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%I #32 May 09 2021 09:53:16
%S 1,4,9,16,21,36,49,40,81,84,101,96,85,196,189,136,145,180,325,336,153,
%T 404,529,216,521,340,729,496,393,756,901,520,509,292,1029,384,685,652,
%U 765,840,801,612,1849,1016,1701,1060,737,504,2401,2084,1305,1360,1405,1476,521,1096,1629,1572
%N The number of pairs of numbers below n that, when generating a Fibonacci-like sequence modulo n, contain zeros.
%C a(n) = n^2 iff n is in A064414, a(n) is not equal to n^2 iff n is in A230457.
%C a(n) + A232357(n) = n^2.
%H B. Avila and T. Khovanova, <a href="http://arxiv.org/abs/1403.4614">Free Fibonacci Sequences</a>, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Avila/avila4.html">J. Int. Seq. 17 (2014) # 14.8.5</a>.
%F Conjecture: a(n) = Sum_{d|n} phi(d)*A001177(d), where phi = Euler's totient function (A000010). - _Logan J. Kleinwaks_, Oct 28 2017
%F Sum_{d|n} phi(d)*A001177(d) = Sum_{k=1..n} A001177(n/gcd(n,k)) = Sum_{k=1..n} A001177(gcd(n,k))phi(gcd(n,k)/phi(n/gcd(n,k)). - _Richard L. Ollerton_, May 09 2021
%e The sequence 2,1,3,4,2,1 is the sequence of Lucas numbers modulo 5. Lucas numbers are never divisible by 5. The 4 pairs (2,1), (1,3), (3,4), (4,2) are the only pairs that can generate a sequence modulo 5 that doesn't contain zeros. Thus, a(5) = 21, as 21 other pairs generate sequences that do contain zeros.
%e Any Fibonacci like sequence contains elements divisible by 2, 3, or 4. Thus, a(2) = 4, a(3) = 9, a(4) = 16.
%t fibLike[list_] := Append[list, list[[-1]] + list[[-2]]]; Table[k^2 -Count[Flatten[Table[Count[Nest[fibLike, {n, m}, k^2]/k, _Integer], {n, k - 1}, {m, k - 1}]], 0], {k, 70}]
%Y Cf. A064414, A230457, A232357.
%K nonn
%O 1,2
%A _Brandon Avila_ and _Tanya Khovanova_, Nov 27 2013
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