[seqfan] Re: PS and SP numbers (bis)

[David Wilson]

Yes, in addition to noting that A062237 is finite, I also found the two 
additional
elements that you found, and updated the sequence accordingly.  Also, I 
noted
that the sequence was base-dependent.

At any rate, the finity argument goes as follows.  Let |n|, s(n), and p(n) 
be the
number, sum, and product of the base-10 digits of n.  For SP numbers, we 
have
n = s(n) concat p(n), so that |n| = |s(n)| + |p(n)|.  But for |n| >= 88, 
|s(n)|+|p(n)| < |n|,
so no number of 88 or more digits is an SP number.

There is also a heuristic argument that we shouldn't find too many SP 
numbers.

Let n be an SP number.  If n has the digit 0, p(n) = 0, and n = s(n) concat 
0.
Thus s(n) = s(s(n) concat 0) = s(s(n)).   If s(n) >= 10, s(s(n)) < s(n), so 
s(n) <= 9.
This gives n = s(n) concat 0 = 10s(n) <= 90.  No SP number > 90 includes the
digit 0.  The rest of this message assumes n > 90.

Suppose n is an SP number.  Suppose p(n) includes the digit 0.  Then
n = s(n) concat p(n) includes 0, and by the previous paragraph, n is not SP.
The contradiction shows p(n) does not include 0.  This p(n) cannot end in 0,
so that 10 does not divide p(n), and p(n) cannot contain both an even digit
and a 5.

Now note that p(n), being the product of decimal digits, is a 7-smooth 
number,
that is, of the form 2^a 3^b 5^c 7^d.  Large numbers of this form will tend 
to
include the digit 0.  Just as it is all but certain that sufficiently large 
powers of 2
include the digit 0, the same can be said of sufficiently large 7-smooth 
numbers.
Thus, in the absence of the proven 87-digit limit, we would be safe to 
conjecture
that there are a finite number of SP numbers.  From a practical standpoint, 
this
phenomenon kicks in way below 88 digits, I would be amazed if there were
an SP number of 20 or more digits.

Final note.  As I said before, p(n) is of the form 2^a 3^b 5^c 7^d.  Since 
p(n)
cannot end in 0, 10 does not divide p(n), so p(n) = 2^a 3^b 7^c or 3^a 5^b 
7^c.
It is computationally feasible to generate every p(n) of this form <= 87 
digits.
Given p(n), s(n) = s(s(n))+s(p(n)) should allow us to find close bounds on
s(n).  This means that computing all necessary candidate pairs (s(n), p(n))
should be computationally feasible.  This would make it possible to find 
every
element of this sequence, just as I was able to find every element of 
A038369
(the sum-product numbers, defined as n = s(n)p(n) as opposed to s(n) concat 
p(n)).

----- Original Message ----- 
From: "Chuck Seggelin" <seqfan at plastereddragon.com>
To: <ham>; "David Wilson" <davidwwilson at comcast.net>; "Eric Angelini" 
<keynews.tv at skynet.be>; <ham at jussieu.fr>; <seqfan at ext.jussieu.fr>
Sent: Thursday, April 28, 2005 4:33 PM
Subject: Re: [seqfan] Re: PS and SP numbers (bis)


>I assume that although it is finite, not all the terms have been given?
>
> I say this because unless I'm mistaken, the 17'th and 18'th terms of this 
> sequence should be:
>
> 3634992 (digits sum to 36, product is 34992)
> 43139968 (digits sum to 43, product is 139968)
>
> Are there any terms beyond 43139968?
>
>            -- Chuck
>
>
> ----- Original Message ----- 
> From: "David Wilson" <davidwwilson at comcast.net>
> To: "Eric Angelini" <keynews.tv at skynet.be>; <ham at jussieu.fr>; 
> <seqfan at ext.jussieu.fr>
> Sent: Thursday, April 28, 2005 3:15 PM
> Subject: [seqfan] Re: PS and SP numbers (bis)
>
>
> BTW, A062237 is finite.
>
>
> ----- Original Message ----- 
> From: "Eric Angelini" <keynews.tv at skynet.be>
> To: <ham>; <seqfan at ext.jussieu.fr>
> Sent: Thursday, April 28, 2005 10:37 AM
> Subject: PS and SP numbers (bis)
>
>
>> Sorry, it is A062237
>>
>> à+
>> É.
>> __________________________________________________
>> http://angelink.be/?CeRaccourciFaitCinquanteSignes
>>
>
>
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