A074848 Number of 4-infinitary divisors of n.

1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 2, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, 2, 4, 4, 4, 4, 9, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 2, 4, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 12, 2, 4, 6, 6, 4, 8, 2, 6, 4, 8, 2, 12, 2, 4, 6, 6, 4, 8, 2, 4, 2, 4, 2, 12, 4, 4, 4, 8, 2, 12, 4, 6, 4, 4, 4, 8, 2, 6, 6, 9, 2, 8, 2, 8, 8

Offset: 1

Author

Yasutoshi Kohmoto, Sep 10 2002

Keywords

• nonn
• mult

Comments

If n = Product p(i)^r(i) and d = Product p(i)^s(i), each s(i) has a digit a<=b in its 4-ary expansion everywhere that the corresponding r(i) has a digit b, then d is a 4-infinitary-divisor of n.

Examples

			2^4*3 is a 4-infinitary-divisor of 2^5*3^2 because 2^4*3 = 2^10*3^1 and 2^5*3^2 = 2^11*3^2 in 4-ary expanded power. All corresponding digits satisfy the condition. 1<=1, 0<=1, 1<=2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Index entries for sequences computed from exponents in factorization of n

Crossrefs

Cf. A037445, A038148, A074847, A268444.

Programs

Maple

A074848 := proc(n) if n= 1 then 1; else ifa := ifactors(n)[2] ; a := 1; for f in ifa do e := convert(op(2,f),base,4) ; a := a*mul(d+1,d=e) ; end do: end if; end proc:
seq(A074848(n),n=1..70) ; # R. J. Mathar, Feb 08 2011

Mathematica

f[p_, e_] := Times @@ (IntegerDigits[e, 4] + 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]  (* Amiram Eldar, Sep 09 2020 *)

PARI

A268444(n) = { my(m=1, d); while(n, d = (n%4); m *= (1+d); n = (n-d)/4); m; };
A074848(n) = factorback(apply(e -> A268444(e), factorint(n)[, 2])) \\ (After A037445) - Antti Karttunen, May 28 2017

Python

from math import prod
from sympy import factorint
from gmpy2 import digits
def A268444(n):
    s = digits(n,4)
    return prod((int(d)+1)**s.count(d) for d in '123')
def A074848(n): return prod(A268444(e) for e in factorint(n).values()) # Chai Wah Wu, Apr 24 2025

Scheme

(definec (A074848 n) (if (= 1 n) n (* (A268444 (A067029 n)) (A074848 (A028234 n))))) ;; Antti Karttunen, May 28 2017

Formula

Multiplicative: If e = sum d_k 4^k, then a(p^e) = prod (d_k+1). - Christian G. Bower, May 19 2005

a(1) = 1; for n > 1, a(n) = A268444(A067029(n)) * a(A028234(n)). [After Christian G. Bower's 2005 formula.] - _Antti Karttunen, May 28 2017

Extensions

More terms from Antti Karttunen, May 28 2017

Name shortened by Amiram Eldar, Sep 09 2020