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A201627 E.g.f. satisfies: A(x) = 1/(1 - sin(x*A(x))).

%I A201627 #11 Jul 07 2012 19:08:04
%S A201627 1,1,4,29,312,4481,80768,1754549,44647040,1303097665,42923116032,
%T A201627 1575332861101,63754405679104,2820829737123841,135469202252333056,
%U A201627 7018336152909163205,390175030207597805568,23169468447962190613121,1463683656780476860989440,98016257612539018485477821
%N A201627 E.g.f. satisfies: A(x) = 1/(1 - sin(x*A(x))).
%C A201627 Coefficients in the expansion of 1/(1-sin(x)) yield the Euler numbers (A000111).
%F A201627 E.g.f. A(x) satisfies: A( x*(1 - sin(x)) ) = 1/(1 - sin(x)).
%F A201627 E.g.f.: (1/x)*Series_Reversion( x*(1 - sin(x)) ).
%F A201627 a(n) = [x^n] 1/(1 - sin(x))^(n+1) / (n+1).
%F A201627 a(n) = A214222(n+1)/(n+1).
%e A201627 E.g.f.: A(x) = 1 + x + 4*x^2/2! + 29*x^3/3! + 312*x^4/4! + 4481*x^5/5! +...
%e A201627 The coefficients in initial powers of G(x) = 1/(1 - sin(x)) begin:
%e A201627 G^1: [(1), 1, 2, 5, 16, 61, 272, 1385, 7936, ..., A000111(n+1), ...];
%e A201627 G^2: [1,(2), 6, 22, 96, 482, 2736, 17302, ...];
%e A201627 G^3: [1, 3,(12), 57, 312, 1923, 13152, 98697, ...];
%e A201627 G^4: [1, 4, 20,(116), 760, 5524, 44000, 380516, ...];
%e A201627 G^5: [1, 5, 30, 205,(1560), 13025, 118080, 1153105, ...];
%e A201627 G^6: [1, 6, 42, 330, 2856,(26886), 272832, 2963850, ...];
%e A201627 G^7: [1, 7, 56, 497, 4816, 50407, (565376), 6754097, ...];
%e A201627 G^8: [1, 8, 72, 712, 7632, 87848, 1078272,(14036392), ...]; ...
%e A201627 where coefficients in parenthesis form initial terms of this sequence:
%e A201627 [1/1, 2/2, 12/3, 116/4, 1560/5, 26886/6, 565376/7, 14036392/8, ...].
%o A201627 (PARI) {a(n)=n!*polcoeff(1/x*serreverse(x*(1-sin(x+x^2*O(x^n)))),n)}
%o A201627 (PARI) {a(n)=n!*polcoeff(1/(1-sin(x+x*O(x^n)))^(n+1)/(n+1), n)}
%Y A201627 Cf. A214222, A201594, A000111.
%K A201627 nonn
%O A201627 0,3
%A A201627 _Paul D. Hanna_, Dec 03 2011