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%I #14 Sep 14 2015 16:53:14
%S 1,1,1,3,24,90,720,2520,10080,120960,604800,5913600,79833600,
%T 691891200,15567552000,65383718400,1307674368000,11115232128000,
%U 66691392768000,1187940433680000,79829597143296000,3568256278659072000
%N a(n) = the largest divisor of n! such that (sum{k=1 to n} a(k)) is a divisor of n!.
%C 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?
%H R. G. Wilson v, <a href="/A156832/b156832.txt">Table of n, a(n) for n = 1..250</a>
%e 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!.
%p 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
%t 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 *)
%Y For n!/a(n) see A145500.
%Y Cf. A145499.
%K nonn
%O 1,4
%A _Leroy Quet_, Feb 16 2009
%E More terms from _Robert G. Wilson v_, _Joshua Zucker_ and _R. J. Mathar_, Feb 16 2009
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