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A037445
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Number of infinitary divisors (or i-divisors) of n.
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53
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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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
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FORMULA
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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(A000079(A000120(A124010(n,k))): k=1..A001221(n)). - 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)
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EXAMPLE
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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.
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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Cf. A000120, A001316, A004607, A007358, A007357, A038148, A049417, A064547, A074848, A124010, A156552, A286575.
Sequence in context: A084718 A154851 A281854 * A318307 A286324 A318472
Adjacent sequences: A037442 A037443 A037444 * A037446 A037447 A037448
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KEYWORD
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nonn,nice,easy,mult
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AUTHOR
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Yasutoshi Kohmoto
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EXTENSIONS
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Corrected and extended by Naohiro Nomoto, Jun 21 2001
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STATUS
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approved
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