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A328202
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a(n) is the greatest common divisor of all the numbers in row n of Pascal's triangle excluding 1 and n.
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2
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6, 10, 5, 7, 14, 6, 3, 11, 11, 13, 13, 1, 2, 34, 17, 19, 19, 1, 1, 23, 23, 5, 5, 3, 3, 29, 29, 31, 62, 2, 1, 1, 1, 37, 37, 1, 1, 41, 41, 43, 43, 1, 1, 47, 47, 7, 7, 1, 1, 53, 53, 1, 1, 1, 1, 59, 59, 61, 61, 1, 2, 2, 1, 67, 67, 1, 1, 71, 71, 73, 73, 1, 1, 1, 1
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OFFSET
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4,1
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COMMENTS
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If p is prime, a(p) is a multiple of p and no term a(n) where n < p is a multiple of p.
If p is prime, a(p^k) and a(p^k+1) are divisible by p for all k>=1.
If p is a Fermat prime > 3, a(p)=2*p.
If p is a Mersenne prime, a(p+1)=2*p.
Conjecture: these are the only cases where a(n)>n. (End)
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LINKS
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FORMULA
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EXAMPLE
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For n=8, take row 8 of Pascal's triangle:
1 8 28 56 70 56 28 8 1,
remove the first and last 2 numbers:
28 56 70 56 28,
the greatest common divisor of 28, 56, 70, 56, 28 is 14, thus a(8)=14.
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MAPLE
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f:= proc(n) local k, g;
g:= binomial(n, 2);
for k from 3 to n/2 do
g:= igcd(g, binomial(n, k));
if g = 1 then return g fi;
od;
g
end proc:
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MATHEMATICA
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a[n_] := GCD @@ Binomial[n, Range[2, n/2]]; a /@ Range[4, 90] (* Giovanni Resta, Oct 08 2019 *)
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PROG
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(PARI) a(n) = gcd(vector((n+1)\2-1, k, binomial(n, k+1))); \\ Michel Marcus, Oct 08 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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