cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-18 of 18 results.

A277628 Positive integers n such that 3^n == 6 (mod n).

Original entry on oeis.org

1, 3, 21, 936340943, 10460353197, 9374251222371, 23326283250291, 615790788171551
Offset: 1

Views

Author

Dmitry Ezhov, Oct 24 2016

Keywords

Comments

No other terms below 10^15. - Max Alekseyev, Sep 12 2017

Crossrefs

Solutions to 3^n == k (mod n): A277340 (k=-11), A277289 (k=-7), A277288 (k=-5), A015973 (k=-2), A015949 (k=-1), A067945 (k=1), A276671 (k=2), A276740 (k=5), this sequence (k=6), A277126 (k=7), A277630 (k=8), A277274 (k=11).

Programs

  • PARI
    isok(n) = Mod(3, n)^n == Mod(6, n);

Extensions

a(6)-a(8) from Max Alekseyev, Sep 12 2017

A277630 Positive integers n such that 3^n == 8 (mod n).

Original entry on oeis.org

1, 5, 2352527, 193841707, 17126009179703, 380211619942943
Offset: 1

Views

Author

Dmitry Ezhov, Oct 24 2016

Keywords

Comments

No other terms below 10^15. - Max Alekseyev, Sep 13 2017

Crossrefs

Solutions to 3^n == k (mod n): A277340 (k=-11), A277289 (k=-7), A277288 (k=-5), A015973 (k=-2), A015949 (k=-1), A067945 (k=1), A276671 (k=2), A276740 (k=5), A277628 (k=6), A277126 (k=7), this sequence (k=8), A277274 (k=11).

Programs

  • PARI
    isok(n) = Mod(3, n)^n == Mod(8, n);

Extensions

a(5)-a(6) established by Max Alekseyev, Sep 13 2017

A328033 Numbers m that divide 7^m + 6.

Original entry on oeis.org

1, 13, 793, 1943, 150341, 183793, 2348789, 26052527, 27982637, 54789869, 1588344433, 3928538029, 8115802931, 16936276919, 17786709541, 47778790033, 973094452518029
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Oct 02 2019

Keywords

Comments

Conjecture: For k > 1, k^m == 1 - k (mod m) has infinite number of positive solutions.
Also includes 2073273696480171732497. - Giovanni Resta, Oct 04 2019

Crossrefs

Solutions to k^m == 1-k (mod m): A006521 (k = 2), A015973 (k = 3), A327840 (k = 4), A123047 (k = 5), A327943 (k = 6), this sequence (k = 7), A327468 (k = 8).
Cf. A253210 (7^n + 6).

Programs

  • Magma
    [1] cat [n: n in [1..10^8] | Modexp(7, n, n) + 6 eq n];

Extensions

a(12)-a(16) from Giovanni Resta, Oct 04 2019
a(17) from Max Alekseyev, Feb 07 2024

A327468 Numbers m that divide 8^m + 7.

Original entry on oeis.org

1, 3, 5, 25, 519, 290502305, 821808425, 979288025, 982989263, 25783323897, 27771237541, 31045665345, 65130752425, 3708883906025, 15079242289703, 973336048301405
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Oct 04 2019

Keywords

Comments

Conjecture: For k > 1, k^m == 1-k (mod m) has an infinite number of positive solutions.
Integer m not divisible by 3 is a term if and only if 3m is a term of A240941. - Max Alekseyev, Feb 07 2024
Also terms 930486448009391617725 and 21036656390681764555645540794214294457925. - Giovanni Resta, Oct 04 2019
Other terms 71245661271703622047, 7093208961478946798805, 7807963392818324067361574236385. - Max Alekseyev, Feb 07 2024

Crossrefs

Solutions to k^m == 1-k (mod m): 1 (k = 1), A006521 (k = 2), A015973 (k = 3), A327840 (k = 4), A123047 (k = 5), A327943 (k = 6), A328033 (k = 7), this sequence (k = 8).
Cf. A240941, A253211.

Programs

  • Magma
    [m: m in [1..7] | (8^m + 7) mod m eq 0] cat [m: m in [8..10^8] | Modexp(8, m, m) + 7 eq m]; // Jon E. Schoenfield, Oct 05 2019
  • PARI
    isok(n) = Mod(8, n)^n==-7; \\ Michel Marcus, Oct 05 2019

Extensions

a(10)-a(13) from Giovanni Resta, Oct 04 2019
a(14)-a(16) from Max Alekseyev, Feb 07 2024

A333413 Positive integers k such that k divides 13^k + 2.

Original entry on oeis.org

1, 3, 5, 185, 2199, 14061, 5672119, 6719547, 192178873, 913591893, 4589621727, 9762178659, 1157052555699
Offset: 1

Views

Author

Seiichi Manyama, Mar 20 2020

Keywords

Comments

a(14) > 6*10^12. - Giovanni Resta, Mar 29 2020

Crossrefs

Solutions to 13^k == m (mod k): this sequence (m = -2), A015963 (m = -1), A116621 (m = 1), A116622 (m = 2), A116629 (m = 3), A116630 (m = 4), A116611 (m = 5), A116631 (m = 6), A116632 (m = 7), A295532 (m = 8), A116636 (m = 9), A116620 (m = 10), A116638 (m = 11), A116639 (k = 15).
Solutions to b^k == -2 (mod k): A015973 (b = 3), A123062 (b = 5), A277370 (b = 7), this sequence (b = 13), A333414 (b = 17).

Programs

  • Mathematica
    Select[Range[100000], Divisible[PowerMod[13, #, #] + 2, #] &] (* Jinyuan Wang, Mar 28 2020 *)
  • PARI
    for(k=1, 1e6, if(Mod(13, k)^k==-2, print1(k", ")))

Extensions

a(13) from Giovanni Resta, Mar 29 2020

A333414 Positive integers k such that k divides 17^k + 2.

Original entry on oeis.org

1, 19, 35, 115, 44095, 211117, 14376053, 43472060395, 561558718915, 2182879071661
Offset: 1

Views

Author

Seiichi Manyama, Mar 20 2020

Keywords

Comments

a(8) > 10^10.
a(11) > 4*10^12. - Giovanni Resta, Mar 22 2020

Crossrefs

Solutions to b^n == -2 (mod n): A015973 (b=3), A123062 (b=5), A277370 (b=7), A333413 (b=13), this sequence (b=17).
Cf. A333269.

Programs

  • PARI
    for(k=1, 1e6, if(Mod(17, k)^k==-2, print1(k", ")))

Extensions

a(8)-a(10) from Giovanni Resta, Mar 22 2020

A370578 Numbers k such that k + 1 divides 3^k + 1.

Original entry on oeis.org

0, 1, 3, 27, 531, 1035, 4635, 6363, 11475, 19683, 40131, 80955, 266475, 280755, 307395, 356643, 490371, 544347, 557955, 565515, 572715, 808227, 1256355, 1695483, 1959075, 1995075, 2771595, 2837835, 3004155, 3208491, 3337635, 3886443, 4670955, 5619411, 6434595, 6942817
Offset: 1

Views

Author

Akiva Weinberger, Feb 22 2024

Keywords

Comments

The sequence is infinite. It contains all numbers of the form 3^(3^m).
After 3, the smallest term that is not a multiple of 9 is a(13) = 266475.
After 1, the smallest term that is not a multiple of 3 is a(36) = 6942817.
After 1, the smallest term that is not 3 (mod 8) is, also, a(36) = 6942817.
No term can be 2 (mod 3). Proof: Otherwise, k + 1 would be a multiple of 3 while 3^k + 1 would not.
All terms after 0 are odd. Proof: Suppose k is even, so that k+1 is odd. Let p be a prime factor of k+1. Then (by definition of k) 3^k == -1 (mod p) and 3^(2k) == 1 (mod p), so the order of 3 (mod p) divides 2k but not k. Thus the order of 3 is a multiple of 2^(v_2(k)+1) where v_2(k) = A007814(k) is the exponent of 2 in the prime factorization of k. But 3^(p-1) == 1 (mod p) by Fermat's little theorem, so p == 1 (mod 2^(v_2(k)+1)). Multiplying this for all prime factors p of k+1 gives k+1 == 1 (mod 2^(v_2(k)+1)), or k == 0 (mod 2^(v_2(k)+1)). But this contradicts the definition of v_2.
Apart from 0, the only possible residues mod 72 are 1, 3, 13, 25, 27, 37, 49, 51, and 61. It is conjectured that all appear eventually. (See John Omielan's answer to the author's question on Mathematics Stack Exchange.)
Empirically, approximately 80% of the terms are 27 (mod 72).

Examples

			(3^27+1)/(27+1) is an integer, so 27 is in the sequence. This can be shown efficiently using a modular exponentiation algorithm to find 3^27 mod 28.

Links

Crossrefs

Cf. A006521, A006517, A007814, A015973, A055777.

Programs

  • Mathematica
    Select[Range[0,7000000],TrueQ[PowerMod[3,#,#+1]==#]&] (* James C. McMahon, Feb 25 2024 *)
  • Python
    for n in range(100_000_000):
  if (pow(3,n,n+1)==n):
    print(n)

A328138 Numbers m that divide 9^m + 8.

Original entry on oeis.org

1, 17, 803, 1241, 20264753, 28214180783393, 228454543831049
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Oct 04 2019

Keywords

Comments

Conjecture: For n > 1, k^n == 1-k (mod n) has an infinite number of positive solutions.
No term can be a multiple of 2, 3, 5, 7, or 13. Also 4879573990210017348077958628152400091281634488825721395187 is a term. - Giovanni Resta, Oct 07 2019
Also 6788776064693081883870036833 is a term. - Max Alekseyev, Dec 27 2024

Crossrefs

Subsequence of A008364.
Solutions to k^m == k-1 (mod m): 1 (k = 1), A006521 (k = 2), A015973 (k = 3), A327840 (k = 4), A123047 (k = 5), A327943 (k = 6), A328033 (k = 7), A327468 (k = 8), this sequence (k = 9).
Cf. A253212 (9^n + 8).

Programs

  • Magma
    [1] cat [n: n in [1..10^8] | Modexp(9, n, n) + 8 eq n];
  • PARI
    isok(n) = Mod(9, n)^n==-8; \\ Michel Marcus, Oct 05 2019

Formula

a(n) > 15n for large enough n. (Surely the sequence grows superlinearly, but I can't prove it.) - Charles R Greathouse IV, Dec 27 2024

Extensions

a(7) from Giovanni Resta confirmed and a(6) added by Max Alekseyev, Dec 27 2024
Previous Showing 11-18 of 18 results.

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