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%I #49 Dec 08 2024 17:16:06
%S 1,2,3,6,12,18,24,30,36,42,48,54,60,72,84,90,96,108,114,120,126,132,
%T 138,144,150,156,162,168,180,192,198,204,210,216,222,228,234,240,252,
%U 264,270,276,288,294,300,306,312,324,330,336,342,348,354,360,372,378,384,390,396,402,408,414,420,432,444,450,456,462,468,480,492,504,510,516,522,528,540,546,552,558,564,570,576,588,594,600,612,624,630,636
%N Numbers k such that Fibonacci(k) is a totient.
%C Respectively, corresponding Fibonacci numbers are 1, 1, 2, 8, 144, 2584, 46368, 832040, 14930352, 267914296, 4807526976, 86267571272, 1548008755920, 498454011879264, 160500643816367088, 2880067194370816120, ...
%C Note that sequence does not contain all the positive multiples of 6, e.g., 66 and 102. See A335976 for a related sequence.
%C Conjecture: Sequence is infinite. - _Altug Alkan_, Jul 05 2020
%C All terms > 2 are multiples of 3, because Fibonacci(k) is odd unless k is a multiple of 3. Are all terms > 3 multiples of 6? If a term k is not a multiple of 6, then since Fibonacci(k) is not divisible by 4, Fibonacci(k)+1 must be in A114871. - _Robert Israel_, Aug 02 2020
%C Unless there is an odd term > 3, this sequence as a set is {1, 2, 3} U 6*(Z^+ \ A335976). - _Max Alekseyev_, Dec 08 2024
%H Max Alekseyev, <a href="/A280681/b280681.txt">Table of n, a(n) for n = 1..665</a>
%e 12 is in the sequence because Fibonacci(12) = 144 is in A000010.
%p select(k -> numtheory:-invphi(combinat:-fibonacci(k))<>[], [1,2,seq(i,i=3..100,3)]); # _Robert Israel_, Aug 02 2020
%o (PARI) isok(k) = istotient(fibonacci(k)); \\ _Altug Alkan_, Jul 05 2020
%Y Cf. A000010, A000045, A114871, A280592, A335976.
%K nonn
%O 1,2
%A _Altug Alkan_, Jan 07 2017
%E a(28)-a(49) from _Jinyuan Wang_, Jul 08 2020
%E Terms a(50) onward from _Max Alekseyev_, Dec 08 2024