zeroless squares

[Ron Knott]

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

>... 
> and as Ignacio Larrosa Ca<F1>estro points out:
> 
> > A(k) = 6[k] = 6(10^k - 1)/9  ===>
> 
> > (A(k))^2 = (6(10^k-1)/9)^2 = (4/9)(10^(2k) - 2*10^k + 1)
> 
> > which is k-1 4's, followed by a 3, k-1 5's and a 6.
> 
> So I will add this
> 
> %S A102794 
> 1,36,4356,443556,44435556,4444355556,444443555556,44444435555556,
> 4444444355555556,
> %T A102794 
> 444444443555555556,44444444435555555556,4444444444355555555556,
> 444444444443555555555556,
> %U A102794 
> 44444444444435555555555556,4444444444444355555555555556,
> 444444444444443555555555555556
> %N A102794 The number (666...6)^2.
> %C A102794 An infinite sequence of squares with no zeros in base 10.
> ....
> 
> as a new sequence.
> 
> NJAS