A368544 The number of divisors of n whose prime factors are all of the form k^2+1.

1, 2, 1, 3, 2, 2, 1, 4, 1, 4, 1, 3, 1, 2, 2, 5, 2, 2, 1, 6, 1, 2, 1, 4, 3, 2, 1, 3, 1, 4, 1, 6, 1, 4, 2, 3, 2, 2, 1, 8, 1, 2, 1, 3, 2, 2, 1, 5, 1, 6, 2, 3, 1, 2, 2, 4, 1, 2, 1, 6, 1, 2, 1, 7, 2, 2, 1, 6, 1, 4, 1, 4, 1, 4, 3, 3, 1, 2, 1, 10, 1, 2, 1, 3, 4, 2, 1

Offset

1,2

Comments

The number of terms of A180252 that divide n.

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

Multiplicative with a(p^e) = e+1 if p is of the form k^2+1, and 1 otherwise.

a(n) >= 1, with equality if and only if all the prime factors of n are in A070303.

a(n) <= A000005(n), with equality if and only if n is in A180252.

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{k in A005574} (1 + 1/k^2) = 2.80986546... .

Mathematica

q[n_] := AllTrue[FactorInteger[n][[;; , 1]], IntegerQ[Sqrt[# - 1]] &]; f[p_, e_] := If[q[p], e + 1, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

Prog

(PARI) a(n) = {my(f=factor(n)); prod(i=1, #f~, if(issquare(f[i, 1]-1), f[i, 2] + 1, 1))};
(Python)
from math import prod
from sympy import factorint
from sympy.ntheory.primetest import is_square
def A368544(n): return prod(e+1 for p, e in factorint(n).items() if is_square(p-1)) # Chai Wah Wu, Dec 30 2023

Crossrefs

Cf. A002496, A005574, A070303, A180252.

Sequence in context: A374998 A375000 A337195 * A260088 A364969 A374999

Adjacent sequences: A368541 A368542 A368543 * A368545 A368546 A368547

Keyword

nonn,easy,mult

Author

Amiram Eldar, Dec 29 2023

Status

approved