A226960 -id:A226960 - cp's OEIS frontend

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

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.

A229289 Primes p of the form p = 2^k * m + 1, where (i) m is squarefree and odd, (ii) all primes that divide m are in the sequence, and (iii) k is 0, 1, or 2.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 23, 29, 31, 43, 47, 53, 59, 61, 67, 71, 79, 107, 131, 139, 157, 173, 211, 263, 269, 277, 283, 311, 317, 331, 347, 349, 367, 373, 421, 431, 461, 463, 547, 557, 599, 643, 659, 661, 683, 691, 709, 733, 743, 787, 827, 853, 859, 863, 911, 941

Offset: 1

José María Grau Ribas, Oct 05 2013

nonn

Taking m=1 in the definition we get the primes 2, 3, 5.
If n is in A226960, then n is a product of terms of this sequence.
If k is only allowed to be 0 or 1, we get 2, 3, 7, 43 and no more. - Jianing Song, Feb 21 2021
Also prime factors of terms in A341858. It is conjectured that this sequence is infinite. - Jianing Song, Feb 22 2021

Ray Chandler, Table of n, a(n) for n = 1..1000

Cf. A227007, A226960, A007814, A229290, A229291, A289355, A341858.
For the complement, see A289355.
Proper subsequence of A066651.

fa = FactorInteger; free[n_] := n == Product[fa[n][[i, 1]], {i, Length[fa[n]]}] ; Os[b_, 1] = True; Os[b_, b_] = True; Os[b_, n_] := Os[b, n] = PrimeQ[n] && free[(n - 1)/b^IntegerExponent[n - 1, b]] &&IntegerExponent[n - 1, b] < 3 && Union@Table[Os[b, fa[n - 1][[i, 1]]], {i, Length[fa[n - 1]]}] == {True};G[b_] := Select[Prime[Range[1000]], Os[b, #] &];G[2]

is(n)=if(!isprime(n),return(0)); if(n<13,return(1)); my(k=valuation(n-1,2), m=n>>k, f); if(k>2,return(0)); f=factor(m); if(lcm(f[,2])>1, return(0)); for(i=1,#f~, if(!is(f[i,1]), return(0))); 1 \\ Charles R Greathouse IV, Oct 28 2013

Revised definition from Charles R Greathouse IV, Nov 13 2013

Terms corrected by José María Grau Ribas, Nov 14 2013

cp's OEIS Frontend

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

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