A037445 Number of infinitary divisors (or i-divisors) of n.

1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 2, 2, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 4, 4, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 8, 4, 8, 4, 4, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 2, 4, 2, 8, 4, 4, 4, 8, 2, 8, 4, 4, 4, 4, 4, 8, 2, 4, 4, 4, 2, 8, 2, 8, 8

Offset: 1

Author

Yasutoshi Kohmoto

Keywords

• nonn
• nice
• easy
• mult

Comments

A divisor of n is called infinitary if it is a product of divisors of the form p^{y_a 2^a}, where p^y is a prime power dividing n and sum_a y_a 2^a is the binary representation of y.

The smallest number m with exactly 2^n infinitary divisors is A037992(n); for these values m, a(m) increases also to a new record. - Bernard Schott, Mar 09 2023

Examples

			For n = 8, n = 2^3 = 2^"11" (writing 3 in binary) so the infinitary divisors are 2^"00" = 1, 2^"01" = 2, 2^"10" = 4 and 2^"11" = 8, so a(8) = 4.
For n = 90, n = 2*5*9 where 2,5,9 are in A050376, so a(90) = 2^3 = 8.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Steven R. Finch, Unitarism and Infinitarism, February 25, 2004. [Cached copy, with permission of the author]
• J. O. M. Pedersen, Tables of Aliquot Cycles [Broken link]
• J. O. M. Pedersen, Tables of Aliquot Cycles [Via Internet Archive Wayback-Machine]
• J. O. M. Pedersen, Tables of Aliquot Cycles [Cached copy, pdf file only]
• Eric Weisstein's World of Mathematics, Infinitary Divisor
• Index entries for sequences computed from exponents in factorization of n

Crossrefs

Cf. A000120, A001316, A004607, A007358, A007357, A037992, A038148, A049417, A064547, A074848, A077609, A124010, A156552, A286575.

Programs

Haskell

a037445 = product . map (a000079 . a000120) . a124010_row
-- Reinhard Zumkeller, Mar 19 2013

Maple

A037445 := proc(n)
    local a,p;
    a := 1 ;
    for p in ifactors(n)[2] do
        a := a*2^wt(p[2]) ;
    end do:
    a ;
end proc: # R. J. Mathar, May 16 2016

Mathematica

Table[Length@((Times @@ (First[it]^(#1 /. z -> List)) & ) /@
Flatten[Outer[z, Sequence @@ bitty /@
Last[it = Transpose[FactorInteger[k]]], 1]]), {k, 2, 240}]
bitty[k_] := Union[Flatten[Outer[Plus, Sequence @@ ({0, #1} & ) /@ Union[2^Range[0, Floor[Log[2, k]]]*Reverse[IntegerDigits[k, 2]]]]]]
y[n_] := Select[Range[0, n], BitOr[n, # ] == n & ] divisors[Infinity][1] := {1}
divisors[Infinity][n_] := Sort[Flatten[Outer[Times, Sequence @@ (FactorInteger[n] /. {p_, m_Integer} :> p^y[m])]]] Length /@ divisors[Infinity] /@ Range[105] (* Paul Abbott (paul(AT)physics.uwa.edu.au), Apr 29 2005 *)
a[1] = 1; a[n_] := Times @@ Flatten[ 2^DigitCount[#, 2, 1]&  /@ FactorInteger[n][[All, 2]] ]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Aug 19 2013, after Reinhard Zumkeller *)

PARI

A037445(n) = factorback(apply(a -> 2^hammingweight(a), factorint(n)[,2])) \\ Andrew Lelechenko, May 10 2014

Python

from sympy import factorint
def wt(n): return bin(n).count("1")
def a(n):
    f=factorint(n)
    return 2**sum([wt(f[i]) for i in f]) # Indranil Ghosh, May 30 2017

Scheme

(define (A037445 n) (if (= 1 n) n (* (A001316 (A067029 n)) (A037445 (A028234 n))))) ;; Antti Karttunen, May 28 2017

Formula

Multiplicative with a(p^e) = 2^A000120(e). - David W. Wilson, Sep 01 2001

Let n = q_1*...*q_k, where q_1,...,q_k are different terms of A050376. Then a(n) = 2^k (the number of subsets of a set with k elements is 2^k). - Vladimir Shevelev, Feb 19 2011.

a(n) = Product_{k=1..A001221(n)} A000079(A000120(A124010(n,k))). - Reinhard Zumkeller, Mar 19 2013

From Antti Karttunen, May 28 2017: (Start)

a(n) = A286575(A156552(n)). [Because multiplicative with a(p^e) = A001316(e).]

a(n) = 2^A064547(n). (End)

a(A037992(n)) = 2^n. - Bernard Schott, Mar 10 2023

Extensions

Corrected and extended by Naohiro Nomoto, Jun 21 2001