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A124796
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Coefficients in expansion of powers of the operator "multiplication by f(x) followed by differentiation", in the prime factorization order.
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0
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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; refs; listen; history; internal format)
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OFFSET
| 1,6
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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.
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
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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.
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CROSSREFS
| Sequence in context: A119624 A119612 A101949 * A065714 A110700 A181875
Adjacent sequences: A124793 A124794 A124795 * A124797 A124798 A124799
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KEYWORD
| nonn
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AUTHOR
| Max Alekseyev (maxale(AT)gmail.com), Nov 29 2006
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