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A126892 a(n) = row of Wythoff's array T(n,j) containing the sequence of values T(n,j-1) + T(n,j+1).

%I A126892 #11 May 02 2023 08:59:10
%S A126892 1,15,8,12,44,19,62,26,30,91,37,109,120,48,138,55,59,167,66,185,73,77,
%T A126892 214,84,88,243,95,261,102,106,290,113,308,319,124,337,131,135,366,142,
%U A126892 384,149,153,413,160,431,442,171,460,178,182,489,189,507,196,200
%N A126892 a(n) = row of Wythoff's array T(n,j) containing the sequence of values T(n,j-1) + T(n,j+1).
%C A126892 Every Fibonacci sequence with positive terms occurs as some row of Wythoff's array (A035513), so a(n) is always defined. There appear to be two possible offsets for the sequence of sums within row a(n); either T(n,j-1) + T(n,j+1) = T(a(n),j-3) for j>=3 or T(n,j-1) + T(n,j+1) = T(a(n),j-1) for j>=1. The first case seems to occur whenever n has a Zeckendorf representation which ends in 1, 10100, 101000100, 1010001000100, 10100010001000100, etc. (each successive ending is obtained by changing the left-hand 1 to 10100). The values of these endings are 1,11,79,545,3739,25631,175681 ... and equal F(i)*F(i+1) + F(i+2)^2 for i = 0,2,4,6,... where F(i) is the i-th Fibonacci number. These values also appear in the table A127561 at a(1,0), a(1,1), a(2,3), a(5,8), ..., a(F(2n-1),F(2n)) for n = 0,1,2,3....
%C A126892 The Zeckendorf representation of n is the unique binary sequence ...,b(4),b(3),b(2) for which n = sum_{i>=2} b(i)F(i) and two consecutive b's cannot both be 1. For example, the Zeckendorf representation of 100 is 1000010100, since 100 = 89+8+3 = F(11)+F(6)+F(4).
%F A126892 Conjecture: If the Zeckendorf representation of n ends in 1, then a(n) = 15 + H(n-H(n))*29 + (n-H(n) - H(n-H(n)))*18, where H(n) is Hofstadter's G sequence A005206. Otherwise, a(n) = 1 + H(H(n))*7 + (H(n) - H(H(n)))*4 unless the Zeckendorf representation of n has one of the 0-endings listed in the first comment line, in which case a(n) = a(n+1) - 11.
%e A126892 a(2)=8 because the sequence of sums T(2,j-1)+T(2,j+1) begins with 6+16=22=T(8,0) and 10+26=36=T(8,1). a(1)=15 because the sequence of sums T(1,j-1)+T(1,j+1) begins with 4+11=15, 7+18=25, 11+29=40=T(15,0) and 18+47=65=T(15,1).
%t A126892 T[i_,j_]:=i*Fibonacci[j+1]+Fibonacci[j+2]*Floor[(i+1)(1+Sqrt[5])/2]; U[i_,j_]:=T[i,j-1]+T[i,j+1]; Tpair[i_,j_]:={T[i,j],T[i,j+1]}; Upair[i_,j_]:={U[i,j],U[i,j+1]}; a[n_]:=a[n]=Module[{v},For[v=0,True,v++,If[Upair[n,1]==Tpair[v,0]||Upair[n,3]==Tpair[v,0],Return[v]]]]
%Y A126892 Cf. A035513, A005206, A003622, A127561.
%K A126892 nonn
%O A126892 0,2
%A A126892 _Kenneth J Ramsey_, Jan 13 2007
%E A126892 Edited by _Dean Hickerson_, Feb 09 2007