A124796 Coefficients in expansion of powers of the operator "multiplication by f(x) followed by differentiation", in the prime factorization order.

1, 1, 1, 1, 0, 3, 0, 1, 1, 1, 0, 6, 0, 0, 0, 1, 0, 7, 0, 4, 0, 0, 0, 10, 0, 0, 1, 1, 0, 4, 0, 1, 0, 0, 0, 25, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 15, 0, 0, 0, 0, 0, 15, 0, 5, 0, 0, 0, 30, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 65, 0, 0, 0, 0, 0, 0, 0, 20, 1, 0, 0, 7, 0, 0, 0, 1, 0, 11, 0, 0, 0, 0, 0, 21, 0, 0, 0, 4, 0

Offset: 1

Author

Max Alekseyev, Nov 29 2006

Keywords

• nonn

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

Examples

			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)
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384
• Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Crossrefs

Cf. A139605, A145271.

Programs

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

Formula

For n=p0^k0*p1^k1*... where 2=p0

a([k0,k1,0,0,...]) = S(k0+k1+1,k0+1), Stirling number of the 2nd kind, see A008277.

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