cp's OEIS Frontend

A191579 Triangular array related to continued fractions of square root of (N^2 - 1) for N>1, apparently containing A004148 and summing to A091964.

%I A191579 #26 Jan 08 2013 10:47:28
%S A191579 1,1,1,1,2,1,2,3,3,1,4,6,6,4,1,8,13,13,10,5,1,17,28,30,24,15,6,1,37,
%T A191579 62,69,59,40,21,7,1,82,140,160,144,105,62,28,8,1,185,320,375,350,271,
%U A191579 174,91,36,9,1,423,740,885,852,690,474,273,128,45,10,1
%N A191579 Triangular array related to continued fractions of square root of (N^2 - 1) for N>1, apparently containing A004148 and summing to A091964.
%C A191579 The row sums of this triangle seems to be A091964 (verified to 12 terms), cf. diagonal sums of the triangle A124428. The 1st column seems to be A004148.  The 2nd and 3rd column seems to be A089735, and A098075 (verified to 10 terms).
%C A191579 As each of these sequence is related to enumeration of RNA molecule structures, but was generated independently by reference to square array A192062 (re continued fractions for square roots of n^2-1 for n>1, see comments in the example below), it could be interesting to check this further for a relationship. As Mathar noted, this triangle appears identical to A097724. - edited by _Kenneth J Ramsey_, Oct 25 2012
%C A191579 Is this (apart from offsets) the same as A097724? - R. J. Mathar, Aug 01 2011
%F A191579 The only way I know to generate this triangle is by reference to the square array A192062. The columns of that array, T(i,j) are such that for any given i>0, each term T(i,2*n) equals the sum as k = 1 to n, T(i-1,2*k)*C_k where C_k is the k th term of the n th row of this triangle. So solving the system of linear equations for each n > 0 gives the n th row of this triangle.
%e A191579 The triangle begins
%e A191579 1;
%e A191579 1, 1;
%e A191579 1, 2, 1;
%e A191579 2, 3, 3, 1;
%e A191579 4, 6, 6, 4, 1;
%e A191579 8, 13, 13, 10, 5, 1;
%e A191579 17, 28, 30, 24, 15, 6, 1;
%e A191579 37, 62, 69, 59, 40, 21, 7, 1;
%e A191579 82, 140, 160, 144, 105, 62, 28, 8, 1;
%e A191579 185, 320, 375, 350, 271, 174, 91, 36, 9, 1;
%e A191579 423, 740, 885, 852, 690, 474, 273, 128, 45, 10, 1;
%e A191579 ...
%e A191579 The 4th row is 2,3,3,1 because the 2nd,4th,6th and 8th terms of columns j = 1-5 of square array T(i,j) A192062  form the 4*5 matrix {{1,3,8,21},{1,4,15,56},{1,5,24,115},{1,6,35,204},{1,7,48,329}}. Solving the resulting system of linear equations results in the identities:
%e A191579 2*1 + 3*3 + 3*8 + 1*21 = 56 = T(8,2) of A192062
%e A191579 2*1 + 3*4 + 3*15+ 1*56 = 115 = T(8,3) of A192062
%e A191579 2*1 + 3*5 + 3*24 + 1*115 = 204 = T(8,4) of A192062
%e A191579 2*1 + 3*6 + 3*35 + 1*204 = 329 = T(8,5) of A192062
%Y A191579 Cf. A091964, A124428, A004148, A089735, A098075, A192062.
%K A191579 nonn,tabl
%O A191579 1,5
%A A191579 _Kenneth J Ramsey_, Jun 07 2011