A112031 Numerator of 3/4 + 1/4 - 3/8 - 1/8 + 3/16 + 1/16 - 3/32 - 1/32 + 3/64 + ....

3, 1, 5, 1, 11, 3, 21, 5, 43, 11, 85, 21, 171, 43, 341, 85, 683, 171, 1365, 341, 2731, 683, 5461, 1365, 10923, 2731, 21845, 5461, 43691, 10923, 87381, 21845, 174763, 43691, 349525, 87381, 699051, 174763, 1398101, 349525, 2796203, 699051, 5592405

Offset

0,1

Comments

Numerator of partial sums of A112030(n)/A016116(n+4), denominators = A112032;

a(n)/A112032(n) - 2/3 = (-1)^floor(n/2) / A112033(n);

lim_{n->infinity} a(n)/A112032(n) = 2/3.

References

G. Pólya and G. Szegő, Problems and Theorems in Analysis I (Springer 1924, reprinted 1972), Part One, Chap. 4, Sect. 1, Problem 148.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..3000

Index entries for linear recurrences with constant coefficients, signature (0,1,0,2).

Formula

a(n) = (2^(floor(n/2) + 2 + (-1)^n) + (-1)^floor(n/2)) / 3.

From Colin Barker, Apr 05 2013: (Start)

a(n) = a(n-2) + 2*a(n-4);

g.f.: (2*x^2+x+3) / ((1+x^2)*(1-2*x^2)). (End)

Mathematica

LinearRecurrence[{0, 1, 0, 2}, {3, 1, 5, 1}, 50] (* Harvey P. Dale, Dec 31 2017 *)

Prog

(Magma) [(2^(Floor(n/2) + 2 + (-1)^n) + (-1)^Floor(n/2)) / 3: n in [0..50]]; // Vincenzo Librandi, Aug 17 2011
(PARI) m=50; v=concat([3, 1, 5, 1], vector(m-4)); for(n=5, m, v[n]=v[n-2] +2*v[n-4]); v \\ G. C. Greubel, Nov 08 2018

Crossrefs

Cf. A016116, A112030, A112032, A112033, A001045 (bisections).

Sequence in context: A360756 A133730 A330773 * A146285 A146059 A046531

Adjacent sequences: A112028 A112029 A112030 * A112032 A112033 A112034

Keyword

nonn,frac

Author

Reinhard Zumkeller, Aug 27 2005

Extensions

a(22) corrected by Vincenzo Librandi, Aug 17 2011

Status

approved