=9), A280041 (m=19), A280043 (m=43), A302343 (m=79), A302344 (m=193).

Sequence in context: A098070 A100042 A061027 * A221784 A052656 A094061

Adjacent sequences: A226959 A226960 A226961 * A226963 A226964 A226965

Keyword

nonn,more

Author

José María Grau Ribas, Jun 24 2013

Extensions

a(1)=1 prepended by Max Alekseyev, Aug 26 2013

Status

approved