OFFSET
1,4
COMMENTS
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?
LINKS
R. G. Wilson v, Table of n, a(n) for n = 1..250
EXAMPLE
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!.
MAPLE
A156832 := proc(n) local dvs, i, largd ; option remember; if n = 1 then 1; else dvs := sort(convert(numtheory[divisors](n!), list)) ; for i from 1 to nops(dvs) do largd := op(-i, dvs) ; if largd+add( procname(i), i=1..n-1) in dvs then RETURN(largd) ; fi; od: error(n) ; fi; end: for n from 1 do printf("%d, \n", A156832(n)) ; od; # R. J. Mathar, Feb 20 2009
MATHEMATICA
f[n_] := f[n] = Block[{d = 1, s = Sum[ f@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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Leroy Quet, Feb 16 2009
EXTENSIONS
STATUS
approved