A007955 Product of divisors of n.

1, 2, 3, 8, 5, 36, 7, 64, 27, 100, 11, 1728, 13, 196, 225, 1024, 17, 5832, 19, 8000, 441, 484, 23, 331776, 125, 676, 729, 21952, 29, 810000, 31, 32768, 1089, 1156, 1225, 10077696, 37, 1444, 1521, 2560000, 41, 3111696, 43, 85184, 91125, 2116, 47, 254803968, 343

Offset: 1

Author

R. Muller

Keywords

• nonn
• nice

Comments

All terms of this sequence occur only once. See the second T. D. Noe link for a proof. - T. D. Noe, Jul 07 2008

Every natural number has a unique representation in terms of divisor products. See the W. Lang link. - Wolfdieter Lang, Feb 08 2011

a(n) = n only if n is prime or 1 (or, if n is in A008578). - Alonso del Arte, Apr 18 2011

Sometimes called the "divisorial" of n. - Daniel Forgues, Aug 03 2012

a(n) divides EulerPhi(x^n-y^n) (see A. Rotkiewicz link). - Michel Marcus, Dec 15 2012

The proof that all the terms of this sequence occur only once (mentioned above) was given by Niven in 1984. - Amiram Eldar, Aug 16 2020

Examples

			Divisors of 10 = [1, 2, 5, 10]. So, a(10) = 2*5*10 = 100. - _Indranil Ghosh_, Mar 22 2017
		

References

• József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 1, p. 57.
• J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 83.

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
• Wolfdieter Lang, Divisor Product Representation for Natural Numbers.
• M. Le, On Smarandache Divisor Products, Smarandache Notions Journal, Vol. 10, No. 1-2-3, 1999, 144-145.
• F. Luca, On the product of divisors of n and sigma(n), J. Ineq. Pure Appl. Math. 4 (2) 2003, Article 46.
• T. D. Noe, The Divisor Product is Unique
• A. Rotkiewicz, On the numbers Phi(a^n +/- b^n), Proc. Amer. Math. Soc. 12 (1961), 419-421.
• Rodica Simon and Frank W. Schmid, Problem E 2946, The American Mathematical Monthly, Vol. 89, No. 5 (1982), p. 333, Ivan Niven, Product of all Positive Divisors of n, solution to problem E 2946, ibid., Vol. 91, No. 10 (1984), p. 650.
• F. Smarandache, Only Problems, Not Solutions!.
• Eric Weisstein's World of Mathematics, Divisor Product.
• Zhu Weiyi, On the divisor product sequences, Smarandache Notions J., Vol. 14 (2004), pp. 144-146.
• OEIS Wiki, Divisorial.

Crossrefs

Cf. A000005, A000196, A001222, A007422, A007956, A027750, A030628, A046523, A069264, A072046, A111398, A162947, A224381, A243103, A280076, A283995.

Cf. A000203 (sums of divisors).

Cf. A000010 (comments on product formulas).

Programs

GAP

List(List([1..50],n->DivisorsInt(n)),Product); # Muniru A Asiru, Feb 17 2019

Haskell

a007955 = product . a027750_row  -- Reinhard Zumkeller, Feb 06 2012

Magma

f := function(n); t1 := &*[d : d in Divisors(n) ]; return t1; end function;

Maple

A007955 := proc(n) mul(d,d=numtheory[divisors](n)) ; end proc: # R. J. Mathar, Mar 17 2011
seq(isqrt(n^numtheory[tau](n)), n=1..50); # Gary Detlefs, Feb 15 2019

Mathematica

Array [ Times @@ Divisors[ # ]&, 100 ]
a[n_] := n^(DivisorSigma[0, n]/2); Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Nov 21 2013 *)

PARI

a(n)=if(issquare(n,&n),n^numdiv(n^2),n^(numdiv(n)/2)) \\ Charles R Greathouse IV, Feb 11 2011

Python

from sympy import prod, divisors
print([prod(divisors(n)) for n in range(1, 51)]) # Indranil Ghosh, Mar 22 2017

Python

from math import isqrt
from sympy import divisor_count
def A007955(n):
    d = divisor_count(n)
    return isqrt(n)**d if d % 2 else n**(d//2) # Chai Wah Wu, Jan 05 2022

Sage

[prod(divisors(n)) for n in (1..100)] # Giuseppe Coppoletta, Dec 16 2014

Sage

[n^(sigma(n,0)/2) for n in (1..49)] # Stefano Spezia, Jul 14 2025

Scheme

;; A naive stand-alone implementation:
(define (A007955 n) (let loop ((d n) (m 1)) (cond ((zero? d) m) ((zero? (modulo n d)) (loop (- d 1) (* m d))) (else (loop (- d 1) m)))))
;; Faster, if A000005 and A000196 are available:
(define (A007955 n) (A000196 (expt n (A000005 n))))
;; Antti Karttunen, Mar 22 2017

Formula

a(n) = n^(d(n)/2) = n^(A000005(n)/2). Since a(n) = Product_(d|n) d = Product_(d|n) n/d, we have a(n)*a(n) = Product_(d|n) d*(n/d) = Product_(d|n) n = n^(tau(n)), whence a(n) = n^(tau(n)/2).

a(p^k) = p^A000217(k). - Enrique Pérez Herrero, Jul 22 2011

a(n) = A078599(n) * A178649(n). - Reinhard Zumkeller, Feb 06 2012

a(n) = A240694(n, A000005(n)). - Reinhard Zumkeller, Apr 10 2014

From Antti Karttunen, Mar 22 2017: (Start)

a(n) = A000196(n^A000005(n)). [From the original formula.]

A001222(a(n)) = A069264(n). [See Geoffrey Critzer's Feb 03 2015 comment in the latter sequence.]

A046523(a(n)) = A283995(n).

(End)

a(n) = Product_{k=1..n} gcd(n,k)^(1/phi(n/gcd(n,k))) = Product_{k=1..n} (n/gcd(n,k))^(1/phi(n/gcd(n,k))) where phi = A000010. - Richard L. Ollerton, Nov 07 2021

From Bernard Schott, Jan 11 2022: (Start)

a(n) = n^2 iff n is in A007422.

a(n) = n^3 iff n is in A162947.

a(n) = n^4 iff n is in A111398.

a(n) = n^5 iff n is in A030628.

a(n) = n^(3/2) iff n is in A280076. (End)

From Amiram Eldar, Oct 29 2022: (Start)

a(n) = n * A007956(n).

Sum_{k=1..n} 1/a(k) ~ log(log(n)) + c + O(1/log(n)), where c is a constant (Weiyi, 2004; Sandor and Crstici, 2004). (End)

a(n) = Product_{k=1..n} (n * (1 - ceiling(n/k - floor(n/k))))/k + ceiling(n/k - floor(n/k)). - Adriano Steffler, Feb 08 2024