OFFSET
1,6
COMMENTS
Let d o f(x) be an operator of multiplication by f(x) followed by differentiation. (d o f)^m = Sum a([k0,k1,...])*((d^0 f)^k0*(d^1 f)^k1*...)*d^(m-k1-2*k2-...) where the sum is taken over all nonnegative integer vectors [k0,k1,...] such that k0+k1+...=m and k1+2*k2+...<=m.
For all k >= 0 it holds that a(2^k) = a(3^k) = 1 and also a(p) = 0 for all primes p > 3. - Alexander Adamchuk, Dec 03 2006 and Antti Karttunen, Feb 28 2023
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
FORMULA
For n=p0^k0*p1^k1*... where 2=p0<p1<... are the sequence of all primes, a(n) = a([k0,k1,...]) satisfy the recurrence a([k0,k1,...]) = a([k0-1,k1,...]) + (k0+1)*a([k0,k1-1,...]) + Sum_{i=2..oo} (k(i-1)+1)*a([k0-1,k1,...,k(i-2),k(i-1)+1,ki-1,k(i+1),...]) with a([0,0,...])=1 and a([k0,k1,...])=0 as soon as some ki<0.
a([k0,k1,0,0,...]) = S(k0+k1+1,k0+1), Stirling number of the 2nd kind, see A008277.
EXAMPLE
From Antti Karttunen, Feb 28 2023: (Start)
For n=6, a(6) = a(2^1 * 3^1) = a([1,1,0,0,0,...]) = a([0,1,0,0,...]) + (1+1)*a([1,0,0,0,...]) + 0 = a(3) + 2*a(1) = 3.
For n=10, a(10) = a(2^1 * 5^1) = a([1,0,1,0,0,0...]) = a([0,0,1,0,0,0,...]) + 2*0 + 1*a([0,1,0,0,0,...]) = a(5) + 0 + 1*a(3) = 1.
For n=20, a(20) = a(2^2 * 5^1) = a([2,0,1,0,0,0...]) = a([1,0,1,0,0,0,...]) + 3*0 + 1*a([1,1,0,0,0,...]) = a(10) + 0 + 1*a(6) = 1+3 = 4.
(End)
PROG
(PARI) A124796(n) = if(1==n, 1, my(u=primepi(vecmax(factor(n)[, 1]))); if(n%3, 0, ((1+valuation(n, 2)) * A124796(n/3))) + if(n%2, 0, (A124796(n/2) + sum(i=3, u, if(n%prime(i), 0, (valuation(n, prime(i-1))+1)*A124796((n/2)*prime(i-1)/prime(i))))))); \\ Antti Karttunen, Feb 28 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
Max Alekseyev, Nov 29 2006
STATUS
approved