

A126892


a(n) = row of Wythoff's array T(n,j) containing the sequence of values T(n,j1) + T(n,j+1).


0



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
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OFFSET

0,2


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,j1) + T(n,j+1) = T(a(n),j3) for j>=3 or T(n,j1) + T(n,j+1) = T(a(n),j1) 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 lefthand 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 ith 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(2n1),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).


LINKS

Table of n, a(n) for n=0..55.


FORMULA

Conjecture: If the Zeckendorf representation of n ends in 1, then a(n) = 15 + H(nH(n))*29 + (nH(n)  H(nH(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 0endings listed in the first comment line, in which case a(n) = a(n+1)  11.


EXAMPLE

a(2)=8 because the sequence of sums T(2,j1)+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,j1)+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).


MATHEMATICA

T[i_, j_]:=i*Fibonacci[j+1]+Fibonacci[j+2]*Floor[(i+1)(1+Sqrt[5])/2]; U[i_, j_]:=T[i, j1]+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]]]]


CROSSREFS

Cf. A035513, A005206, A003622, A127561.
Sequence in context: A094501 A320572 A090636 * A195035 A245624 A182165
Adjacent sequences: A126889 A126890 A126891 * A126893 A126894 A126895


KEYWORD

nonn


AUTHOR

Kenneth J Ramsey, Jan 13 2007


EXTENSIONS

Edited by Dean Hickerson, Feb 09 2007


STATUS

approved



