A232656 The number of pairs of numbers below n that, when generating a Fibonacci-like sequence modulo n, contain zeros.
1, 4, 9, 16, 21, 36, 49, 40, 81, 84, 101, 96, 85, 196, 189, 136, 145, 180, 325, 336, 153, 404, 529, 216, 521, 340, 729, 496, 393, 756, 901, 520, 509, 292, 1029, 384, 685, 652, 765, 840, 801, 612, 1849, 1016, 1701, 1060, 737, 504, 2401, 2084, 1305, 1360, 1405, 1476, 521, 1096, 1629, 1572
Offset: 1
Keywords
Examples
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. Any Fibonacci like sequence contains elements divisible by 2, 3, or 4. Thus, a(2) = 4, a(3) = 9, a(4) = 16.
Links
- B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5.
Programs
-
Mathematica
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}]
Formula
Conjecture: a(n) = Sum_{d|n} phi(d)*A001177(d), where phi = Euler's totient function (A000010). - Logan J. Kleinwaks, Oct 28 2017
a(n) = 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
Comments