%I #40 Oct 25 2019 01:18:24
%S 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,
%T 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,
%U 59,59,61,61,1,2,2,1,67,67,1,1,71,71,73,73,1,1,1,1
%N a(n) is the greatest common divisor of all the numbers in row n of Pascal's triangle excluding 1 and n.
%C If p is prime, a(p) is a multiple of p and no term a(n) where n < p is a multiple of p.
%C From _Robert Israel_, Oct 16 2019: (Start)
%C If p is prime, a(p^k) and a(p^k+1) are divisible by p for all k>=1.
%C If p is a Fermat prime > 3, a(p)=2*p.
%C If p is a Mersenne prime, a(p+1)=2*p.
%C Conjecture: these are the only cases where a(n)>n. (End)
%H Robert Israel, <a href="/A328202/b328202.txt">Table of n, a(n) for n = 4..10000</a>
%F a(n) = A014963(n) * A014963(n - 1) (conjecture). - _Jon Maiga_, Oct 08 2019
%e For n=8, take row 8 of Pascal's triangle:
%e 1 8 28 56 70 56 28 8 1,
%e remove the first and last 2 numbers:
%e 28 56 70 56 28,
%e the greatest common divisor of 28, 56, 70, 56, 28 is 14, thus a(8)=14.
%p f:= proc(n) local k,g;
%p g:= binomial(n,2);
%p for k from 3 to n/2 do
%p g:= igcd(g,binomial(n,k));
%p if g = 1 then return g fi;
%p od;
%p g
%p end proc:
%p map(f, [$4..100]);# _Robert Israel_, Oct 16 2019
%t a[n_] := GCD @@ Binomial[n, Range[2, n/2]]; a /@ Range[4, 90] (* _Giovanni Resta_, Oct 08 2019 *)
%o (PARI) a(n) = gcd(vector((n+1)\2-1, k, binomial(n, k+1))); \\ _Michel Marcus_, Oct 08 2019
%Y Cf. A007318 (Pascal's triangle), A014963.
%K nonn,look
%O 4,1
%A _Joel Kaufmann_, Oct 07 2019
%E More terms from _Giovanni Resta_, Oct 08 2019