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A226960
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Numbers n such that 1^n + 2^n + 3^n +...+ n^n == 2 (mod n).
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12
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OFFSET
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1,2
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COMMENTS
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Also, numbers n such that B(n)*n == 2 (mod n), where B(n) is the n-th Bernoulli number. Equivalently, SUM[prime p, (p-1) divides n] n/p == -2 (mod n). - Max Alekseyev, Aug 25 2013
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LINKS
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Table of n, a(n) for n=1..5.
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]
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MATHEMATICA
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Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 2 &]
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PROG
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(PARI) is(n)=if(n%2, return(n==1)); Mod(sumdiv(n/2, d, if(isprime(2*d+1), n/(2*d+1)))+n/2, n)==-2 \\ Charles R Greathouse IV, Nov 13 2013
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CROSSREFS
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Cf. A031971, A014117.
Subsequence of A124240.
Solutions to 1^n+2^n+...+n^n == m (mod n): A005408 (m=0), A014117 (m=1), this sequence (m=2), A226961 (m=3), A226962 (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: A205337 A263866 A208802 * A081214 A331087 A194004
Adjacent sequences: A226957 A226958 A226959 * A226961 A226962 A226963
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KEYWORD
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nonn,fini,full
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AUTHOR
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José María Grau Ribas, Jun 24 2013
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EXTENSIONS
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a(1)=1 prepended by Max Alekseyev, Aug 25 2013
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STATUS
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approved
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