|
|
A329382
|
|
Product of exponents of prime factors of A108951(n), where A108951 is fully multiplicative with a(prime(i)) = prime(i)# = Product_{i=1..i} A000040(i).
|
|
15
|
|
|
1, 1, 1, 2, 1, 2, 1, 3, 4, 2, 1, 3, 1, 2, 4, 4, 1, 6, 1, 3, 4, 2, 1, 4, 8, 2, 9, 3, 1, 6, 1, 5, 4, 2, 8, 8, 1, 2, 4, 4, 1, 6, 1, 3, 9, 2, 1, 5, 16, 12, 4, 3, 1, 12, 8, 4, 4, 2, 1, 8, 1, 2, 9, 6, 8, 6, 1, 3, 4, 12, 1, 10, 1, 2, 18, 3, 16, 6, 1, 5, 16, 2, 1, 8, 8, 2, 4, 4, 1, 12, 16, 3, 4, 2, 8, 6, 1, 24, 9, 16, 1, 6, 1, 4, 18
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
COMMENTS
|
Also the product of parts of the conjugate of the integer partition with Heinz number n, where the Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). For example, the partition (3,2) with Heinz number 15 has conjugate (2,2,1) with product a(15) = 4. - Gus Wiseman, Mar 27 2022
|
|
LINKS
|
|
|
FORMULA
|
If n = p(k1)^e(k1) * p(k2)^e(k2) * p(k3)^e(k3) * ... * p(kx)^e(kx), with p(n) = A000040(n) and k1 > k2 > k3 > ... > kx, then a(n) = e(k1)^(k1-k2) * (e(k1)+e(k2))^(k2-k3) * (e(k1)+e(k2)+e(k3))^(k3-k4) * ... * (e(k1)+e(k2)+...+e(kx))^kx.
(End)
|
|
MATHEMATICA
|
Table[Times @@ FactorInteger[Times @@ Map[#1^#2 & @@ # &, FactorInteger[n] /. {p_, e_} /; e > 0 :> {Times @@ Prime@ Range@ PrimePi@ p, e}]][[All, -1]], {n, 105}] (* Michael De Vlieger, Jan 21 2020 *)
|
|
PROG
|
(PARI)
A034386(n) = prod(i=1, primepi(n), prime(i));
(PARI) A329382(n) = if(1==n, 1, my(f=factor(n), e=0, m=1); forstep(i=#f~, 1, -1, e += f[i, 2]; m *= e^(primepi(f[i, 1])-if(1==i, 0, primepi(f[i-1, 1])))); (m)); \\ Antti Karttunen, Jan 14 2020
|
|
CROSSREFS
|
This is the conjugate version of A003963 (product of prime indices).
The Heinz number of the conjugate partition is given by A122111.
Cf. A000701, A000720, A001221, A046682, A175508, A290822, A303975, A316524, A324850, A352486-A352491, A353570.
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|