[seqfan] Re: Factorials And Their Divisors

[franktaw at netscape.net]

It might be easier to look at this sequence in terms of the cumulative 
sum:
1, 2, 3, 6, 30, 120, 840, 3360, 13440, 134400, 739200, 6652800, 
86486400, 778377600
a(n) is the largest divisor of n! such that a(n) - a(n-1) also divides 
n!.

Looking at it this way, it seems to me that I can almost construct a 
proof that the sequence is infinite, but I haven't quite been able to 
put it all together.  I'll post it if I manage to work it all out.

Franklin T. Adams-Watters


-----Original Message-----
From: Robert G. Wilson, v <rgwv at rgwv.com>

Leroy,

     I can not answer your question, but I can extend your sequence out
to 250 terms.

Sequentially yours, Bob.

%S A156832
1,1,1,3,24,90,720,2520,10080,120960,604800,5913600,79833600,691891200,
%T A156832
15567552000,65383718400,1307674368000,11115232128000,66691392768000,
%U A156832
1187940433680000,79829597143296000,3568256278659072000,802857662698291200
00
%N A156832 a(n) = the largest divisor of n! such that (sum{k=1 to n}
a(k)) is a divisor of n!.
%C A156832 Is this sequence finite; or is there always a divisor of n!
where the sum of the first n terms of the sequence divides n!, for 
every
positive integer n? %C A156832 n!/a(n) for n=1: 1, 2, 6, 8, 5, 8, 7, 
16,
36, 30, 66, 81, 78, 126, 84, 320, 272, 576, 1824, 2048, 640, 315, 322,
231, 525, 195, 648, 256, 261, 216, 217, 336, 330, 680, ..., . - Robert
G. Wilson v, Feb 16 2009.
%e A156832 For n = 5 we check the divisors of 5!=120, from the largest
downward: a(1)+a(2)+a(3)+a(4) + 120 = 126, which is not a divisor of
120. 1+1+1+3 + 60 = 66, which is not a divisor of 120. 1+1+1+3 + 40 =
46, which is not a divisor of 120. 1+1+1+3 + 30 = 36, which is not a
divisor of 120. But 1+1+1+3 + 24 = 30, which is a divisor of 120. So,
a(5) = 24 = the largest divisor of 5! such that 
a(1)+a(2)+a(3)+a(4)+a(5)
also divides 5!.
%H A156832 Robert G. Wilson, Table of n, a(n) for n = 0..150 
<b156832.txt>.
%t A156832 f[n_] := f[n] = Block[{d = 1, s = Sum[ f at i, {i, n - 1}]},
While[ Mod[n!, d] > 0 || Mod[n!, n!/d + s] > 0, d++ ]; n!/d]; Array[f,
23] (Robert G. Wilson, v Feb 16 2009).
%K A156832 nonn
%O A156832 1,4 %A A156832 Leroy Quet, Feb 16 2009.
%E A156832 a(8)-a(250) More terms from Robert G. Wilson v
(rgwv at rgwv.com), Feb 16 2009.

Leroy Quet wrote:

>Sorry for the faux pas. I meant to submit this with a title. (And I 
might be
more offensive to some by re-sending this.)
>
>-----
>
>I just submitted this sequence, an old sequence I just now got around 
to
submitting:
>
>%S A156832 1,1,1,3,24,90,720
>%N A156832 a(n) = the largest divisor of n! such that (sum{k=1 to n} 
a(k)) is a
divisor of n!.
>%C A156832 Is this sequence finite; or is there always a divisor of n! 
where
the sum of the first n terms of the sequence divides n!, for every 
positive
integer n?
>%e A156832 For n = 5 we check the divisors of 5!=120, from the largest
downward: a(1)+a(2)+a(3)+a(4) + 120 = 126, which is not a divisor of 
120.
1+1+1+3 + 60 = 66, which is not a divisor of 120. 1+1+1+3 + 40 = 46, 
which is
not a divisor of 120. 1+1+1+3 + 30 = 36, which is not a divisor of 120. 
But
1+1+1+3 + 24 = 30, which is a divisor of 120. So, a(5) = 24 = the 
largest
divisor of 5! such that a(1)+a(2)+a(3)+a(4)+a(5) also divides 5!.
>%K A156832 more,nonn
>%O A156832 1,4
>
>The question in the comment line is the issue here.
>Is this sequence finite; or is there always a divisor of n! where the 
sum of
the first n terms of the sequence divides n!, for every positive 
integer n?
>
>Thanks,
>Leroy Quet
>
>
>
>
>
>
>
>
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>
>Seqfan Mailing list - http://list.seqfan.eu/
>
>
>

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