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

1, 15, 8, 12, 44, 19, 62, 26, 30, 91, 37, 109, 120, 48, 138, 55, 59, 167, 66, 185, 73, 77, 214, 84, 88, 243, 95, 261, 102, 106, 290, 113, 308, 319, 124, 337, 131, 135, 366, 142, 384, 149, 153, 413, 160, 431, 442, 171, 460, 178, 182, 489, 189, 507, 196, 200

Offset: 0

Author

Kenneth J Ramsey, Jan 13 2007

Keywords

• nonn

Comments

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....

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).

Examples

			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).
		

Crossrefs

Cf. A035513, A005206, A003622, A127561.

Programs

Mathematica

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]]]]

Formula

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.

Extensions

Edited by Dean Hickerson, Feb 09 2007