

A126865


a(n) = gcd(Product_{pn} (p+1)^b(p,n), Product_{pn} (p1)^b(p,n)), where the products are over the distinct primes, p, that divide n and p^b(p,n) is the highest power of p dividing n.


2



1, 1, 2, 1, 2, 2, 2, 1, 4, 2, 2, 2, 2, 6, 8, 1, 2, 4, 2, 2, 4, 2, 2, 2, 4, 6, 8, 6, 2, 8, 2, 1, 4, 2, 24, 4, 2, 6, 8, 2, 2, 12, 2, 2, 16, 2, 2, 2, 4, 4, 8, 6, 2, 8, 8, 6, 4, 2, 2, 8, 2, 6, 8, 1, 12, 4, 2, 2, 4, 24, 2, 4, 2, 6, 16, 18, 12, 24, 2, 2, 16, 2, 2, 12, 4, 6, 8, 2, 2, 16, 8, 2, 4, 2, 24, 2, 2, 12, 8
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OFFSET

1,3


COMMENTS

First occurrence of k or 0 if not possible (including all odd primes k): 2, 1, 0, 9, 0, 14, 0, 15, 0, 0, 0, 42, 0, 0, 0, 45, 0, 76, 0, 589, 0, 0, 0, 35, 0, 0, 0, 4381, 0, 0, ..., .  Robert G. Wilson v, Sep 08 2007


LINKS

Antti Karttunen, Table of n, a(n) for n = 1..12441
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000


FORMULA

From Antti Karttunen, Dec 17 2018: (Start)
a(n) = gcd(A003958(n), A003959(n)).
a(A007947(n)) = A066086(n).
(End)


EXAMPLE

400 = 2^4 * 5^2. So a(400) = gcd((2+1)^4 * (5+1)^2, (21)^4 * (51)^2) = gcd(2916, 16) = 4.


MATHEMATICA

f[n_] := Block[{fi = FactorInteger@n}, GCD[Times @@ ((First /@ fi  1)^Last /@ fi), Times @@ ((First /@ fi + 1)^Last /@ fi)]]; Array[f, 99]  Robert G. Wilson v, Sep 08 2007


PROG

(PARI)
A003958(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1]); factorback(f); };
A003959(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1]++); factorback(f); };
A126865(n) = gcd(A003958(n), A003959(n)); \\ Antti Karttunen, Dec 17 2018


CROSSREFS

Cf. A003958, A003959, A066086, A322362.
Sequence in context: A167865 A218654 A054571 * A104640 A193335 A016727
Adjacent sequences: A126862 A126863 A126864 * A126866 A126867 A126868


KEYWORD

nonn


AUTHOR

Leroy Quet, Mar 15 2007


EXTENSIONS

More terms from Robert G. Wilson v, Sep 08 2007


STATUS

approved



