%I A127561
%S A127561 0,5,1,20,11,4,45,31,19,9,80,61,44,29,16,125,101,79,59,41,25,180,151,
%T A127561 124,99,76,55,36
%N A127561 Lattice table of Fibonacci characteristic values from Wechsler's J determinant 
               sequence A022344 uniquely position such that the row and column determine 
               starting a,b values of a Fibonacci sequence having the same characteristic 
               value.
%C A127561 A vector from (0,0) to any prime value P in the array does not pass through 
               any other lattice point. If that vector is extended it passes through 
               lattice points having sucessively the values 0, P*1^2, P*2^2, P*3^2, 
               P*4^2 ... All primes ending in 1, 5 and 9 or the product thereof 
               appear in the array, no prime ending in 3 or 7 appears in the array 
               except in a square product which may be multiplied by a square-free 
               product of primes ending in 1, 5 or 9.
%C A127561 The table can be expanded by allowing negative arguments in the formula, 
               but any positive value obtained can be expressed with nonnegative 
               arguments.
%C A127561 The second row is the sequence A062786. The term in every succeeding 
               row is 2* the term immediately above minus the next above term plus 
               2.
%C A127561 If the table is rearrange by shifting each column down by twice the column 
               number, then the terms in second column would be equal to the row 
               number squared plus the row number minus 1 and every succeding term 
               to the right would be equal to twice the lefthand term minus the 
               next fefthand term minus 2.
%C A127561 It appears that any prime ending in 1,5, or 9 or any such prime times 
               5 appears only once in the table and that every power of such a prime 
               or product thereof has one and only one nonnegative row and column 
               position such that the row and column positions are coprime. A method 
               for finding a coprime row and column position of the 2^n th power 
               of any prime ending in 1,5,or 9, or of the product thereof, from 
               the coprime row and column position of that prime or product is suggested 
               by the discussion in the link titled "Wythoff Array, Pythagorean 
               Triples, Primes".
%C A127561 It seems that if you stack the row and column positions of two numbers 
               in the array that the determinant gives a column in which the product 
               appears. Thus since the row and column position of 29 and 41 are 
               3,1 and 4,1 respectively then the product (41*29) appears in column 
               1*4 - 3*1 or column 1. The same value appears also in column -1 so 
               3*1-1*4 is a valid answer also. For our purposes however we choose 
               the order that gives a positive value. Once the column number of 
               the product is known it is easy to find the row number. There may 
               be new determinant based math to find the row directly, but I don't 
               know of any. It may happen that the row is negative, in which case 
               the following transformation works a(r,c) = a(-r,c+r). Applied twice 
               this transformation gives the original starting pair. I have yet 
               to find any case in which one starts out with positive values for 
               the row and column of each factor of a number appearing in the table 
               and using the above determinant math can not find positive values 
               for the row and column of the product. I posted a few interesting 
               results in the Cut-the-knot forum. Use the link given previously.
%H A127561 K. J. Ramsey, <a href="http://www.cut-the-knot.org/htdocs/dcforum/DCForumID4/
               745">Wythoff Array, Pythagorean triplets, Primes</a>
%F A127561 C(a,b) = (a+2b)^2 + b(a+2b) - b^2
%e A127561 C(0,1) = 5 because (0+2*1)^2 + 1*(0+2*1) - 1^2 = 5 and also because the 
               Fibonacci sequence having the Horadam ID {a,b,1,1} with a = 0+2*1 
               and b = 1 has the characteristic value a^2 + b*a - b^2
%Y A127561 Cf. A035513, A022344.
%Y A127561 Sequence in context: A145372 A145373 A088577 this_sequence A144879 A049411 
               A070729
%Y A127561 Adjacent sequences: A127558 A127559 A127560 this_sequence A127562 A127563 
               A127564
%K A127561 nonn,tabl,uned
%O A127561 0,2
%A A127561 Kenneth Ramsey (RamseyKK2(AT)aol.com), Jan 18 2007, Feb 05 2007, Feb 
               06 2007