A022811 Number of terms in n-th derivative of a function composed with itself 3 times.
1, 1, 3, 6, 13, 23, 44, 74, 129, 210, 345, 542, 858, 1310, 2004, 2996, 4467, 6540, 9552, 13744, 19711, 27943, 39452, 55172, 76865, 106200, 146173, 199806, 272075, 368247, 496642, 666201, 890602, 1184957, 1571417, 2075058, 2731677, 3582119, 4683595, 6102256
Offset: 0
Keywords
Examples
From _Gus Wiseman_, Jul 19 2018: (Start) Using the chain rule, we compute the second derivative of f(f(f(x))) to be the following sum of a(2) = 3 terms. d^2/dx^2 f(f(f(x))) = f'(f(x)) f'(f(f(x))) f''(x) + f'(x)^2 f'(f(f(x))) f''(f(x)) + f'(x)^2 f'(f(x))^2 f''(f(f(x))). (End)
References
- W. C. Yang, Derivatives of self-compositions of functions, preprint, 1997.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..3000 (terms n = 501..959 from Vaclav Kotesovec)
- W. C. Yang, Derivatives are essentially integer partitions, Discrete Mathematics, 222(1-3), July 2000, 235-245.
Crossrefs
Programs
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Maple
A022811 := proc(n) local a,P,p,lp ; a := 0 ; P := combinat[partition](n) ; for p in P do lp := nops(p) ; a := a+combinat[numbpart](lp) ; od: RETURN(a) ; end: for n from 1 do print(n,A022811(n)) ; od: # R. J. Mathar, Aug 12 2008
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Mathematica
a[n_] := Total[PartitionsP[Length[#]]& /@ IntegerPartitions[n]]; Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, 80}] (* Jean-François Alcover, Apr 28 2017 *) Table[Length[1+D[f[f[f[x]]],{x,n}]]-1,{n,10}] (* Gus Wiseman, Jul 19 2018 *)
Formula
If a(n,m) = number of terms in m-derivative of a function composed with itself n times, p(n,k) = number of partitions of n into k parts, then a(n,m) = Sum_{i=0..m} p(m,i)*a(n-1,i).
G.f.: Sum_{k>=0} p(k) * x^k / Product_{j=1..k} (1 - x^j), where p(k) = number of partitions of k. - Ilya Gutkovskiy, Jan 28 2020
Extensions
Typo corrected by Neven Juric, Mar 25 2013
Comments