OFFSET
0,3
COMMENTS
This can be viewed as an irregular table: after the initial zero on row 0, start each row n with k = F(n+2)-1 and subtract repeatedly the number of "1-fibits" (number of terms in Zeckendorf expansion of k) from k to get successive terms, until the number that has already been listed (which is always (F(n+1)-1) is encountered, which is not listed second time, but instead, the current row is finished and the next row starts with (F(n+3))-1, with the same process repeated. Here F(n) = the n-th Fibonacci number, A000045(n).
LINKS
Indranil Ghosh, Python program to generate the sequence
FORMULA
EXAMPLE
As an irregular table, the sequence looks like:
0;
1;
2;
4;
7, 5;
12, 9;
20, 17, 14;
33, 29, 27, 24, 22;
54, 50, 47, 45, 42, 40, 37, 35;
...
After zero, each row n is A261091(n) elements long.
PROG
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Antti Karttunen, Aug 09 2015
STATUS
approved