# [seqfan] Re: (no subject)

Joshua Zucker joshua.zucker at gmail.com
Tue Feb 17 08:51:36 CET 2009

```My (possibly buggy!) program says that the sequence begins
1 1 1 3 24 90 720 2520 10080 120960 604800 5913600 79833600 691891200
15567552000 65383718400 1307674368000 11115232128000 66691392768000
1187940433680000 79829597143296000 3568256278659072000
80285766269829120000 2685923817027010560000 29545161987297116160000
2068161339110798131200000 16803810880275234816000000
1190970096139507267584000000 33876482734634873389056000000
1228022499130514160353280000000 37893265687455865519472640000000
783127490874087887402434560000000 26313083693369353016721801216000000
434165880940594324775909720064000000
4920546650660069014126976827392000000
98410933013201380282539536547840000000
8856983971188124225428558289305600000000
72966324981389664029018864969318400000000
4587917697075448348771993193860300800000000

And so on.  My utterly stupid brute-force program generates terms up
to a(100) in 15 seconds or so.

Perhaps a more manageable list:
a(n) = n! / A156832(n)
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 2100 3780 1554 7168
4446 1680 15744 7380 560 572 1170 3588 14664 5031 5096 3900 768 780
1590 4080 2673 385 1216 1305 6400 6000 12200 37200 25488 7200 7280
14784 45024 182784 188416 17664 17640 17640 35770 36778 18500 18620
37730 14700 61620 7500 23328 70848 69056 139776 169728 70520 70992
48246 48416 16896 2145 2898 4284 24769 23560 24064 24056 26112 58212
42160

(That is how I wrote my brute-force program: loop through k, and test
that n! is divisible by k and n! is divisible by (sum of sequence so
far + n!/k), then output n!/k.  Since it never had to go up to very
large k, it ran reasonably quickly.)

--Joshua Zucker

On Mon, Feb 16, 2009 at 2:42 PM, Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:
> 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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>

```