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Square root of v(2n)/v(2n-1), where v=A203748.
3

%I #15 Jul 13 2021 01:54:30

%S 1,14,741,87024,18068505,5845458528,2718866959893,1719570636306432,

%T 1419543579377755377,1482454643117692608000,1910657530214126188243749,

%U 2978927846824451394372304896,5526241720077994999033052180169

%N Square root of v(2n)/v(2n-1), where v=A203748.

%C See A203748.

%F Define a sequence f(n) by means of the double product f(n) = |Product_{1 <= a, b <= n} (a - b*w)|, where w = exp(2*Pi*i/3) is a primitive cube root of unity. So f(n) is a sort of 2-dimensional analog of n!. Then a(n) = f(n)/(f(1)*f(n-1)) is the first column of the triangle ( f(n)/(f(k)*f(n-k)) ) 0<=k<=n, an analog of Pascal's triangle. - _Peter Bala_, Sep 21 2013

%e Triangle ( f(n)/(f(k)*f(n-k)) ), 0 <= k <= n, begins

%e 1;

%e 1, 1;

%e 1, 14, 1;

%e 1, 741, 741, 1;

%e 1, 87024, 4606056, 87024, 1;

%e ... - _Peter Bala_, Sep 21 2013

%t (See A203748.)

%Y Cf. A203748, A203774.

%K nonn

%O 1,2

%A _Clark Kimberling_, Jan 05 2012