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A156832
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a(n) = the largest divisor of n! such that (sum{k=1 to n} a(k)) is a divisor of n!.
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3
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1, 1, 1, 3, 24, 90, 720, 2520, 10080, 120960, 604800, 5913600, 79833600, 691891200, 15567552000, 65383718400, 1307674368000, 11115232128000, 66691392768000, 1187940433680000, 79829597143296000, 3568256278659072000
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OFFSET
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1,4
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COMMENTS
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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?
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LINKS
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EXAMPLE
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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!.
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MAPLE
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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
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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