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A080924 Jacobsthal gap sequence. 3
0, 1, 3, 1, 15, 1, 63, 1, 255, 1, 1023, 1, 4095, 1, 16383, 1, 65535, 1, 262143, 1, 1048575, 1, 4194303, 1, 16777215, 1, 67108863, 1, 268435455, 1, 1073741823, 1, 4294967295, 1, 17179869183, 1, 68719476735, 1, 274877906943, 1, 1099511627775, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Inverse binomial transform of A080925

From Peter Bala, Dec 26 2012: (Start)

Let F(x) = product {n >= 0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number F(1/4) = 0.79761 68651 30459 16010 ... = 1/(1 + 1/(3 + 1/(1 + 1/(15 + 1/(1 + 1/(63 + 1/(1 + 1/(255 + ...)))))))). See A111317. (End)

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..300

FORMULA

a(2n) = 3*A001045(2n) = 3*A002450(n) = 4^n-1, a(2n+1)=1.

a(n) = (2^n-2*(-1)^n+(-2)^n)/2.

G.f.: x*(1+4*x)/((1+x)*(1+2*x)*(1-2*x)).

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

MATHEMATICA

CoefficientList[Series[x (1 + 4 x) / ((1 + x) (1 + 2 x) (1 - 2 x)), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 05 2013 *)

CROSSREFS

Cf. A001045, A002450, A080926, A080927, A111317.

Sequence in context: A214073 A141459 A176727 * A232179 A128042 A108083

Adjacent sequences:  A080921 A080922 A080923 * A080925 A080926 A080927

KEYWORD

nonn,easy

AUTHOR

Paul Barry, Feb 26 2003

STATUS

approved

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Last modified September 2 10:05 EDT 2014. Contains 246350 sequences.