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Odd-indexed terms of the binomial transform equals 1 and the even-indexed terms of the second binomial transform equals 1.
3

%I #9 Feb 13 2016 09:47:28

%S 1,0,-3,9,-15,15,-63,399,-255,-7425,-1023,355839,-4095,-22360065,

%T -16383,1903790079,-65535,-209865211905,-262143,29088885637119,

%U -1048575,-4951498051026945,-4194303,1015423886515240959,-16777215,-246921480190174429185

%N Odd-indexed terms of the binomial transform equals 1 and the even-indexed terms of the second binomial transform equals 1.

%C Compare the first and 2nd binomial transforms of this sequence:

%C first binomial={1,1,-2,1,4,1,-62,1,1384,1,-50522,1,2702764,..};

%C 2nd binomial={1,2,1,-1,1,17,1,-271,1,7937,1,-353791,..};

%C to that of the first and 2nd binomial transforms of A090145:

%C first binomial of A090145={1,0,1,-3,1,15,1,-273,1,7935,1,..};

%C 2nd binomial of A090145={1,1,2,1,-4,1,62,1,-1384,1,50522,..}.

%C Comparison reveals this e.g.f. relation of the two sequences:

%C e.g.f.: exp(x)*G090158(x) + exp(2x)*G090145(x) = 2 + 2*sinh(x);

%C e.g.f.: exp(2*x)*G090158(x) - exp(x)*G090145(x) = 2*sinh(x);

%C thus G090158(x) = 2*(1+sinh(x) + exp(x)*sinh(x))/(exp(x)*(1+exp(2*x)))

%C G090145(x) = 2*((1+sinh(x))*exp(x) - sinh(x))/(exp(x)*(1+exp(2*x))).

%F E.g.f.: 2*(1 + sinh(x) + exp(x)*sinh(x)) / (exp(x)*(1 + exp(2*x))).

%F a(2n) = 1 - 2^(2n);

%F 1 = sum_{k=0..2n-1} C(2n-1, k)*a(k);

%F 1 = sum_{k=0..2n} 2^(2n-k)*C(2n, k)*a(k).

%t With[{nn=30},CoefficientList[Series[2 (1+Sinh[x]+Exp[x]Sinh[x])/ (Exp[x] (1+ Exp[2x])),{x,0,nn}],x] Range[0,nn]!] (* _Harvey P. Dale_, Feb 13 2016 *)

%Y Cf. A090145.

%K sign

%O 0,3

%A _Paul D. Hanna_, Nov 22 2003