Jens Rasmussen - cp's OEIS frontend

User: Jens Rasmussen

Jens Rasmussen has authored 2 sequences.

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

Jens Rasmussen, Feb 11 2021

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

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

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).

Cf. A000032, A000045, A001906, A130893.

Bisection of A003893.

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

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

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).

A341616 Table read by ascending antidiagonals: T(n,j) = Fibonacci(n)*Lucas(n+j), product of the n-th term in the Fibonacci sequence (with F(1)=1 and F(2)=1) and the (n+j)-th term in the Lucas sequence (with L(1)=1 and L(2)=3 and j=0,1,2,...).

Original entry on oeis.org

1, 3, 3, 8, 4, 4, 21, 14, 7, 7, 55, 33, 22, 11, 11, 144, 90, 54, 36, 18, 18, 377, 232, 145, 87, 58, 29, 29, 987, 611, 376, 235, 141, 94, 47, 47, 2584, 1596, 988, 608, 380, 228, 152, 76, 76, 6765, 4182, 2583, 1599, 984, 615, 369, 246, 123, 123

Offset: 1

Jens Rasmussen, Feb 16 2021

j is the offset when combining terms from the two initial sequences.

			T(4,3) = Fibonacci(4)*Lucas(4+3) = 3*29 = 87.
Square array showing T(n,j) begins:
      j=0 j=1 j=2 j=3 j=4  ..
  n=1   1   3   4   7  11  ..
  n=2   3   4   7  11  18  ..
  n=3   8  14  22  36  58  ..
  n=4  21  33  54  87 141  ..
  ...  ..  ..  ..  ..  ..  ..

For j=0 the resulting sequence is used as input in A341414.

Cf. A000045, A000032, A001622, A094214, A104457.

T(n,j) = fibonacci(2*n+j) - (-1)^n*fibonacci(j);
matrix(7,7,n,k, T(n,k-1)) \\ Michel Marcus, Mar 02 2021

For phi=(1+sqrt(5))/2 and tau=(1-sqrt(5))/2:
T(n,j) = Fibonacci(n)*Lucas(n+j).
T(n,j) = (phi^n - tau^n)*(phi^(n+j) + tau^(n+j))/sqrt(5).
T(n,j) = Fibonacci(2n+j) - (-1)^n*Fibonacci(j).
Lim_{n, j -> oo} T(n+1,j)/T(n,j) = phi^2 (A104457).
Lim_{n, j -> oo} T(n,j+1)/T(n,j) = phi (A001622).

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User: Jens Rasmussen

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

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