A350969 Let phi^(k) denote the k-th iterate of phi (A000010); a(n) is smallest positive k such that phi^(k)(Fibonacci(n)) = 1.
1, 1, 1, 2, 3, 3, 4, 4, 5, 6, 7, 6, 8, 8, 8, 9, 9, 10, 11, 12, 11, 13, 13, 13, 15, 16, 15, 16, 17, 17, 19, 18, 19, 19, 20, 20, 21, 22, 22, 23, 26, 23, 25, 25, 26, 27, 28, 27, 28, 30, 28, 31, 32, 30, 34, 32, 33, 34, 35, 34, 38, 37, 36, 37, 39, 38, 40, 39, 40, 40
Offset: 1
Keywords
Examples
Iterating phi, F_7 = 13 -> 12 -> 4 -> 2 -> 1 takes 4 steps to reach 1, so a(7) = 4.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..450
- Douglas Lind (Proposer), Problem B-51, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 2, No. 3 (1964), p. 232; Solution, ibid., Vol. 60, No. 1 (2022), pp. 83-84.
- S. Sivasankaranarayana Pillai, On a function connected with phi(n), Bull. Amer. Math. Soc., Vol. 35, No. 6 (1929), pp. 837-841.
Programs
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Maple
a:= proc(n) uses numtheory; local f, k; f:= phi((<<0|1>, <1|1>>^n)[1, 2]); for k while f>1 do f:= phi(f) od; k end: seq(a(n), n=1..70); # Alois P. Heinz, Mar 03 2022 -
Mathematica
a[1] = a[2] = 1; a[n_] := Length@NestWhileList[EulerPhi, Fibonacci[n], # > 1 &] - 1; Array[a, 100] (* Amiram Eldar, Mar 03 2022 *)
Comments