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%I #8 Dec 11 2019 23:40:36
%S 1,1,2,1,6,3,1,12,16,4,1,20,50,32,5,1,30,120,140,55,6,1,42,245,448,
%T 316,86,7,1,56,448,1176,1284,622,126,8,1,72,756,2688,4170,3102,1113,
%U 176,9,1,90,1200,5544,11550,12122
%N Bisection of triangle A060098: odd-indexed members of column sequences of A060098 (not counting leading zeros).
%C Row sums give A060557. Column sequences without leading zeros give for m=0..5: A000012 (powers of 1), A002378 = 2*A000217, A004320, 4*A040977, A060558, 2*A060559.
%C Companion triangle (even-indexed members) A060102.
%C With offset 1 for n and k, T(n,k) is the number of (1-2-3)-avoiding trapezoidal words of length n that contain n+1-k 1s. A trapezoidal word (following Riordan) is a sequence (a_1,a_2,...,a_n) of integers with 1 <= a_i <= 2i-1. For example, T(3,3)=3 counts 122, 132, 133 and T(4,2)=12 counts 1112, 1113, 1114, 1115, 1116, 1117, 1121, 1131, 1141, 1151, 1211, 1311. - _David Callan_, Aug 25 2009
%F a(n, m)= A060098(2*n+1-m, m).
%F G.f. for column m: (x^m)*Po(m+1, x)/(1-x)^(2*m+1), with Po(n, x) = Sum_{j=0..floor(n/2)} binomial(n, 2*j+1)*x^j (odd members of row n of Pascal triangle A007318).
%F a(n, m) = Sum_{j=0..floor((m+1)/2)} binomial(n-j+m, 2*m)*binomial(m+1, 2*j+1), n >= m >= 0, otherwise zero.
%e {1}; {1,2}; {1,6,3}; {1,12,16,4}; ...; Po(3,x) = 3 + x.
%K nonn,easy,tabl
%O 0,3
%A _Wolfdieter Lang_, Apr 06 2001
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