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.

Showing 1-3 of 3 results.

A372771 Numbers m such that the congruence x^(m+1) == m (mod m+1) is solvable.

Original entry on oeis.org

1, 2, 4, 6, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 25, 26, 28, 30, 32, 33, 34, 36, 38, 40, 42, 44, 46, 48, 49, 50, 52, 54, 56, 57, 58, 60, 62, 64, 66, 68, 70, 72, 73, 74, 76, 78, 80, 81, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 105, 106, 108, 110
Offset: 1

Views

Author

Juri-Stepan Gerasimov, May 12 2024

Keywords

Comments

Includes all positive even numbers. - Robert Israel, Mar 14 2025

Examples

			9 is a term because x^(9+1) == 9 (mod 9+1) for x = 3 and x = 7, i.e., 3^10 = 59049 == 9 (mod 10) and 7^10 = 282475249 == 9 (mod 10).
		

Crossrefs

Programs

  • Maple
    select(m -> traperror(NumberTheory:-ModularRoot(m,m+1,m+1))::integer, [$1..200]); # Robert Israel, Mar 14 2025
  • Mathematica
    Select[Range[1, 110], With[{m = #}, AnyTrue[Range[1, m + 1], PowerMod[#, m + 1, m + 1] == m &]] &] (* Robert P. P. McKone, May 14 2024 *)

Extensions

Terms corrected by Robert P. P. McKone, May 14 2024

A375347 a(n) is the number of nonnegative numbers k < n such that the congruence x^k == k (mod n) is solvable.

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 4, 4, 4, 6, 5, 7, 5, 8, 9, 8, 10, 10, 8, 11, 10, 13, 17, 13, 13, 14, 13, 15, 17, 18, 17, 16, 17, 22, 16, 17, 16, 19, 18, 21, 20, 22, 20, 22, 22, 32, 36, 25, 25, 28, 30, 23, 34, 28, 28, 25, 24, 36, 38, 33, 27, 31, 29, 32, 31, 39, 35, 36, 44, 35, 42, 31, 39, 36, 38
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Aug 12 2024

Keywords

Examples

			a(1) = 1 because x^0 == 0 (mod 1) is solvable where x: 0, 1, 2, 3, 4,.. A001477;
a(2) = 1 because x^0 == 0 (mod 2) is unsolvable,
                 x^1 == 1 (mod 2) is solvable where x: 1, 3, 5, 7, 9,.. A005408;
a(3) = 1 because x^0 == 0 (mod 3) is unsolvable,
                 x^1 == 1 (mod 3) is solvable where x: 1, 4, 7, 10, 13,.. A016777,
                 x^2 == 2 (mod 3) is unsolvable;
a(4) = 2 because x^0 == 0 (mod 4) is unsolvable,
                 x^1 == 1 (mod 4) is solvable where x: 1, 5, 9, 13, 16,.. A016813,
                 x^2 == 2 (mod 4) is unsolvable,
                 x^3 == 3 (mod 4) is solvable where x: 3, 7, 11, 15, 19,.. A004767.
		

Crossrefs

Programs

  • PARI
    is(k, n) = for (i=0, n-1, if (Mod(i, n)^k == k, return(1)));
    a(n) = sum(k=0, n-1, is(k, n)); \\ Michel Marcus, Aug 13 2024

Extensions

More terms from Michel Marcus, Aug 13 2024

A373899 Semiprimes q*p such that q^p == p (mod (q - p)), where q > p.

Original entry on oeis.org

6, 15, 21, 33, 35, 55, 65, 77, 85, 91, 133, 143, 145, 155, 161, 187, 209, 217, 221, 247, 253, 265, 299, 301, 323, 341, 377, 391, 403, 415, 437, 451, 481, 493, 533, 545, 551, 553, 559, 581, 589, 611, 629, 667, 671, 689, 697, 703, 713, 781, 793, 799, 817, 893, 899, 901
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Jun 22 2024

Keywords

Examples

			15 = 3*5 is a term because 5^3 == 3 (mod 2).
		

Crossrefs

Subsequence of A001358.
Cf. A037074 (subsequence), A046388, A371811.

Programs

  • Mathematica
    seqQ[n_] := Module[{f = FactorInteger[n], p, q}, If[f[[;; , 2]] == {1, 1}, p = f[[1, 1]]; q = f[[2, 1]]; PowerMod[q, p, q - p] == Mod[p, q - p], False]]; Select[Range[1000], seqQ] (* Amiram Eldar, Jun 26 2024 *)
Showing 1-3 of 3 results.