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 A261082 a(n) = number of steps required to reach 0 from F(n+2) by repeatedly subtracting from a natural number the number of ones in its Zeckendorf representation. Here F(n) = the n-th Fibonacci number, F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2, ... 4
 1, 2, 3, 4, 6, 8, 11, 16, 24, 35, 52, 77, 114, 170, 255, 385, 584, 889, 1358, 2081, 3199, 4932, 7625, 11818, 18357, 28568, 44530, 69504, 108607, 169869, 265899, 416507, 652845, 1023946, 1607064, 2524042, 3967246, 6240680, 9825202, 15481988, 24416684, 38539840, 60880090 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 LINKS FORMULA a(n) = A219642(A000045(n+2)). a(0) = 1; for n >= 1, a(n) = A261091(n) + a(n-1). Other identities. For all n >= 0: a(n) = A261081(n)+1. PROG (Scheme, two alternatives, the other one using memoizing definec-macro) (define (A261082 n) (A219642 (A000045 (+ 2 n)))) (definec (A261082 n) (if (zero? n) 1 (+ (A261091 n) (A261082 (- n 1))))) CROSSREFS One more than A261081. First differences give A261091. Cf. A000045, A219642, A261102. Sequence in context: A057048 A281094 A054782 * A271487 A211397 A173542 Adjacent sequences:  A261079 A261080 A261081 * A261083 A261084 A261085 KEYWORD nonn AUTHOR Antti Karttunen, Aug 08 2015 STATUS approved

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Last modified February 24 00:51 EST 2020. Contains 332195 sequences. (Running on oeis4.)