A090158 Odd-indexed terms of the binomial transform equals 1 and the even-indexed terms of the second binomial transform equals 1.

1, 0, -3, 9, -15, 15, -63, 399, -255, -7425, -1023, 355839, -4095, -22360065, -16383, 1903790079, -65535, -209865211905, -262143, 29088885637119, -1048575, -4951498051026945, -4194303, 1015423886515240959, -16777215, -246921480190174429185

Offset: 0

Author

Paul D. Hanna, Nov 22 2003

Keywords

• sign

Comments

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

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

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

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

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

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

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

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

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

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

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

Crossrefs

Cf. A090145.

Programs

Mathematica

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 *)

Formula

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

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

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

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