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A345022
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Smallest number divisible by all numbers 1 through n+1 except n, or 0 if impossible.
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1
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3, 4, 30, 12, 0, 120, 1260, 840, 0, 2520, 0, 27720, 0, 0, 6126120, 720720, 0, 12252240, 0, 0, 0, 232792560, 0, 5354228880, 0, 26771144400, 0, 80313433200, 0, 4658179125600, 72201776446800, 0, 0, 0, 0, 144403552893600, 0, 0, 0, 5342931457063200, 0
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OFFSET
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2,1
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COMMENTS
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a(n) > a(n+1) > 0 iff n > 2 and n is a power of 2 and n+1 is a prime or prime power. Does this occur only for n in {4, 8, 16, 256, 65536}? - Jon E. Schoenfield, Jun 07 2021
Probably yes. Those are the Fermat numbers minus 1 and the number 8 (which is the only power of 2 that is one less than a square number). - J. Lowell, Jun 08 2021
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LINKS
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FORMULA
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EXAMPLE
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a(5)=12 because 12 is divisible by 1, 2, 3, 4, and 6; but not 5.
a(6)=0 because it's impossible for a number to be divisible by 1, 2, 3, 4, 5, and 7; but not 6. Any number divisible by both 2 and 3 is also divisible by 6.
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MATHEMATICA
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Table[If[Mod[l=LCM@@Join[Range[n-1], {n+1}], n]==0, 0, l], {n, 2, 50}] (* Giorgos Kalogeropoulos, Jun 25 2021 *)
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PROG
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(Magma) a:=[]; L:=1; for n in [2..42] do t:=Lcm(L, n+1); if t mod n eq 0 then a[n-1]:=0; else a[n-1]:=t; end if; L:=Lcm(L, n); end for; a; // Jon E. Schoenfield, Jun 05 2021
(Python) # generates initial segment of sequence
from math import gcd
from itertools import accumulate
def lcm(a, b): return a * b // gcd(a, b)
def aupton(nn):
lcm1 = accumulate(range(1, nn), lcm)
lcm2 = [lcm(k, n+1) for n, k in enumerate(lcm1, start=2)]
return [m*(m%n != 0) for n, m in enumerate(lcm2, start=2)]
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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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