A115092 -id:A115092 - cp's OEIS frontend

A073944 a(n) is the smallest m such that n-th prime divides m! + 1.

1, 2, 4, 3, 5, 12, 16, 9, 14, 18, 30, 36, 40, 21, 23, 52, 15, 8, 18, 7, 72, 23, 13, 88, 96, 100, 6, 106, 86, 112, 63, 65, 16, 16, 50, 150, 156, 81, 166, 172, 89, 180, 95, 102, 196, 99, 210, 222, 61, 228, 64, 210, 240, 97, 31, 131, 9, 93, 40, 280, 282, 45, 63, 220, 312

Offset: 1

Jason Earls, Nov 13 2002

Essentially the same as A072937. - R. J. Mathar, Sep 23 2008

By Wilson's theorem, a(n) < prime(n). Sequence A115092 gives the number of m such that prime(n) divides m!+1. - T. D. Noe, Mar 01 2006, Jan 10 2009

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000 (first 2000 terms from T. D. Noe)

Cf. A038507.

Cf. A072937 (same sequence without a(1)).

Table[p=Prime[i]; m=1; While[m0, m++ ]; m, {i,100}] (* T. D. Noe, Mar 01 2006 *)
Module[{sm=Table[{m,m!+1},{m,400}]},Table[SelectFirst[sm,Mod[#[[2]],p]==0&],{p,Prime[ Range[70]]}]][[;;,1]] (* Harvey P. Dale, Sep 15 2023 *)

A115091 Primes p such that p^2 divides m!+1 for some integer m

Original entry on oeis.org

5, 11, 13, 47, 71, 563, 613

Offset: 1

T. D. Noe, Mar 01 2006

By Wilson's theorem, we know that there is an m=p-1 such that p divides m!+1. Sequence A115092 gives the number of m for each prime. Occasionally p^2 also divides m!+1. These primes seem to be only slightly more plentiful than Wilson primes (A007540). No other primes < 10^6.
There is no prime p < 10^8 such that p^2 divides m!+1 for some m <= 1200. [From F. Brunault (brunault(AT)gmail.com), Nov 23 2008]
For a(n), m = p-A259230(n). - Felix Fröhlich, Jan 24 2016

R. K. Guy, Unsolved Problems in Number Theory, 3rd Ed., New York, Springer-Verlag, 2004, Section A2.

Cf. A064237 (n!+1 is divisible by a square), A259230.

nn=1000; lst={}; Do[p=Prime[i]; p2=p^2; f=1; m=1; While[mMichael De Vlieger, Jan 24 2016, Version 10 *)

forprime(p=1, , for(k=1, p-1, if(Mod((p-k)!, p^2)==-1, print1(p, ", "); break({1})))) \\ Felix Fröhlich, Jan 24 2016

A154554 Primes p such that m=p-1 is the least number such that p divides m!+1.

Original entry on oeis.org

2, 3, 5, 13, 17, 31, 37, 41, 53, 73, 89, 97, 101, 107, 113, 151, 157, 167, 173, 181, 197, 211, 223, 229, 241, 281, 283, 313, 331, 337, 349, 353, 373, 409, 421, 433, 439, 457, 487, 509, 541, 547, 587, 617, 643, 653, 659, 677, 701, 751, 757, 761, 769, 773, 821

Offset: 1

T. D. Noe, Jan 12 2009

nonn

The graph of A073944 shows two prominent curves. This sequence gives the primes on the upper curve. Primes on the lower curve are in sequence A154555. Note that the terms of A115092 are 1 for these primes.

T. D. Noe, Table of n, a(n) for n=1..1000

Rest[Reap[Do[p=Prime[i]; f=1; m=1; While[f=Mod[f*m,p]; f+1

A160245 a(n) = index of the n-th prime in A051301 (least prime factor of m!+1).

Original entry on oeis.org

2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 3, 2, 2, 6, 1, 3, 3, 1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 4, 1, 2, 1, 1, 3, 3, 2, 2, 3, 1, 1, 1, 5, 3, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 4, 2, 2, 5, 1, 2, 1, 2, 1, 1, 2, 2, 1, 2, 3, 3, 3, 1, 2, 1, 3, 1, 1, 2, 1, 1, 2, 4, 2, 4

Offset: 1

Frederick Magata (frederick.magata(AT)web.de), May 05 2009

nonn

Because of Wilson's theorem A051301(p-1)=p for every prime p. Hence a(n)>0, and since A051301(k)>k, a(n) is actually finite.

The first 18 values of the sequence were calculated with Maple. The others were derived from T. D. Noe's b-file for b051301.txt.

			a(17)=3 because A051301(15)=A051301(43)=A051301(58)=59, and there are no other occurrences of 59=17th prime number in A051301.

Cf. A051301, A115092.

a:=proc(n) option remember; local k,l,p: p:=ithprime(n): l:=0: for k from 0 to p-2 do if A051301(k)=p then l:=l+1; fi; od; l+1; end;

prev={}; Table[p=Prime[n]; s=Select[Complement[Range[0,p-1],prev], Mod[ #!+1,p]==0&]; prev=Union[s,prev]; Length[s], {n,100}] (* T. D. Noe, May 12 2009 *)

Extended by T. D. Noe, May 12 2009

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A073944 a(n) is the smallest m such that n-th prime divides m! + 1.

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A115091 Primes p such that p^2 divides m!+1 for some integer m

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A154554 Primes p such that m=p-1 is the least number such that p divides m!+1.

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A160245 a(n) = index of the n-th prime in A051301 (least prime factor of m!+1).

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