A341414 a(n) = (Fibonacci(n)*Lucas(n)) mod 10.

0, 1, 3, 8, 1, 5, 4, 7, 7, 4, 5, 1, 8, 3, 1, 0, 9, 7, 2, 9, 5, 6, 3, 3, 6, 5, 9, 2, 7, 9, 0, 1, 3, 8, 1, 5, 4, 7, 7, 4, 5, 1, 8, 3, 1, 0, 9, 7, 2, 9, 5, 6, 3, 3, 6, 5, 9, 2, 7, 9, 0, 1, 3, 8, 1, 5, 4, 7, 7, 4, 5, 1, 8, 3, 1, 0, 9, 7, 2, 9, 5, 6, 3, 3, 6, 5, 9, 2, 7, 9

Offset: 0

Author

Jens Rasmussen, Feb 11 2021

Keywords

• easy
• nonn

Comments

Fibonacci starting with 0,1 and Lucas starting with 2,1.

Blocks of 30 numbers with 10 even and 20 uneven numbers.

Symmetric as a(7-i)=a(8+i) for i=1,2,...,6, and a(22-j)=a(23+j) for j=1..21.

Decimal expansion of 13801675776055042253380279/999000999000999000999000999. - Jianing Song, Apr 04 2021

Examples

			For n=5: a(5) = (Fibonacci(5)*Lucas(5)) mod 10 = (5*11) mod 10 = 55 mod 10 = 5.
		

Links

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

Crossrefs

Cf. A000032, A000045, A001906, A130893.

Bisection of A003893.

Programs

Mathematica

Table[Mod[Fibonacci@n*LucasL@n, 10], {n, 0, 100}] (* Giorgos Kalogeropoulos, Mar 31 2021 *)

PARI

a(n) = fibonacci(2*(n%30)) % 10 \\ Jianing Song, Apr 04 2021

Formula

a(n) = (Fibonacci(n)*Lucas(n)) mod 10 = Fibonacci(2*n) mod 10 using Binet's formula for Fibonacci and corresponding formula for Lucas.

a(n) = a(n-30).

a(n) = a(n-3) - a(n-6) + a(n-9) - a(n-12) + a(n-15) - a(n-18) + a(n-21) - a(n-24) + a(n-27).

a(n) = A003893(2*n).