

A350969


Let phi^(k) denote the kth iterate of phi (A000010); a(n) is smallest positive k such that phi^(k)(Fibonacci(n)) = 1.


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
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OFFSET

1,4


COMMENTS

a(n) <= n.
The Fibonacci Quarterly asks what the range of a(n) is. For example, is a(n) ever equal to 14 or 24?


LINKS

Douglas Lind (Proposer), Problem B51, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 2, No. 3 (1964), p. 232; Solution, ibid., Vol. 60, No. 1 (2022), pp. 8384.


FORMULA



EXAMPLE

Iterating phi, F_7 = 13 > 12 > 4 > 2 > 1 takes 4 steps to reach 1, so a(7) = 4.


MAPLE

a:= proc(n) uses numtheory; local f, k;
f:= phi((<<01>, <11>>^n)[1, 2]);
for k while f>1 do f:= phi(f) od; k
end:


MATHEMATICA

a[1] = a[2] = 1; a[n_] := Length@NestWhileList[EulerPhi, Fibonacci[n], # > 1 &]  1; Array[a, 100] (* Amiram Eldar, Mar 03 2022 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



