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%I #23 Dec 10 2016 17:16:35
%S 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,
%T 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,
%U 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
%N 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).
%C 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.
%C 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).
%C 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)).
%C 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.
%H Kenneth J Ramsey,<a href="http://tech.groups.com/group/Triangular_and_Fibonacci_Numbers/messages">Triangular and Fibonacci Numbers</a>
%F 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).
%F 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.
%F T(2*i,j) = T(i,j)*A(i,j) where A(i,j) is from the table A(i,j) of A191971.
%F T(4*i,j) = (T(2*i+1)^2 - T(2*i-1)^2)/j
%F T(4*i+2,j) = T(2*i+2,j)^2 - T(2*i,j)^2
%e Array as meant by the definition
%e First column has index j=0
%e 0 0 0 0 0 0 0 ...
%e 1 1 1 1 1 1 1 ...
%e 1 1 1 1 1 1 1 ...
%e 1 2 3 4 5 6 7 ...
%e 2 3 4 5 6 7 8 ...
%e 1 5 11 19 29 41 55 ...
%e 3 8 15 24 35 48 63 ...
%e 1 13 41 91 169 281 433 ...
%e 4 21 56 115 204 329 496 ...
%e .
%e .
%e .
%Y Cf. A191579, A191971.
%K nonn,tabl
%O 0,14
%A _Kenneth J Ramsey_, Jun 21 2011
%E Corrected and edited by _Olivier Gérard_, Jul 05 2011
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