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 1-2 of 2 results.

A178812 (2^(p-1) - 1)/p^2 modulo prime p, if p^2 divides 2^(p-1) - 1.

Original entry on oeis.org

487, 51
Offset: 1

Views

Author

Jonathan Sondow, Jun 16 2010

Keywords

Comments

(2^(p-1) - 1)/p^2 modulo p, where p is a Wieferich prime A001220.
(2^(p-1) - 1)/p^2 modulo p, if prime p divides the Fermat quotient (2^(p-1) - 1)/p.
See A001220 for references, links, and additional comments.

Examples

			a(1) = 487 as the first Wieferich prime is 1093 and (2^1092 - 1)/1093^2 == 487 (mod 1093).
The 2nd Wieferich prime is 3511 and (2^3510 - 1)/3511^2 == 51 (mod 3511), so a(2) = 51.
		

Crossrefs

Formula

a(n) = (2^(p-1) - 1)/p^2 modulo p, where p = A001220(n).
a(1) = A178813(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.