zeroless squares

[Ignacio Larrosa Cañestro]

Sunday, February 27, 2005 11:04 PM [GMT+1=CET],
Ron Knott <ron at ronknott.com> escribió:

> A slightly simpler infinite series of zeroless squares is
> ((10^n-1)/3+1)^2 which, for n=1..8 gives
> 16, 1156, 111556, 11115556, 1111155556, 111111555556,
> 11111115555556, 1111111155555556
>
> Do all such sequences end with 5....56?
>
> Ron Knott

A(n) = ((10^n-1)/3+1)^2 = (10^(2n) + 4*10^n + 4)/9

A(n) = (10^(2n) - 1)/9 + 4*(10^n - 1)/9 + 1

A(n) = 1[2n] + 4[n] + 1 = 1[n]5[n-1]6

I.e., A(n)  = n 1's, followed by (n-1) 4's and a 6.

Best regards,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosa at mundo-r.com