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A099925 a(n) = Lucas(n) + (-1)^n. 3
3, 0, 4, 3, 8, 10, 19, 28, 48, 75, 124, 198, 323, 520, 844, 1363, 2208, 3570, 5779, 9348, 15128, 24475, 39604, 64078, 103683, 167760, 271444, 439203, 710648, 1149850, 1860499, 3010348, 4870848, 7881195, 12752044, 20633238, 33385283, 54018520, 87403804 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

Let phi = 1/2*(1 + sqrt(5)) denote the golden ratio and put c = sum {n = 1..inf} 1/2^floor(n*(phi + 2)). The bicimal expansion of the constant c begins 0.001000100100010001001.... The binary digits are the generalized Fibonacci word A221150.

The sequence 2^a(n) for n >= 1 gives the partial quotients, apart from the first, in the simple continued fraction expansion of the constant 1/2*c = 0.06692 72114 83804 90296 ... = 1/(14 + 1/(2^0 + 1/(2^4 + 1/(2^3 + 1/(2^8 + 1/(2^10 + 1/(2^19 + ...))))))). Cf. A008346. - Peter Bala, Nov 06 2013

LINKS

Colin Barker, Table of n, a(n) for n = 0..1000

YĆ¼ksel Soykan, Generalized Pell-Padovan Numbers, Asian Journal of Advanced Research and Reports (2020) Vol. 11, No. 2, 8-28, Article No. 57839.

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

FORMULA

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

a(0) = 3, a(1) = 0, a(2) = 4 and a(n) = 2*a(n-2) + a(n-3) for n >= 3. - Peter Bala, Nov 06 2013

a(n) = A068397(n) - 1 for n>2.

a(n) = ((-1)^n+(1/2*(1-sqrt(5)))^n+(1/2*(1+sqrt(5)))^n). - Colin Barker, Jun 03 2016

MATHEMATICA

CoefficientList[Series[(3 - 2 x^2)/((1 + x) (1 - x - x^2)), {x, 0, 38}], x] (* Michael De Vlieger, Sep 16 2020 *)

PROG

(PARI) Vec((3-2*x^2)/((1+x)*(1-x-x^2)) + O(x^40)) \\ Colin Barker, Jun 03 2016

CROSSREFS

Cf. A000032, A008346, A098600, A068397.

Sequence in context: A344905 A181839 A100939 * A351002 A201563 A016644

Adjacent sequences:  A099922 A099923 A099924 * A099926 A099927 A099928

KEYWORD

nonn,easy

AUTHOR

Ralf Stephan, Nov 02 2004

STATUS

approved

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Last modified September 29 18:39 EDT 2022. Contains 357090 sequences. (Running on oeis4.)