A182816 Number of values b in Z/nZ such that b^n = b.
1, 2, 3, 2, 5, 4, 7, 2, 3, 4, 11, 4, 13, 4, 9, 2, 17, 4, 19, 4, 9, 4, 23, 4, 5, 4, 3, 8, 29, 8, 31, 2, 9, 4, 9, 4, 37, 4, 9, 4, 41, 8, 43, 4, 15, 4, 47, 4, 7, 4, 9, 8, 53, 4, 9, 4, 9, 4, 59, 8, 61, 4, 9, 2, 25, 24, 67, 4, 9, 16, 71, 4, 73, 4, 9, 8, 9, 8, 79, 4, 3, 4, 83, 8, 25, 4, 9, 4, 89, 8, 49, 4, 9, 4, 9, 4, 97, 4, 9, 4, 101, 8, 103, 4, 45, 4, 107, 4, 109, 8, 9, 8, 113
Offset: 1
Keywords
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A063994.
Programs
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Maple
f:= n -> mul(1+igcd(n-1,p[1]-1), p = ifactors(n)[2]): map(f, [$1..200]); # Robert Israel, Sep 05 2018
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Mathematica
Table[Times @@ Map[(1 + GCD[n - 1, # - 1]) &, FactorInteger[n][[All, 1]] ], {n, 113}] (* Michael De Vlieger, Sep 01 2020 *)
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PARI
A182816(n)=sum(a=1,n,Mod(a,n)^n==a);
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PARI
{ A182816(n) = my(p=factor(n)[,1]); prod(j=1,#p,1+gcd(n-1,p[j]-1)); } \\ Max Alekseyev, Dec 06 2010
Formula
a(n) = Product_{i=1..m} (1 + gcd(n-1, p_i-1)), where p_1, p_2, ..., p_m are all distinct primes dividing n. - Max Alekseyev, Dec 06 2010
a(p^k) = p for prime p with k > 0. - Thomas Ordowski, Sep 05 2018
Comments