A134680 a(n) = length (or lifetime) of the meta-Fibonacci sequence {f(1) = ... = f(n) = 1; f(k)=f(k-f(k-1))+f(k-f(k-n))} if that sequence is only defined for finitely many terms, or 0 if that sequence is infinite.

6, 0, 164, 0, 60, 2354, 282, 1336, 100, 1254, 366, 419, 498, 483, 778, 1204, 292, 373, 845, 838, 1118, 2120, 815, 2616, 686, 1195, 745, 1112, 2132, 1588, 754, 1227, 1279, 1661, 716, 2275, 784, 2341, 1874, 1463, 1122, 2800, 1350, 1613, 2279, 1557, 1532

Offset

1,1

Comments

Such a sequence has finite length when the k-th term becomes greater than k.

The term a(2) = 0 is only conjectural - see A005185. a(4) = 0 is a theorem of Balamohan et al. (2007). - N. J. A. Sloane, Nov 07 2007, Apr 18 2014.

Links

Seiichi Manyama, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)

B. Balamohan, A. Kuznetsov and S. Tanny, On the behavior of a variant of Hofstadter's Q-sequence, J. Integer Sequences, Vol. 10 (2007), #07.7.1.

D. R. Hofstadter, Curious patterns and non-patterns in a family of meta-Fibonacci recursions, Lecture in Doron Zeilberger's Experimental Mathematics Seminar, Rutgers University, April 10 2014; Part 1, Part 2.

D. R. Hofstadter, Graph of first 21500 terms.

Index entries for Hofstadter-type sequences

Example

a(1) = 6: the f-sequence is defined by f(1) = 1, f(n) = 2f(n-f(n-1)), which gives 1,2,2,4,2,8 but f(7) = 2f(-1) is undefined, so the length is 6.

Mathematica

Table[Clear[a]; a[n_] := a[n] = If[n<=k, 1, a[n-a[n-1]]+a[n-a[n-k]]]; t={1}; n=2; While[n<10000 && a[n-1]<n, AppendTo[t, a[n]]; n++ ]; len=Length[t]; If[len==9999, 0, len], {k, 100}]

Crossrefs

Cf. A005185, A046700, A063882, A132172, A134679 (sequences for n=2..6).

See A240810 for another version.

A diagonal of the triangle in A240813.

Sequence in context: A052679 A266218 A240818 * A362794 A336303 A111372

Adjacent sequences: A134677 A134678 A134679 * A134681 A134682 A134683

Keyword

nonn

Author

T. D. Noe, Nov 06 2007

Status

approved