login
A056191
Characteristic cube divisor of n: cube of g = gcd(K,F), where K is the largest square root divisor of n (A000188) and F = n/(K*K) = A007913(n) is its squarefree part; g^2 divides K^2 = A008833(n) = g^2*L^2 and g divides F = gf.
6
1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 27, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 27, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1
OFFSET
1,8
COMMENTS
This is not the largest cube which divides n. It is canonical, since the decomposition n = KKgggf is unique (factors are defined above and dependent on n).
LINKS
FORMULA
a(n) = A055229(n)^3 = g^3 = ggg; n = (KK)*(ggg)*f = K^2*g^3*f = KK*a(n)^3*f.
Multiplicative with a(p^e)=1 for even e, a(p)=1, a(p^e)=p^3 for odd e > 1. - Vladeta Jovovic, May 01 2002
EXAMPLE
If n=24, largest square divisor is 4, squarefree part is 6, g=2, a(24)=8; n=81, largest square divisor is 81, both F and g is 1, a(81)=1.
MATHEMATICA
a[n_]:=With[{sf=Times@@Power@@@({#[[1]], Mod[#[[2]], 2]}&/@FactorInteger[n])}, GCD[sf, n/sf]]; Table[a[n]^3, {n, 1, 100}] (* Vincenzo Librandi, Oct 08 2017 *)
PROG
(Scheme, with memoization-macro definec, after Vladeta Jovovic's multiplicative formula)
(definec (A056191 n) (cond ((= 1 n) n) ((or (= 1 (A067029 n)) (even? (A067029 n))) (A056191 (A028234 n))) (else (* (A000578 (A020639 n)) (A056191 (A028234 n)))))) ;; Antti Karttunen, Oct 07 2017
(PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] == 1 || !(f[i, 2]%2), 1, f[i, 1]^3)); } \\ Amiram Eldar, Sep 05 2023
CROSSREFS
KEYWORD
nonn,mult
AUTHOR
Labos Elemer, Aug 02 2000
STATUS
approved