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A157315
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G.f.: A(x) = sin( Sum_{n>=0} 2^((2n+1)^2) * C(2n,n)/4^n * x^(2n+1)/(2n+1) ); alternating zeros omitted.
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2
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OFFSET
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1,1
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COMMENTS
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Compare g.f. to the expansion of the inverse sine of x:
arcsin(x) = Sum_{n>=0} C(2n,n)/4^n * x^(2n+1)/(2n+1).
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LINKS
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EXAMPLE
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G.f.: A(x) = 2*x + 84*x^3 + 2516412*x^5 + 25131689308776*x^7 + ...
The inverse sine of A(x) begins:
arcsin(A(x)) = 2*x + 2^9*(2/4)*x^3/3 + 2^25*(6/4^2)*x^5/5 + 2^49*(20/4^3)*x^7/7 + 2^81*(70/4^4)*x^9/9 + ...
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MAPLE
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m := 30;
S := series( sin(add(2^(4*j^2+2*j+1)*binomial(2*j, j)*x^(2*j+1)/(2*j+1), j = 0..m+2)), x, m+1);
seq(coeff(S, x, 2*j+1), j = 0..m/2); # G. C. Greubel, Mar 16 2021
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MATHEMATICA
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With[{m = 30}, CoefficientList[Series[Sin[Sum[2^(4*n^2+2*n+1)*((n+1)/(2*n+1)) *CatalanNumber[n]*x^(2*n+1), {n, 0, m+2}]], {x, 0, m}], x]][[2 ;; ;; 2 ]] (* G. C. Greubel, Mar 16 2021 *)
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PROG
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(PARI) {a(n)=polcoeff(sin(sum(m=0, n\2, 2^((2*m+1)^2)*binomial(2*m, m)/4^m*x^(2*m+1)/(2*m+1))+x*O(x^n)), n)}
(Magma)
m:=30;
R<x>:=PowerSeriesRing(Rationals(), m);
b:=Coefficients(R!( Sin( (&+[2^(4*j^2+2*j+1)*Binomial(2*j, j)*x^(2*j+1)/(2*j+1): j in [0..m+2]]) ) ));
[b[2*n-1]: n in [1..Floor((m-2)/2)]]; // G. C. Greubel, Mar 16 2021
(Sage)
m=30
P.<x> = PowerSeriesRing(QQ, prec)
return P( sin( sum(2^(4*j^2+2*j+1)*binomial(2*j, j)*x^(2*j+1)/(2*j+1) for j in [0..m+2])) ).list()
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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