

A359411


a(n) is the number of divisors of n that are both infinitary and exponential.


8



1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
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OFFSET

1,8


COMMENTS

If e > 0 is the exponent of the highest power of p dividing n (where p is a prime), then for each divisor d of n that is both an infinitary and an exponential divisor, the exponent of the highest power of p dividing d is a number k such that k  e and the bitwise AND of e and k is equal to k.
The least term that is higher than 2 is a(216) = 4.
The position of the first appearance of a prime p in this sequence is 2^A359081(p), if A359081(p) > 1. E.g., 2^39 = 549755813888 for p = 3, 2^175 = 4.789...*10^52 for p = 5, and 2^1275 = 6.504...*10^383 for p = 7.


LINKS



FORMULA

Multiplicative with a(p^e) = A246600(e).
a(n) = 1 if and only if n is in A138302.
Asymptotic mean: Limit_{m>oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + Sum_{k>=1} A246600(k)/p^k) = 1.135514937... .


EXAMPLE

a(8) = 2 since 8 has 2 divisors that are both infinitary and exponential: 2 and 8.


MATHEMATICA

s[n_] := DivisorSum[n, 1 &, BitAnd[n, #] == # &]; f[p_, e_] := s[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]


PROG

(PARI) s(n) = sumdiv(n, d, bitand(d, n)==d);
a(n) = {my(f = factor(n)); prod(i = 1, #f~, s(f[i, 2])); }
(Python)
from math import prod
from sympy import divisors, factorint
def A359411(n): return prod(sum(1 for d in divisors(e, generator=True) if ed == e) for e in factorint(n).values()) # Chai Wah Wu, Sep 01 2023


CROSSREFS



KEYWORD

nonn,mult


AUTHOR



STATUS

approved



