In OEIS terms, is this sequence interesting? I guess "nonn" and "easy". Is it "nice"? Name(?): Smallest term in Zeckendorf representation of n http://en.wikipedia.org/wiki/Zeckendorf's_theorem 1, 2, 3, 1, 5, 1, 2, 8, 1, 2, 3, 1, 13, 1, 2, 3, 1, 5, 1, 2... Example: The Zeckendorf representation of 7 = 5 + 2, so a(7) = 2. Formulas: a(n) = n if n is a Fibonacci number, else a( n - (largest Fibonacci number < n) ) a(n) = the value of the (exactly one) digit that turns on between the Fibonacci-base representations of n-1 and n. E.g. from 6 (1001) to 7 (1010), the two's digit turns on. a(n) = top element of the column of the Wythoff array that contains n http://mathworld.wolfram.com/WythoffArray.html a(n) = Fibonacci( A035614(n) + 1 ) Related to http://www.research.att.com/~njas/sequences/A035614 : 0, 1, 2, 0, 3, 0, 1, 4, 0, 1, 2, 0, 5, 0, 1, 2, 0, 3, 0, 1... "which column of Wythoff array (starting column count at 0) contains n." a(n) is also (Google translation:) "the numbers of matches to withdraw to win up to 55 matches (which is the first number of following Fibonacci greater than or equal to 50). " from "La stratgie des allumettes" ("The 'Matches' Strategy") http://mathematiques.ac-bordeaux.fr/elv/clubs/mej/mej2005/allumettes.pdf http://tinyurl.com/5sfgw8 (Google French pdf-->html-->English) It's also called a "Fibonacci fractal" on this page, in a section apparently about a stone-taking game: http://gifted.msquare.or.kr/lib/2003/2003bcamp.pdf http://tinyurl.com/5cy28c (Google Korean pdf-->html-->English) The frequencies of occurrences of the values in both these sequences are related by the golden ratio. You can see that the part of a() shown has eight 1's, five 2's, three 3's, two 5's, one 8, and one 13. I want to clock some processes at roughly golden-ratio-related frequencies, and I'm looking for the simplest way to generate A035614, basically. I was thinking boolean operations on long words, but maybe just generating Fibonacci numbers is easiest. A035614 is more useful but a(n) seems prettier and more mnemonic. I started with this picture: 1111111111111111111111111111111111 34 x 1 = 34 1 11 1 11 11 1 11 1 11 11 1 11 11 21 x 1 = 21 2 2 2 2 2 2 2 2 2 2 2 2 2 13 x 2 = 26 3 3 3 3 3 3 3 3 8 x 3 = 24 5 5 5 5 5 5 x 5 = 25 8 8 8 3 x 8 = 24 13 13 2 x 13 = 26 21 1 x 21 = 21 34 1 x 34 = 34 Maybe this way is better: 1 11 1 11 11 1 11 1 11 11 1 11 11 21 x 1 = 21 2 2 2 2 2 2 2 2 2 2 2 2 2 13 x 2 = 26 3 3 3 3 3 3 3 3 8 x 3 = 24 5 5 5 5 5 5 x 5 = 25 8 8 8 3 x 8 = 24 13 13 2 x 13 = 26 21 1 x 21 = 21 --Steve reminding myself to see also http://www.research.att.com/~njas/sequences/A107017 "Second largest term in Zeckendorf representation of n, a(n)=0 if n itself is a Fibonacci number." http://www.research.att.com/~njas/sequences/A014417 base fibo numbers http://www.research.att.com/~njas/sequences/A006519 highest power of two dividing n