[seqfan] A175388(?)

[Robert Munafo]

I don't get the two comments, which seem to not only contradict each other
but actually call attention to it: If a(n) <= A002275(n), then a(17) would
have to be <= A002275(17) = (10^17-1)/9, but then the second comment clearly
says a(17) >> 10^17.

I suggest that the writer of the first comment probably meant something like
a(n) <= A002275(a(n-1)), because for all A002275(x), the sum of digits
squared is x, and therefore a solution of length n can be made into a
solution of length n+1 by prepending A002275(x) where x is the first item in
the existing length n solution.

On Sat, Apr 24, 2010 at 23:43, zak seidov <zakseidov at yahoo.com> wrote:

> Not yet submitted -
> pending for your comments/critics/praise;)
>
> Thanks, Zak
>
> %I A175388
> %S A175388 1,2,11,23,19,7,29,16,5,8,9,3,36,6,88,269,15999
> %N A175388 a(n) = smallest integer with exactly n iterations "summing  of
> digits squared" ending with 1 or 4.
> %C A175388 Sequence is infinite: for any n, a(n) <= R_n = A002275(n).
> %C A175388 Actually, a(17) = 46*10^198 - 1, 45 concatenated (or
> concatenating?) with 197 nines.
> %e A175388 a(0)=1 because 1 -> 1 (0 iterations)
> %e A175388 a(1)=2 because 2 -> 4 (1 iteration)
> %e A175388 a(2)=11 because 11 -> 2 -> 4 (2 iterations)
> %e A175388 a(3)=23 because 23 -> 13 -> 10 -> 1 (3 iterations)
> %e A175388 a(4)=19 because 19 -> 82 -> 20 -> 4 (4 iterations).
> %Y A175388 Cf. A002275, A176709.
> %K A175388 nonn
> %O A175388 0,2
> %A A175388 Zak Seidov (zakseidov(AT)yahoo.com), Apr 24 2010
>

-- 
 Robert Munafo  --  mrob.com