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Let S_n be the set {n!/(i!*j!*k!) | i, j, k > 0, i+j+k = n} (i.e., trinomial coefficients that involve all three monomials). Then a(n) is the smallest gcd of any three members of S_n.
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%I #5 Jul 28 2017 23:39:00

%S 6,12,10,30,7,28,6,30,11,66,13,91,6,12,34,102,19,38,12,22,23,46,15,65,

%T 6,12,29,435,62,124,6,34,10,36,37,703,6,24,41,82,86,43,20,46,47,94,21,

%U 70,6,12,53,159,10,35,21,58,59,177,61,1891,14,28,10,30,67,134,12,14,142,142

%N Let S_n be the set {n!/(i!*j!*k!) | i, j, k > 0, i+j+k = n} (i.e., trinomial coefficients that involve all three monomials). Then a(n) is the smallest gcd of any three members of S_n.

%C Are there any 1's in this sequence?

%e S_7 = {42, 105, 140, 210}, gcd(42, 105, 140) = 7, gcd(42, 105, 210) = 21, gcd(42, 140, 210) = 14, gcd(105, 140, 210) = 35. So a(7) is the smallest of these, 7.

%Y Cf. A046816, A091963.

%K nonn,less

%O 3,1

%A _David Wasserman_, Mar 25 2004