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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.
1
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
Eric Weisstein's World of Mathematics, Pisano Number
Wikipedia, Pisano period
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
KEYWORD
nonn
AUTHOR
STATUS
approved