[seqfan] (no subject)
Leroy Quet
q1qq2qqq3qqqq at yahoo.com
Mon Feb 16 23:42:09 CET 2009
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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