A368543 The number of divisors of n whose prime factors are all of the form 2^k + 1 (A092506).
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, 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, 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
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
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]
-
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))};
Formula
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.
a(n) >= 1, with equality if and only if all the prime factors of n are not of the form 2^k + 1.
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).
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.