A292616 a(n) = 3*a(n-2) - a(n-4) for n > 3, with a(0)=4, a(1)=3, a(2)=a(3)=6, a sequence related to bisections of Fibonacci numbers.

4, 3, 6, 6, 14, 15, 36, 39, 94, 102, 246, 267, 644, 699, 1686, 1830, 4414, 4791, 11556, 12543, 30254, 32838, 79206, 85971, 207364, 225075, 542886, 589254, 1421294, 1542687, 3720996, 4038807, 9741694, 10573734, 25504086, 27682395, 66770564, 72473451, 174807606, 189737958

Offset

0,1

Comments

The "Pisano" periods, i.e., the periods of this sequence mod n, begin: 1, 6, 8, 6, 20, 24, 16, 12, 24, 60, 5, 24, ..., a sequence which is not in the OEIS. Period 5 for mod 11 seems to be the only odd period > 1 (checked up to 1000 terms).

Links

Table of n, a(n) for n=0..39.

Eric Weisstein's World of Mathematics, Pisano Number

Wikipedia, Pisano period

Index entries for linear recurrences with constant coefficients, signature (0, 3, 0, -1).

Formula

G.f.: (4 + 3*x - 6*x^2 - 3*x^3)/(1 - 3*x^2 + x^4).

a(n) = A291660(n) + A292521(n).

a(n) = (1/20)*((25 - 13*sqrt(5))*((1/2)*(-1 - sqrt(5)))^n + (5 - 7*sqrt(5)) *((1/2)*(1 - sqrt(5)))^n + ((1/2)*(1 + sqrt(5)))^n*(5 + 7*sqrt(5)) + ((1/2)*(-1 + sqrt(5)))^n*(25 + 13*sqrt(5))).

a(2n+1) = 3*A001519(n+1).

a(2n+1) = a(2n) + A001906(n-1).

Maple

A292616 := gfun:-rectoproc({a(n) = 3 * a(n-2) - a(n-4), a(0) = 4, a(1) = 3, a(2) = 6, a(3) = 6}, a(n), remember):  map(A292616, [$0..10^3]);  # Muniru A Asiru, Oct 16 2017

Mathematica

LinearRecurrence[{0, 3, 0, -1}, {4, 3, 6, 6}, 40]
(* Or, recomputing from Fibonacci numbers: *)
Join[{4, -1, 3}, Flatten[Table[{Fibonacci[2*n], Fibonacci[2*n+6]}, {n, 0, 18} ]]] // Accumulate

Prog

(GAP) a := [4, 3, 6, 6];; for n in [5..10^2] do a[n] := 3 * a[n-2] - a[n-4]; od; A292616 := a; #  Muniru A Asiru, Oct 31 2017

Crossrefs

Cf. A000045, A001519, A001906, A291660, A292521.

Sequence in context: A021233 A352922 A290612 * A071901 A266576 A103476

Adjacent sequences: A292613 A292614 A292615 * A292617 A292618 A292619

Keyword

nonn

Author

Jean-François Alcover and Paul Curtz, Sep 20 2017

Status

approved