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%I #11 Dec 30 2023 09:33:24
%S 1,2,2,3,2,4,1,4,3,4,1,6,1,2,4,5,2,6,1,6,2,2,1,8,3,2,4,3,1,8,1,6,2,4,
%T 2,9,1,2,2,8,1,4,1,3,6,2,1,10,1,6,4,3,1,8,2,4,2,2,1,12,1,2,3,7,2,4,1,
%U 6,2,4,1,12,1,2,6,3,1,4,1,10,5,2,1,6,4,2,2,4,1,12,1,3,2,2,2,12,1,2,3,9
%N The number of divisors of n whose prime factors are all of the form 2^k + 1 (A092506).
%H Amiram Eldar, <a href="/A368543/b368543.txt">Table of n, a(n) for n = 1..10000</a>
%F Multiplicative with a(p^e) = e+1 if p is in A092506 (i.e., p is either 2 or a Fermat prime), and 1 otherwise.
%F a(n) >= 1, with equality if and only if all the prime factors of n are not of the form 2^k + 1.
%F a(n) <= A000005(n), with equality if and only if all the prime factors of n are in A092506 (n is in A143513 assuming that there are only 5 Fermat primes).
%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/Product_{k>=1} (1 - 1/A092506(k)) = 3.99999999906867742538... . This value is exactly 4294967295/1073741824 if there are only 5 Fermat primes.
%t q[n_] := AllTrue[FactorInteger[n][[;; , 1]], # - 1 == 2^IntegerExponent[# - 1, 2] &]; f[p_, e_] := If[q[p], e + 1, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
%o (PARI) a(n) = {my(f=factor(n)); prod(i=1, #f~, if((f[i,1]-1) >> valuation(f[i,1]-1, 2) == 1 , f[i,2] + 1, 1))};
%Y Cf. A019434, A092506, A143513.
%K nonn,easy,mult
%O 1,2
%A _Amiram Eldar_, Dec 29 2023