A200743 Divide integers 1..n into two sets, minimizing the difference of their products. This sequence is the smaller product.

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

Offset

1,3

Links

Max Alekseyev, Table of n, a(n) for n = 1..140 (terms for n=1..35 from Michael S. Branicky)

Formula

a(n) = A127180(n) - A200744(n) = A200744(n) - A038667(n) = (A127180(n) - A038667(n)) / 2. - Max Alekseyev, Jun 18 2022

Example

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.

Maple

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

Mathematica

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]];
Table[Print[n, " ", a[n]]; a[n], {n, 1, 25}] (* Jean-François Alcover, Nov 03 2020 *)

Prog

(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)
# Christian Perfect, Feb 04 2015
(Python)
from math import prod, factorial
from itertools import combinations
def A200743(n):
    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

Crossrefs

Cf. A060776, A038667, A127180, A200744.

Sequence in context: A372123 A148089 A363065 * A061055 A060776 A148090

Adjacent sequences: A200740 A200741 A200742 * A200744 A200745 A200746

Keyword

nonn

Author

Franklin T. Adams-Watters, Nov 21 2011

Extensions

a(24)-a(30) from Alois P. Heinz, Nov 22 2011

a(31) from Michael S. Branicky, May 21 2021

Status

approved