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A210184 Number of distinct residues of all factorials mod prime(n). 5
2, 3, 4, 5, 6, 10, 12, 12, 17, 19, 21, 26, 29, 26, 31, 35, 37, 41, 42, 39, 44, 49, 55, 59, 59, 65, 71, 75, 63, 73, 80, 82, 90, 90, 104, 86, 103, 104, 107, 111, 113, 114, 120, 125, 120, 115, 139, 149, 132, 141, 147, 150, 147, 164, 166, 172, 172, 170, 172, 180 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Conjecture: a(n)/p_n > 1/2.

The standard (folklore?) conjecture is that a(n)/prime(n) = 1 - 1/e = 0.63212.... - Charles R Greathouse IV, May 11 2015

LINKS

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1...1000 from Alois P. Heinz)

Yong-Gao Chen and Li-Xia Dai, Congruences with factorials modulo p, INTEGERS 6 (2006), #A21.

EXAMPLE

Let n=4, p_4=7. We have modulo 7: 1!==1, 2!==2, 3!==6, 4!==3, 5!==1, 6!==6 and for m>=7, m!==0, such that we have 5 distinct residues 0,1,2,3,6. Therefore a(4) = 5.

MAPLE

a:= proc(n) local p, m, i, s;

      p:= ithprime(n);

      m:= 1;

      s:= {};

      for i to p do m:= m*i mod p; s:=s union {m} od;

      nops(s)

    end:

seq(a(n), n=1..100);  # Alois P. Heinz, Mar 19 2012

MATHEMATICA

Table[Length[Union[Mod[Range[Prime[n]]!, Prime[n]]]], {n, 100}] (* T. D. Noe, Mar 18 2012 *)

PROG

(PARI) apply(p->#Set(vector(p, n, n!)%p), primes(100)) \\ Charles R Greathouse IV, May 11 2015

(PARI) a(n, p=prime(n))=my(t=1); #Set(vector(p, n, t=(t*n)%p)) \\ Charles R Greathouse IV, May 11 2015

CROSSREFS

Cf. A000040, A000142.

Sequence in context: A102570 A113007 A113325 * A015853 A018642 A116998

Adjacent sequences:  A210181 A210182 A210183 * A210185 A210186 A210187

KEYWORD

nonn

AUTHOR

Vladimir Shevelev, Mar 18 2012

STATUS

approved

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Last modified December 1 20:25 EST 2021. Contains 349435 sequences. (Running on oeis4.)