A192062 Square Array T(ij) read by antidiagonals (from NE to SW) with columns 2j being the denominators of continued fraction convergents to square root of (j^2 + 2j).

0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 2, 0, 1, 1, 3, 3, 1, 0, 1, 1, 4, 4, 5, 3, 0, 1, 1, 5, 5, 11, 8, 1, 0, 1, 1, 6, 6, 19, 15, 13, 4, 0, 1, 1, 7, 7, 29, 24, 41, 21, 1, 0, 1, 1, 8, 8, 41, 35, 91, 56, 34, 5, 0, 1, 1, 9, 9, 55, 48, 169, 115, 153, 55, 1, 0, 1, 1, 10, 10, 71, 63, 281, 204, 436, 209, 89, 6

Offset: 0

Author

Kenneth J Ramsey, Jun 21 2011

Keywords

• nonn
• tabl

Comments

Column j=1 is the Fibonacci sequence A000045. Column 2 is A002530; column 4 is A041011; column 6 is A041023; column 8 is A041039, column 10 is A041059, column 12 is A041083, column 14 is A041111 corresponding the denominators of continued fraction convergents to square root of 3,8,15,24,35,48 and 63.

T(2*i-1,j)*T(2*i,j)^2*T(2*i+1,j)*j/2 appears to be always a triangular number, T(j*T(2*i,j)^2).

T(2*i,j)*T(2*i+1,j)^2*T(2*i+2)*j/2 appears to always equal a triangular number, T(j*T(2*i,j)*T(2*i+2,j)).

Conjecture re relation of A192062 to the sequence of primes: T(2*n,j) = A(n,j)*T(n,j) where A(n,j) is from the square array A191971. There, A(3*n,j) = A(n,j)*B(n,j) where B(n,j) are integers. It appears further that B(5*n,j)=B(n,j)*C(n,j); C(7*n,j)= C(n,j)*D(n,j); D(11*n,j) = D(n,j)*E(n,j); E(13*n,j) = E(n,j)*F(n,j) and F(17*n,j) = F(n,j)*G(n,j) where C(n,j), D(n,j) etc. are all integers. My conjecture is that this property continues indefinitely and follows the sequence of primes.

Examples

			Array as meant by the definition
First column has index j=0
0  0  0   0   0   0   0 ...
1  1  1   1   1   1   1 ...
1  1  1   1   1   1   1 ...
1  2  3   4   5   6   7 ...
2  3  4   5   6   7   8 ...
1  5 11  19  29  41  55 ...
3  8 15  24  35  48  63 ...
1 13 41  91 169 281 433 ...
4 21 56 115 204 329 496 ...
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Links

• Kenneth J Ramsey,Triangular and Fibonacci Numbers

Crossrefs

Cf. A191579, A191971.

Formula

Each column j is a recursive sequence defined by T(0,j)=0, T(1,j) = 1, T(2i,j)= T(2i-2,j)+T(2i-1,j) and T(2i+1,j) = T(2i-1,j)+j*T(2i,j). Also, T(n+2,j) = (j+2)*T(n,j)-T(n-2,j).

T(2n,j) = Sum(k=1 to n) C(k)*T(2*k,j-1) where the C(k) are the n-th row of the triangle A191579.

T(2*i,j) = T(i,j)*A(i,j) where A(i,j) is from the table A(i,j) of A191971.

T(4*i,j) = (T(2*i+1)^2 - T(2*i-1)^2)/j

T(4*i+2,j) = T(2*i+2,j)^2 - T(2*i,j)^2

Extensions

Corrected and edited by Olivier Gérard, Jul 05 2011