|
Search: id:A124796
|
|
|
| A124796 |
|
Coefficients in expansion of powers of the operator "multiplication by f(x) followed by differentiation", in the prime factorization order. |
|
+0
1
|
|
| 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
(list; graph; listen)
|
|
|
OFFSET
|
1,6
|
|
|
COMMENT
|
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.
It appears that a(2^k) = 1 and a(3^k) = 1 and that a(p) = 0 for prime p>3. - Alexander Adamchuk (alex(AT)kolmogorov.com), Dec 03 2006
|
|
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([k0-1,k1,...]) + 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.
|
|
CROSSREFS
|
Sequence in context: A119624 A119612 A101949 this_sequence A065714 A110700 A051908
Adjacent sequences: A124793 A124794 A124795 this_sequence A124797 A124798 A124799
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Max Alekseyev (maxale(AT)gmail.com), Nov 29 2006
|
|
|
Search completed in 0.002 seconds
|
