A156832 a(n) = the largest divisor of n! such that (sum{k=1 to n} a(k)) is a divisor of n!.

1, 1, 1, 3, 24, 90, 720, 2520, 10080, 120960, 604800, 5913600, 79833600, 691891200, 15567552000, 65383718400, 1307674368000, 11115232128000, 66691392768000, 1187940433680000, 79829597143296000, 3568256278659072000

Offset: 1

Author

Leroy Quet, Feb 16 2009

Keywords

• nonn

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?

Examples

			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!.
		

Links

• R. G. Wilson v, Table of n, a(n) for n = 1..250

Crossrefs

For n!/a(n) see A145500.

Cf. A145499.

Programs

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 *)

Extensions

More terms from Robert G. Wilson v, Joshua Zucker and R. J. Mathar, Feb 16 2009