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A200743
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Divide integers 1..n into two sets, minimizing the difference of their products. This sequence is the smaller product.
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12
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1, 1, 2, 4, 10, 24, 70, 192, 576, 1890, 6300, 21600, 78624, 294840, 1140480, 4561920, 18849600, 79968000, 348566400, 1559376000, 7147140000, 33522128640, 160745472000, 787652812800, 3938264064000, 20080974513600, 104348244639744, 552160113120000, 2973491173785600, 16286186592000000, 90678987245246400
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OFFSET
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1,3
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LINKS
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FORMULA
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EXAMPLE
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For n=1, we put 1 in one set and the other is empty; with the standard convention for empty products, both products are 1.
For n=13, the central pair of divisors of n! are 78975 and 78848. Since neither is divisible by 10, these values cannot be obtained. The next pair of divisors are 79200 = 12*11*10*6*5*2*1 and 78624 = 13*9*8*7*4*3, so a(13) = 78624.
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MAPLE
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a:= proc(n) local l, ll, g, p, i; l:= [i$i=1..n]; ll:= [i!$i=1..n]; g:= proc(m, j, b) local mm, bb, k; if j=1 then m else mm:= m; bb:= b; for k to 2 while (mm<p) do if j=2 or k=2 or k=1 and ll[j-1]*mm>bb then bb:= max(bb, g(mm, j-1, bb)) fi; mm:= mm*l[j] od; bb fi end; Digits:= 700; p:= ceil(sqrt(ll[n])); g(1, nops(l), 1) end: seq(a(n), n=1..23); # Alois P. Heinz, Nov 22 2011
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MATHEMATICA
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a[n_] := a[n] = Module[{s, t}, {s, t} = MinimalBy[{#, Complement[Range[n], #]}& /@ Subsets[Range[n]], Abs[Times @@ #[[1]] - Times @@ #[[2]]]&][[1]]; Min[Times @@ s, Times @@ t]];
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PROG
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(Python)
from itertools import combinations
def prod(l):
t=1
for x in l:
t *= x
return t
def a200743(n):
nums = list(range(1, n+1))
widths = combinations(nums, n//2)
dimensions = [(prod(width), prod(x for x in nums if x not in width)) for width in widths]
best = min(dimensions, key=lambda x:max(*x)-min(*x))
return min(best)
(Python)
from math import prod, factorial
from itertools import combinations
m = factorial(n)
return min((abs((p:=prod(d))-m//p), min(p, m//p)) for l in range(n, n//2, -1) for d in combinations(range(1, n+1), l))[1] # Chai Wah Wu, Apr 07 2022
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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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