# [seqfan] Re: Comment on A151750 [erratum]

Hagen von Eitzen hagen at von-eitzen.de
Sun Aug 2 21:24:12 CEST 2009

```Max Alekseyev schrieb:
> On Sun, Aug 2, 2009 at 6:14 AM,
> peter.luschny<peter.luschny at googlemail.com> wrote:
>
>
>> The problems in my original posting are somewhat harder.
>> Ron Graham for example offers 1000\$ for one of them.
>>
>
> As far as I know, Ron offered \$1000 for each of the following proofs:
> 1) infiniteness of A030979
> 2) finiteness of A151750
>
> btw, does anybody know how the bound 10^10000 was obtained in A151750 ?
>
I guess the following algorithm will make your candidate numbers grow
very fast

Start with n = 3161 in one of the bases, 11 say, that is 2414_11
This is ok, so transform to base 7: 12134_7
Here the last digit is bad, so count up to the next "rollover" of the
preceeding digit: 12140_7
There is still a bad digit, so repeat the step: 12200_7
(One might have collected all digits (p-1)/2 before the most significant
bad digit right away)
Transform to the next base, 5: 100220_5 (OK)
Transform to the next base, 3: 11100222_3 -> increase to 11101000_3
Transform to next base, 11: 2437_11 -> 2440_11
Transform to base 7: 12205_7 -> 12210_7
= 100232_5 -> 101000_5
= 11110101_3 (OK)
= 2495_11 -> 2500_11
= 12345_7 (a remarkable coincidence) -> 13000_7
= 102210_5 (OK)
= 11201001_3 -> 100000000_3
= 4A25_11 -> 5000_11
etc.

Only if a number was found with four "OK" in a row, a next member of
A151750 would have been found

These initial steps may not look too promising, but one might expect
only a few small leading digits
for a "random" integer, hence the next number will usually be a few
permille bigger and it will
take a feasible number of rounds (maybe a few million) to reach 10^10000.

Hagen

```

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