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A182590
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Number of distinct prime factors of 2^n - 1 of the form k*n + 1.
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13
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1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 1, 1, 2, 3, 2, 1, 2, 2, 1, 2, 3, 3, 1, 2, 1, 2, 2, 3, 2, 2, 3, 2, 2, 4, 3, 3, 2, 3, 3, 3, 1, 4, 4, 3, 3, 2, 3, 2, 3, 5, 2, 2, 1, 2, 3, 4, 2, 3, 2, 3, 1, 4, 3, 3, 3, 4, 5, 3, 1, 5, 3, 2, 3, 4, 2, 3, 2, 4, 3
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OFFSET
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2,9
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COMMENTS
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By Bang's theorem, a(n) > 0 for all n > 1, see A186522.
Primes p such that a(p) = 1 are the Mersenne exponents A000043.
Composite numbers m for which a(m) = 1 are A292079.
a(n) >= A086251(n), where equality is for all prime numbers and for some composite numbers (among others for all odd prime powers p^k with k > 1).
Theorem: if n is prime, then a(n) = A046800(n).
Conjecture: if a(n) = A046800(n), then n is prime.
Problem: is a(n) < A046800(n) for every composite n? (End)
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LINKS
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EXAMPLE
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For n=10 the prime factors of 2^n - 1 = 1023 are 3, 11 and 31, and 11 = n+1, 31 = 3n+1. Thus a(10)=2.
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MATHEMATICA
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m = 2; n = 2; nmax = 200;
While[n <= nmax, {l = FactorInteger[m^n - 1]; s = 0;
For[i = 1, i <= Length[l],
i++, {p = l[[i, 1]];
If[IntegerQ[(p - 1)/n] == True, s = s + l[[i, 2]]]; }];
a[n] = s; } n++; ];
Table[a[n], {n, 2, nmax}]
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PROG
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(PARI) a(n) = my(f = factor(2^n-1)); sum(k=1, #f~, ((f[k, 1]-1) % n)==0); \\ Michel Marcus, Sep 10 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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