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.

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A178813 a(n) = (prime(n)^(p-1) - 1)/p^2 mod p, where p is the first prime that divides (prime(n)^(p-1) - 1)/p.

Original entry on oeis.org

487, 4, 1, 1, 46, 1, 0, 1, 11, 1, 2, 1, 0, 2
Offset: 1

Views

Author

Jonathan Sondow, Jun 17 2010

Keywords

Comments

a(n) = (prime(n)^(p-1) - 1)/p^2 mod p, where p = A174422(n) is the first Wieferich prime base prime(n).
a(n) = (prime(n)^(p-1) - 1)/p^2 mod p, where p is the first prime such that p^2 divides prime(n)^(p-1) - 1.
See references and additional comments, links, and cross-refs in A001220 and A039951.
a(15) > 2451011, a(16) = 1, a(17) = 4, a(18) = 1, a(19) = 5, a(20) = 2, a(21) = 0, a(22) = 6, a(23) = 1186, a(24) = 0, a(25) = 0, a(26) = 1, a(27) > 10^5, a(28) = 0, a(29) = 1, a(30) = 0, a(31) = 1, a(32) = 7, a(33) = 0, a(35) = 1, a(36) = 4, a(37) = 1, a(38) = 0, a(40) = 1, a(41) = 2, a(42) = 1, a(43) = 2, a(44) = 0, a(45) = 1, a(46) = 2, a(48) = 30, a(49) = 3, a(50) = 1. - J.W.L. (Jan) Eerland, Sep 27 2024

Examples

			Prime(2) = 3 and the first prime p that divides (3^(p-1) - 1)/p is 11, so a(2) = (3^10 - 1)/11^2 mod 11 = 488 mod 11 = 4.
		

Crossrefs

Programs

  • Mathematica
    Table[If[IntegerQ[s[[2]]],s,{s[[1]], "no solution in range 1 <= k <= 10^5"}], {s,Table[k = 1;Monitor[Parallelize[While[k <= 10^5,If[IntegerQ[((Prime[n]^(Prime[k] - 1) - 1)/Prime[k])/Prime[k]],Break[]]; k++];{n, Mod[(Prime[n]^(Prime[k] - 1) - 1)/Prime[k]^2, Prime[k]]}],k], {n, 1, 10}]}] (* J.W.L. (Jan) Eerland, Sep 27 2024 *)

Formula

a(n) = k mod 2, if prime(n) = 4k+1.
a(n) = A178814(prime(n)) .
a(1) = A178812(1).

A178814 (n^(p-1) - 1)/p^2 mod p, where p is the first prime that divides (n^(p-1) - 1)/p.

Original entry on oeis.org

0, 487, 4, 974, 1, 30384, 1, 1, 0, 2, 46, 1571, 1, 17, 24160, 855, 0, 4, 1, 189, 1, 5, 11, 1, 0, 0, 1, 0, 1, 3, 2, 3, 0, 19632919407, 1, 60768, 1, 11, 1435, 8, 0, 0, 2, 2, 1, 1
Offset: 1

Views

Author

Jonathan Sondow, Jun 17 2010

Keywords

Comments

(n^(p-1) - 1)/p^2 mod p, where p is the first prime such that p^2 divides n^(p-1) - 1.
See references and additional comments, links, and cross-refs in A001220 and A039951.

Examples

			The first prime p that divides (3^(p-1) - 1)/p is 11, so a(3) = (3^10 - 1)/11^2 mod 11 = 488 mod 11 = 4.
		

Crossrefs

a(2) = A178812(1) = A178813(1). Cf. A001220, A039951, A174422.

Formula

a(n) = (n^(p-1) - 1)/p^2 mod p, where p = A039951(n).
a(n) = k mod 2, if n = 4k+1.
a(prime(n)) = A178813(n).
Showing 1-2 of 2 results.