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A037445
Number of infinitary divisors (or i-divisors) of n.
112
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
1,2
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
LINKS
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
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
EXAMPLE
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.
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 *)
PROG
(PARI) A037445(n) = factorback(apply(a -> 2^hammingweight(a), factorint(n)[, 2])) \\ Andrew Lelechenko, May 10 2014
(Haskell)
a037445 = product . map (a000079 . a000120) . a124010_row
-- Reinhard Zumkeller, Mar 19 2013
(Scheme) (define (A037445 n) (if (= 1 n) n (* (A001316 (A067029 n)) (A037445 (A028234 n))))) ;; Antti Karttunen, May 28 2017
(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
KEYWORD
nonn,nice,easy,mult
EXTENSIONS
Corrected and extended by Naohiro Nomoto, Jun 21 2001
STATUS
approved