

A191579


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


2



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, 62, 69, 59, 40, 21, 7, 1, 82, 140, 160, 144, 105, 62, 28, 8, 1, 185, 320, 375, 350, 271, 174, 91, 36, 9, 1, 423, 740, 885, 852, 690, 474, 273, 128, 45, 10, 1
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OFFSET

1,5


COMMENTS

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).
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^21 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
Is this (apart from offsets) the same as A097724?  R. J. Mathar, Aug 01 2011


LINKS

Table of n, a(n) for n=1..66.


FORMULA

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(i1,2*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.


EXAMPLE

The triangle begins
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, 62, 69, 59, 40, 21, 7, 1;
82, 140, 160, 144, 105, 62, 28, 8, 1;
185, 320, 375, 350, 271, 174, 91, 36, 9, 1;
423, 740, 885, 852, 690, 474, 273, 128, 45, 10, 1;
...
The 4th row is 2,3,3,1 because the 2nd,4th,6th and 8th terms of columns j = 15 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:
2*1 + 3*3 + 3*8 + 1*21 = 56 = T(8,2) of A192062
2*1 + 3*4 + 3*15+ 1*56 = 115 = T(8,3) of A192062
2*1 + 3*5 + 3*24 + 1*115 = 204 = T(8,4) of A192062
2*1 + 3*6 + 3*35 + 1*204 = 329 = T(8,5) of A192062


CROSSREFS

Cf. A091964, A124428, A004148, A089735, A098075, A192062.
Sequence in context: A052250 A333878 A099569 * A097724 A091836 A291980
Adjacent sequences: A191576 A191577 A191578 * A191580 A191581 A191582


KEYWORD

nonn,tabl


AUTHOR

Kenneth J Ramsey, Jun 07 2011


STATUS

approved



