A136655 Product of odd divisors of n.
1, 1, 3, 1, 5, 3, 7, 1, 27, 5, 11, 3, 13, 7, 225, 1, 17, 27, 19, 5, 441, 11, 23, 3, 125, 13, 729, 7, 29, 225, 31, 1, 1089, 17, 1225, 27, 37, 19, 1521, 5, 41, 441, 43, 11, 91125, 23, 47, 3, 343, 125, 2601, 13, 53, 729, 3025, 7, 3249, 29, 59, 225, 61, 31, 250047, 1, 4225, 1089
Offset: 1
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Haskell
a136655 = product . a182469_row -- Reinhard Zumkeller, May 01 2012
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Maple
with(numtheory); f:=proc(n) local t1,i,k; t1:=divisors(n); k:=1; for i in t1 do if i mod 2 = 1 then k:=k*i; fi; od; k; end; # N. J. A. Sloane, Jul 14 2008
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Mathematica
Array[Times @@ Select[Divisors@ #, OddQ] &, 66] (* Michael De Vlieger, Aug 03 2017 *) a[n_] := (oddpart = n/2^IntegerExponent[n, 2])^(DivisorSigma[0, oddpart]/2); Array[a, 100] (* Amiram Eldar, Jun 26 2022 *)
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PARI
a(n) = my(d=divisors(n)); prod(k=1, #d, if (d[k]%2, d[k], 1)); \\ Michel Marcus, Aug 04 2017
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Python
from math import isqrt from sympy import divisor_count def A136655(n): d = divisor_count(m:=n>>(~n&n-1).bit_length()) return isqrt(m)**d if d&1 else m**(d>>1) # Chai Wah Wu, Jun 27 2025
Formula
a(p) = p if p noncomposite; a(2^n) = 1; a(pq) = p^2 * q^2 when p, q are odd primes.
a(n) = sqrt(n^od(n)/2^ed(n)), where od(n) = number of odd divisors of n = tau(2*n)-tau(n) and ed(n) = number of even divisors of n = 2*tau(n)-tau(2*n). - Vladeta Jovovic, Jun 25 2008
a(n) = Product_{h == 1 mod 4 and h | n}*Product_{i == 3 mod 4 and i | n}.
a(n) = Product_{j == 1 mod 6 and j | n}*Product_{k == 5 mod 6 and k | n}.
Extensions
More terms from N. J. A. Sloane, Jul 14 2008
Edited by N. J. A. Sloane, Aug 29 2008 at the suggestion of R. J. Mathar
Comments