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A001028
E.g.f. satisfies A'(x) = 1 + A(A(x)), A(0)=0.
10
1, 1, 2, 7, 37, 269, 2535, 29738, 421790, 7076459, 138061343, 3089950076, 78454715107, 2238947459974, 71253947372202, 2511742808382105, 97495087989736907, 4145502184671892500, 192200099033324115855, 9676409879981926733908, 527029533717566423156698
OFFSET
1,3
COMMENTS
The e.g.f. is diverging (see the Math Overflow link). - Pietro Majer, Jan 29 2017
REFERENCES
This functional equation (for f(x)=1+A(x-1)) was the subject of problem B5 of the 2010 Putnam exam.
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 1..320 (first 100 terms from Alois P. Heinz)
P. J. Cameron, Sequence operators from groups, Linear Alg. Applic., 226-228 (1995), 109-113.
Math Overflow, f' = exp(f^(-1)), again, January 2017.
FORMULA
E.g.f. satisfies: A(x) = Series_Reversion( Integral 1/(1 + A(x)) dx ). - Paul D. Hanna, Jun 27 2015
MAPLE
A:= proc(n) option remember; local T; if n=0 then 0 else T:= A(n-1); unapply(convert(series(Int(1+T(T(x)), x), x, n+1), polynom), x) fi end: a:= n-> coeff(A(n)(x), x, n)*n!: seq(a(n), n=1..22); # Alois P. Heinz, Aug 23 2008
MATHEMATICA
terms = 21; A[_] = 0; Do[A[x_] = x + Integrate[A[A[x]], x] + O[x]^(n+1) // Normal, {n, terms}];
Rest[CoefficientList[A[x], x]]*Range[terms]! (* Jean-François Alcover, Dec 07 2011, updated Jan 10 2018 *)
PROG
(Maxima) Co(n, k, a):= if k=1 then a(n) else sum(a(i+1)*Co(n-i-1, k-1, a), i, 0, n-k); a(n):= if n=1 then 1 else (1/n)*sum(Co(n-1, k, a)*a(k), k, 1, n-1); makelist(n!*a(n), n, 1, 7); /* Vladimir Kruchinin, Jun 30 2011 */
(PARI) {a(n) = my(A=x); for(i=1, n, A = serreverse(intformal(1/(1+A) +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=1, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Jun 27 2015
CROSSREFS
Sequence in context: A036247 A083659 A107877 * A116481 A367494 A102743
KEYWORD
nonn,eigen,nice
EXTENSIONS
More terms from Christian G. Bower, Oct 15 1998
Corrected by Alois P. Heinz, Aug 23 2008
Two more terms from Sean A. Irvine, Feb 22 2012
STATUS
approved