[seqfan] Re: As much as I hate "base" sequences...

[David Wilson]

Sorry, no can do.

I don't have a real arbitrary precision arithmetic package, so I am using
ActivePerl Math::Bignum, which is somewhat slow.

I started running my program, hoping to duplicate your results, but

(a) Our results disagree starting at n = 121. I checked our values, and they
both reach 1 after 121 iterations. However, yours is smaller, so the bug is
in my program. I suspect my upper bound on digit sum of the predecessor is
too low, so I am missing some possible predecessors.

(b) My program started to seriously bog around n = 150. So even if I fixed
the bug, I don't think I could get out to n = 441 in reasonable time.

The good news is that our sequences do agree up to n = 120, so they are
undoubtedly correct to that point.


However, I do believe you that the sequence ends.

I tested other targets besides 1. It seems that in most cases, the sequence
ends. The following table shows the largest numbers that reach n for small n
(module possible aforementioned bugs in my program):

b(1) >= your a(440), 440 iterations, assuming your program is correct.
b(2) = 259737222559661262443050083680256000, 20 iterations
b(3) = 118098, 4 iterations
b(4) = 9838850899029636080, 12 iterations
b(5) = 9183300480, 7 iterations
b(6) = 972, 2 iterations
b(7) = huge but finite(?), > 100 iterations
b(8) = 839808, 4 iterations
b(9) = 4252141329999822344449363357449447653634578841600000, 27 iterations
b(10) = 10*b(1), 440 iterations (if n reaches 1 after k iterations, 10*n
reaches 10 after k iterations).
b(11) = 1643636270306863675880576630784000000, 20 iterations
b(12) = 1417176, 4 iterations
b(13) = 24564384, 5 iterations
b(14) = 5129317578803998439819357798400, 17 iterations
b(15) = 430467210, 6 iterations
b(16) = 7346640384000, 8 iterations
b(17) = 2007666, 4 iterations
b(18) = b(2), 19 iterations (if n reaches 18 after k iterations, it reaches
2 after k+1 iterations).
b(19) = infinity (1026 * 10^(n-2) reaches 19 in n iterations for n >= 2)
b(20) = 10*b(2) (if n reaches 2 after k iterations, 10*n reaches 20 after k
iterations).

If the digit sum of n is a power of 10, then b(n) = infinity (because
n*10^(k*digit_sum(n)) reaches n after k iterations). Given your (believable)
result about the finity of b(1), I suspect that the converse may be true as
well.

> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Hans
> Havermann
> Sent: Tuesday, January 21, 2014 9:42 AM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: As much as I hate "base" sequences...
> 
> David Wilson: "For a = A235601, let S(0) = {1}, then for each n >= 1,
compute
> set S(n) of possible predecessors of elements of S(n-1). Then a(n) is the
> smallest element of S(n). Using this approach, I was able to compute up to
> a(100), and could have gone further, but the elements were getting very
> large."
> 
> Not too large. I found that the size of S fluctuates wildly, never
exceeding 120
> and dropping to 0 for A235601(441). I've asked David to confirm (or
refute)
> this computation.
> 
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