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A117658 Number of solutions to x^(k+1)=x^k mod n for some k>=1. 2
1, 2, 2, 3, 2, 4, 2, 5, 4, 4, 2, 6, 2, 4, 4, 9, 2, 8, 2, 6, 4, 4, 2, 10, 6, 4, 10, 6, 2, 8, 2, 17, 4, 4, 4, 12, 2, 4, 4, 10, 2, 8, 2, 6, 8, 4, 2, 18, 8, 12, 4, 6, 2, 20, 4, 10, 4, 4, 2, 12, 2, 4, 8, 33, 4, 8, 2, 6, 4, 8, 2, 20, 2, 4, 12 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

If n is prime, then the solutions are x = 0, 1, and so a(n) = 2. - Michael B. Porter, Jul 08 2016

The set of solutions is independent of the choice of k. - Michael B. Porter, Jul 08 2016

LINKS

Robert Israel, Table of n, a(n) for n = 1..10000

S. R. Finch, Idempotents and Nilpotents Modulo n, arXiv:math/0605019 [math.NT], 2006.

FORMULA

a(n) = Sum_{k=1..n} floor((k^n-k^(n-1))/n)-floor((k^n-k^(n-1)-1)/n). - Anthony Browne, Jul 06 2016

Multiplicative with a(p^e) = 1 + p^(e-1)) for primes p. - Robert Israel, Jul 06 2016

EXAMPLE

For n = 10, using k = 1, the solutions are x = 0, 1, 5, and 6, so a(10) = 4. - Michael B. Porter, Jul 08 2016

MAPLE

f:= proc(n) local F, f;

    F:= ifactors(n)[2];

    mul(1 + f[1]^(f[2]-1), f = F)

end proc:

map(f, [$1..100]); # Robert Israel, Jul 06 2016

MATHEMATICA

a[n_] := Module[{F, f}, F = FactorInteger[n]; Product[1 + f[[1]]^(f[[2]] - 1), {f, F}]]; a[1] = 1; Array[a, 100] (* Jean-Fran├žois Alcover, Nov 05 2016, after Robert Israel *)

CROSSREFS

Cf. A117659.

Sequence in context: A237433 A236515 A129294 * A289435 A067540 A218701

Adjacent sequences:  A117655 A117656 A117657 * A117659 A117660 A117661

KEYWORD

mult,nonn

AUTHOR

Steven Finch, Apr 11 2006

STATUS

approved

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Last modified July 22 03:22 EDT 2017. Contains 289648 sequences.