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A226965
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Numbers n such that 1^n + 2^n + 3^n +...+ n^n == 7 (mod n).
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10
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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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, #}], #] == 7&]
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PROG
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CROSSREFS
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Solutions to 1^n+2^n+...+n^n == m (mod n): A005408 (m=0), A014117 (m=1), A226960 (m=2), A226961 (m=3), A226962(m=4), A226963 (m=5), A226964 (m=6), this sequence (m=7), A226966 (m=8), A226967 (m=9), A280041 (m=19), A280043 (m=43), A302343 (m=79), A302344 (m=193).
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KEYWORD
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nonn,full,fini
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AUTHOR
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EXTENSIONS
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Corrected and keywords full,fini added by Max Alekseyev, Aug 25 2013
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STATUS
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approved
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