cp's OEIS Frontend

Let n = 2*m-1 where m is 1,2,3,...; then a(m) = F(3*n)*(L(n)+1) with F(n) a Fibonacci number and L(n) a Lucas number (A000032); also a(n) = F(n)*(L(n)^3+L(n)^2+L(n)+1) which is a repdigit in base L(n), made of four digits F(n).

For n = 1, unary representation must be used to give a repdigit, then 4(10) = 1111(1).

For n = 3, 170(10) = 2222(4).

For n = 5, 7320(10) = 5555(11).

Starting from n = 5, the repdigit is the 4th term of a Fibonacci-type sequence of 5 palindromes.

For example this sequence for n=5 in base 11 is: 1331, 2112, 3443, 5555, 8998 which are the 5 2-digit Fibonacci numbers in base 10 concatenated with their reversed forms.

If the repdigit F(3*n)*(L(n)+1) is the most noticeable result in base L(n), there are other recurrences; naming f the digit F(n) and g the digit F(n)*2, we find

F(0*n)*(L(n)+1) = 0

F(1*n)*(L(n)+1) = ff

F(2*n)*(L(n)+1) = ff0

F(3*n)*(L(n)+1) = ffff

F(4*n)*(L(n)+1) = ffgg0

The base L(n) gives a visual aspect to the formula

F(n*k) = L(n)*F(n*(k-1)) + F(n*(k-2)) with n odd, which is a particular case of the general formula for any integers n,k,m:

F(n*(k)+m) = L(n) * F(n*(k-1)+m) - (-1)^n * F(n*(k-2)+m)