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%I #11 Jun 17 2014 02:22:33
%S 2,3,2,5,2,7,2,3,2,11,3,13,2,3,2,17,2,19,4,3,2,23,3,5,2,3,4,29,6,31,2,
%T 3,2,5,4,37,2,3,5,41,6,43,4,3,2,47,3,7,2,3,4,53,2,5,7,3,2,59,4,61,2,7,
%U 2,5,6,67,4,3,10,71,4,73,2,3,4,7,2,79,5,3,2,83,12,5,2,3,4,89,9,7,4,3,2,5,3
%N a(n) is the smallest gcd of two interior numbers on row n of Pascal's triangle ("interior" means that the 1's at the ends of the rows are excluded).
%C The reference contains a simple proof that there are no 1's in this sequence.
%D R. K. Guy, Unsolved Problems in Number Theory, Sections B31, B33.
%H Robert Israel, <a href="/A091963/b091963.txt">Table of n, a(n) for n = 2..10000</a>
%e In row 8, the interior numbers 8, 28, 56 and 70; gcd(8, 28) = 4; gcd(8, 56) = 8; gcd(8, 70) = 2; gcd(28, 56) = 28; gcd(28, 70) = 14; gcd(56, 70) = 14. The smallest of these is 2, so a(8) = 2.
%p seq(min(seq(igcd(n,binomial(n,k)),k=1..floor(n/2))), n=2..100); # _Robert Israel_, Jun 17 2014
%o (PARI) a(n) = {v = vector(n\2, i, binomial(n, i)); mgcd = n; for (i=1, #v, for (j=i+1, #v, mgcd = min(gcd(v[i], v[j]), mgcd););); return (mgcd);} \\ _Michel Marcus_, Jun 16 2013
%Y Cf. A014410.
%K nonn
%O 2,1
%A _David Wasserman_, Mar 13 2004
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