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A232656 The number of pairs of numbers below n that, when generating a Fibonacci-like sequence modulo n, contain zeros. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
a(n) = n^2 iff n is in A064414, a(n) is not equal to n^2 iff n is in A230457.
a(n) + A232357(n) = n^2.
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.
FORMULA
Conjecture: a(n) = Sum_{d|n} phi(d)*A001177(d), where phi = Euler's totient function (A000010). - Logan J. Kleinwaks, Oct 28 2017
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
EXAMPLE
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.
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}]
CROSSREFS
Sequence in context: A313343 A261849 A246336 * A364699 A010460 A313344
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified June 17 15:48 EDT 2024. Contains 373459 sequences. (Running on oeis4.)