0,2

Comments from Kenneth Ramsey (RamseyKK2(AT)aol.com), Jan 06 2007 (Start) "a(n) = the characteristic value of the row T(n,i) of the Wythoff array A035513 which is the absolute value of T(n,i)^2 - T(n,i-1)*T(n,i+1). Only the number 5 or prime factors ending in 1 or 9 form the square-free portion of a(n). All other factors of a(n) appear only as squares.

"Moreover, the square-free portion (less the factor 5) squared is the characteristic value of the Fibonacci sequence whose bijection relates to c term of the Horadam "Fibonacci Number Triples" Amer. Math. Monthly 68(1961) 751-753. That paper showed that if F(0), F(1), F(2), F(3) are 4 sequential numbers in a row of the Wythoff array, then P = (2F(1)*F(2),F(0)*F(1),2F(1)*F(2) + F(0)^2) is a Pythagorean triple (a,b,c) i.e. a^2 + b^2 = c^2.

"If i varies and c(n,2i-1) = F(n,i)^2 + 2F(n,i+1)*F(n,i+2) and C(n,2i) is set to equal C(n,2i+1)-C(n,2i-1) then, the sequence F(x,i) = C(n,i)/G, where G is the greatest common divisor of the adjacent terms C(n,i), is a Fibonacci sequence having the characteristic value which is the square of the square-free portion of a(n) except without the factor of 5.

"For example the Lucus sequence or the second row of the Wythoff array has the characteristic value of A(2) = 5 and the C(n,i) terms are each 5 times the sequential terms 34,89,233,... which is a bijection of the terms in the 1st row of the Wythoff array which row has the characteristic value of 1. This is so even though adjacent terms of the Lucus sequence are coprime." (End)

Conjecture: Every pair of Fibonacci sequences, F1 and F2, appear in rows n and m of Wythoff's Array, respectively and have respective characteristics a(n) and a(m). Also, there is a third Fibonacci sequence F3, defined by F3(i) = F1(i) * F2(j+1) - F1(i+1)*F2(j) where j is held constant. The sequence F3 appears in row p of Wythoff's array and has the characteristic a(p) = a(n)*a(m). - Kenneth Ramsey (RamseyKK2(AT)aol.com), Feb 11 2007

Allan Wechsler (acw(AT)alum.mit.edu), posting to math-fun mailing list Dec 04 1996.

[ (n+1)*tau ]^2 - n*[ (n+1)*tau ] - n^2.

Digits := 50: t := evalf((1+sqrt(5))/2): f := n->floor( n*t)^2-(n-1)*floor(n*t)-(n-1)^2:

Cf. A035513.

Cf. A127561.

Sequence in context: A110617 A102081 A068397 this_sequence A046588 A086654 A152064

Adjacent sequences: A022341 A022342 A022343 this_sequence A022345 A022346 A022347

nonn

N. J. A. Sloane (njas(AT)research.att.com).