[seqfan] Re: Can more of these terms be found?

[Hans Havermann]

> Previously in this thread I had opined that "if I take T.D. Noe's Mathematica formula for A006886 and change h = 10^k-1 to h = 10^k+1, I can generate many of the terms in my b-file. There are additional terms and missing ones as well, so that alteration is just a first step."
> 
> I had gone back to this formulation and noticed that the additional terms were proxies for the (non-powers-of-ten) missing ones.

They are not proxies. For k = 3 to 8, the divisors of 10^k+1 generated exactly half of the correct terms and (because each correct term corresponds to a missing one) I jumped to the wrong conclusion. For k = 9, twenty of the thirty output numbers are correct.

Sorting the output from a particular k, it is always the initial (smallest) ones that are correct. This led me to look for a constant that separates the correct terms from the incorrect ones. It wasn't long before I noticed that the correct numbers never exceeded 2*10^k/(1+sqrt(5)).

Up to now I had been squaring the numbers and checking every possible split to determine correctness. This new observation will allow me to do much better. For example, I can generate all (182381) solutions for k = 211 to 230 in about two minutes. (Doing just k = 210 will take significantly longer because of the large number of divisors.)

> Excluding powers-of-ten and powers-of-ten-plus-one, each solution generates another one by prepending it with 100.. where the number of zeros needs to be determined.

I now know which power of ten gets prepended. I had mentioned that while many k-digit solution terms derive from the divisors of 10^k+1, a few may come from the divisors of 10^(k+m)+1. The power-of-ten prepend appears to be 10^m. See < http://chesswanks.com/seq/a118938.txt > for a 12-MB file of A118938 terms and their 10^m prepend counterparts.