A056623 If n=LLgggf (see A056192) and a(n) = LL, then its complementary divisor n/LL = gggf and gcd(L^2, n/LL) = 1.

1, 1, 1, 4, 1, 1, 1, 1, 9, 1, 1, 4, 1, 1, 1, 16, 1, 9, 1, 4, 1, 1, 1, 1, 25, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 36, 1, 1, 1, 1, 1, 1, 1, 4, 9, 1, 1, 16, 49, 25, 1, 4, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 9, 64, 1, 1, 1, 4, 1, 1, 1, 9, 1, 1, 25, 4, 1, 1, 1, 16, 81, 1, 1, 4, 1, 1, 1, 1, 1, 9, 1, 4, 1, 1, 1, 4, 1, 49, 9

Offset: 1

Author

Labos Elemer, Aug 08 2000

Keywords

• nonn
• mult
• easy

Comments

The part of the name "Largest unitary square divisor of n" was removed because it is correct only for numbers whose odd exponents in their prime factorization are all smaller than 5. For the correct largest unitary square divisor of n see A350388. - Amiram Eldar, Jul 26 2024

Examples

			a(200) = A008833(200)/A055229(200)^2 = 100/2^2 = 25.
a(250) = A008833(250)/A055229(250)^2 = 25/5^2 = 1.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384

Crossrefs

Cf. A008833, A055229, A000188, A046951, A034444, A056192, A056622, A350388.

Programs

Mathematica

f[p_, 1] := 1; f[p_, e_] := If[EvenQ[e], p^e, p^(e-3)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 11 2020 *)

PARI

A055229(n) = { my(c=core(n)); gcd(c, n/c); }; \\ Charles R Greathouse IV, Nov 20 2012
A008833(n) = n/core(n) \\ Michael B. Porter, Oct 17 2009
A056623(n) = (A008833(n)/(A055229(n)^2)); \\ Antti Karttunen, Nov 19 2017

PARI

a(n) = {my(f = factor(n), p, e); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; if(e == 1, 1, if(e%2, p^(e-3), p^e)));} \\ Amiram Eldar, Dec 25 2023

Scheme

(definec (A056623 n) (if (= 1 n) n (let ((e (A067029 n)) (rest (A056623 (A028234 n)))) (cond ((even? e) (* (A028233 n) rest)) ((= 1 e) rest) (else (* (expt (A020639 n) (- e 3)) rest)))))) ;; After Jovovic's multiplicative formula, using memoization-macro definec - Antti Karttunen, Nov 19 2017

Formula

a(n) = A008833(n)/A055229(n)^2 = K^2/g^2, which coincides with the largest square divisor iff the g-factor is 1.

Multiplicative with a(p^e)=p^e for even e, a(p)=1, a(p^e)=p^(e-3) for odd e > 1. - Vladeta Jovovic, Apr 30 2002

From Amiram Eldar, Dec 25 2023 (Start)

Dirichlet g.f.: zeta(2*s-2) * Product_{p prime} (1 + 1/p^s - 1/p^(3*s-2) + 1/p^(3*s)).

Sum_{k=1..n} a(k) ~ c * n^(3/2) / 3, where c = Product_{p prime} (1 + 1/p^(3/2) - 1/p^(5/2) + 1/p^(9/2)) = 1.81133051934001073532... . (End)

a(n) = A056622(n)^2. - Amiram Eldar, Jul 26 2024

Extensions

Name edited by Amiram Eldar, Jul 26 2024