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

Author

Jean-François Alcover and Paul Curtz, Sep 20 2017

Keywords

• nonn

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

• Eric Weisstein's World of Mathematics, Pisano Number
• Wikipedia, Pisano period
• Index entries for linear recurrences with constant coefficients, signature (0, 3, 0, -1).

Crossrefs

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

Programs

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

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

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).