A211411 -id:A211411 - cp's OEIS frontend

A063980 Pillai primes: primes p such that there exists an integer m such that m! + 1 == 0 (mod p) and p != 1 (mod m).

23, 29, 59, 61, 67, 71, 79, 83, 109, 137, 139, 149, 193, 227, 233, 239, 251, 257, 269, 271, 277, 293, 307, 311, 317, 359, 379, 383, 389, 397, 401, 419, 431, 449, 461, 463, 467, 479, 499, 503, 521, 557, 563, 569, 571, 577, 593, 599, 601, 607

Offset: 1

R. K. Guy, Sep 08 2001

Hardy & Subbarao prove that this sequence is infinite. An upper bound can be extracted from their proof: a(n) < e^e^...^e^O(n log n) with e appearing n times. This tetrational bound could be improved with results on the disjointness of the factorizations of numbers of the form k! + 1. - Charles R Greathouse IV, Sep 15 2015

Named after the Indian mathematician Subbayya Sivasankaranarayana Pillai (1901-1950). - Amiram Eldar, Jun 16 2021

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
G. E. Hardy and M. V. Subbarao, A modified problem of Pillai and some related questions, Amer. Math. Monthly, Vol. 109, No. 6 (2002), pp. 554-559; alternative link.
Wikipedia, Pillai prime.

Smallest m is given in A063828, largest in A211411.

ok[p_] := (r = False; Do[If[Mod[m! + 1, p] == 0 && Mod[p, m] != 1, r = True; Break[]], {m, 2, p}]; r); Select[Prime /@ Range[111], ok] (* Jean-François Alcover, Apr 22 2011 *)
nn=1000; fact=1+Rest[FoldList[Times,1,Range[nn]]]; t={}; Do[p=Prime[i]; m=2; While[mT. D. Noe, Apr 22 2011 *)

is(p)=my(t=Mod(5040,p)); for(m=8, p-2, t*=m; if(t==-1 && p%m!=1, return(isprime(p)))); 0 \\ Charles R Greathouse IV, Feb 10 2013

More terms from David W. Wilson, Sep 08 2001

A063828 Smallest m associated with n-th Pillai prime (A063980).

Original entry on oeis.org

14, 18, 15, 8, 18, 9, 23, 13, 86, 16, 16, 50, 102, 61, 64, 210, 97, 31, 9, 93, 40, 45, 63, 220, 91, 122, 35, 85, 198, 93, 128, 316, 366, 74, 300, 151, 290, 15, 400, 282, 22, 188, 167, 191, 360, 426, 274, 271, 456, 278, 229, 324, 135, 498, 189

Offset: 1

N. J. A. Sloane, Sep 23 2001

nonn

			14! + 1 == 0 (mod 23), while 23 != 1 (mod 14), where 23=A063980(1) so a(1)=14.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
G. E. Hardy and M. V. Subbarao, A modified problem of Pillai and some related questions, Amer. Math. Monthly 109 (2002), no. 6, 554-559.

Cf. A063980, A211411.

nn=1000; fact=1+Rest[FoldList[Times,1,Range[nn]]]; t={}; Do[p=Prime[i]; m=2; While[mT. D. Noe, Apr 22 2011 *)

first(p)=my(t=Mod(5040, p)); for(m=8, p, t*=m; if(t==-1 && p%m!=1, return(m))); 0
Pillai(p)=my(t=Mod(5040, p)); for(m=8, p-2, t*=m; if(t==-1 && p%m!=1, return(1))); 0
apply(first, select(Pillai, primes(300))) \\ Charles R Greathouse IV, Feb 10 2013

More terms from Vladeta Jovovic, Sep 27 2001

cp's OEIS Frontend

A063980 Pillai primes: primes p such that there exists an integer m such that m! + 1 == 0 (mod p) and p != 1 (mod m).

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