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A007955
Product of divisors of n.
229
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
1,2
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
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)
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.
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.
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.
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
EXAMPLE
Divisors of 10 = [1, 2, 5, 10]. So, a(10) = 2*5*10 = 100. - Indranil Ghosh, Mar 22 2017
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 *)
PROG
(Magma) f := function(n); t1 := &*[d : d in Divisors(n) ]; return t1; end function;
(PARI) a(n)=if(issquare(n, &n), n^numdiv(n^2), n^(numdiv(n)/2)) \\ Charles R Greathouse IV, Feb 11 2011
(Haskell)
a007955 = product . a027750_row -- Reinhard Zumkeller, Feb 06 2012
(Sage) [prod(divisors(n)) for n in (1..100)] # Giuseppe Coppoletta, Dec 16 2014
(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
(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
(GAP) List(List([1..50], n->DivisorsInt(n)), Product); # Muniru A Asiru, Feb 17 2019
CROSSREFS
Cf. A000203 (sums of divisors).
Cf. A000010 (comments on product formulas).
Sequence in context: A140651 A190997 A184392 * A324502 A170826 A162537
KEYWORD
nonn,nice
AUTHOR
R. Muller
STATUS
approved