A280041 -id:A280041 - cp's OEIS frontend

A226960 Numbers n such that 1^n + 2^n + 3^n +...+ n^n == 2 (mod n).

1, 4, 12, 84, 3612

Offset: 1

José María Grau Ribas, Jun 24 2013

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

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]

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).

Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 2 &]

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

a(1)=1 prepended by Max Alekseyev, Aug 25 2013

A226961 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 3 (mod n).

Original entry on oeis.org

1, 2, 3, 18, 126, 5418

Offset: 1

José María Grau Ribas, Jun 24 2013

Equivalently, numbers n such that B(n)*n == 3 (mod n), where B(n) is the n-th Bernoulli number. Equivalently, SUM[prime p, (p-1) divides n] n/p == -3 (mod n). - Max Alekseyev, Aug 25 2013

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]

Cf. A031971.

Solutions to 1^n+2^n+...+n^n == m (mod n): A005408 (m=0), A014117 (m=1), A226960 (m=2), this sequence (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).

Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 3 &]

is(n)=Mod(sumdiv(n, d, if(isprime(d+1), n/(d+1))), n)==-3 \\ Charles R Greathouse IV, Nov 13 2013

1, 2, 3 prepended by Max Alekseyev, Aug 25 2013

A226963 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 5 (mod n).

Original entry on oeis.org

1, 2, 5, 10, 30, 210, 9030, 235290, 11072512110

Offset: 1

José María Grau Ribas, Jun 24 2013

Also, numbers n such that B(n)*n == 5 (mod n), where B(n) is the n-th Bernoulli number. Equivalently, SUM[prime p, (p-1) divides n] n/p == -5 (mod n). - Max Alekseyev, Aug 26 2013

There are no other terms below 10^31. - Max Alekseyev, Apr 04 2018

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]

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), A226962 (m=4), this sequence (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).

Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 5 &]

is(n)=Mod(sumdiv(n, d, if(isprime(d+1), n/(d+1))), n)==-5 \\ Charles R Greathouse IV, Nov 13 2013

Terms 1,2,5 prepended and a(9) added by Max Alekseyev, Aug 26 2013

A226967 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 9 (mod n).

Original entry on oeis.org

1, 2, 3, 9, 54, 378, 16254

Offset: 1

José María Grau Ribas, Jun 24 2013

Also, numbers n such that B(n)*n == 9 (mod n), where B(n) is the n-th Bernoulli number. Equivalently, SUM[prime p, (p-1) divides n] n/p == -9 (mod n). There are no other terms below 10^30. - Max Alekseyev, Aug 26 2013

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), A226962(m=4), A226963 (m=5), A226964 (m=6), A226965 (m=7), A226966 (m=8), this sequence (m=9), A280041 (m=19), A280043 (m=43), A302343 (m=79), A302344 (m=193).

Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == Mod[9,#] &]

is(n)=Mod(sumdiv(n, d, if(isprime(d+1), n/(d+1))), n)==-9 \\ Charles R Greathouse IV, Nov 13 2013

1,2,9 prepended by Max Alekseyev, Aug 26 2013

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

Original entry on oeis.org

1, 8, 24, 168, 7224

Offset: 1

José María Grau Ribas, Jun 24 2013

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

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).

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

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

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

A226964 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 6 (mod n).

Original entry on oeis.org

1, 3, 4, 20, 36, 252, 10836

Offset: 1

José María Grau Ribas, Jun 24 2013

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

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), A226962 (m=4), A226963 (m=5), this sequence (m=6), A226965 (m=7), A226966 (m=8), A226967 (m=9), A280041 (m=19), A280043 (m=43), A302343 (m=79), A302344 (m=193).

Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 6 &]

is(n)=Mod(sumdiv(n, d, if(isprime(d+1), n/(d+1))), n)==-6 \\ Charles R Greathouse IV, Nov 13 2013

1,3,4 prepended by Max Alekseyev, Aug 26 2013

A226965 Numbers n such that 1^n + 2^n + 3^n +...+ n^n == 7 (mod n).

Original entry on oeis.org

1, 2, 6, 7, 14, 294, 12642

Offset: 1

José María Grau Ribas, Jun 24 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

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]

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), 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).

Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 7&]

is(n)=Mod(sumdiv(n, d, if(isprime(d+1), n/(d+1))), n)==-7 \\ Charles R Greathouse IV, Nov 13 2013

Corrected and keywords full,fini added by Max Alekseyev, Aug 25 2013

A226966 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 8 (mod n).

Original entry on oeis.org

1, 16, 48, 336, 14448

Offset: 1

José María Grau Ribas, Jun 24 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

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), A226962(m=4), A226963 (m=5), A226964 (m=6), A226965 (m=7), this sequence (m=8), A226967 (m=9), A280041 (m=19), A280043 (m=43), A302343 (m=79), A302344 (m=193).

Select[Range[10000], Mod[Sum[PowerMod[i, #, #], {i, #}], #] == 8 &]

is(n)=Mod(sumdiv(n, d, if(isprime(d+1), n/(d+1))), n)==-8 \\ Charles R Greathouse IV, Nov 13 2013

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

A302343 Solutions to the congruence 1^n + 2^n + ... + n^n == 79 (mod n).

Original entry on oeis.org

1, 2, 6, 79, 158, 474, 3318, 142674

Offset: 1

Max Alekseyev, Apr 05 2018

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.

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]

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), A226965 (m=7), A226966 (m=8), A226967 (m=9), A280041 (m=19), A280043 (m=43), this sequence (m=79), A302344 (m=193).

Cf. A302345.

A302344 Solutions to the congruence 1^n + 2^n + ... + n^n == 193 (mod n).

Original entry on oeis.org

1, 2, 6, 193, 386, 1158, 8106, 348558

Offset: 1

Max Alekseyev, Apr 05 2018

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.

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]

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), A226965 (m=7), A226966 (m=8), A226967 (m=9), A280041 (m=19), A280043 (m=43), A302343 (m=79), this sequence (m=193).

Cf. A302345.

cp's OEIS Frontend

A226960 Numbers n such that 1^n + 2^n + 3^n +...+ n^n == 2 (mod n).

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A226961 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 3 (mod n).

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A226963 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 5 (mod n).

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A226967 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 9 (mod n).

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

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A226964 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 6 (mod n).

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A226965 Numbers n such that 1^n + 2^n + 3^n +...+ n^n == 7 (mod n).

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A226966 Numbers n such that 1^n + 2^n + 3^n + ... + n^n == 8 (mod n).

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