A233248 -id:A233248 - cp's OEIS frontend

A233246 Sum of squares of cycle lengths for different cycles in Fibonacci-like sequences modulo n.

1, 10, 65, 82, 417, 650, 769, 658, 1793, 4170, 1151, 3026, 4705, 7690, 7137, 5266, 10369, 7562, 6319, 19218, 6977, 11510, 25345, 12818, 52417, 47050, 48449, 35410, 11565, 71370, 28351, 42130, 39615, 41482, 81057, 30674, 103969, 25282, 80033

Offset: 1

Brandon Avila and Tanya Khovanova, Dec 06 2013

nonn

Here Fibonacci-like means a sequence following the Fibonacci recursion: b(n)=b(n-1)+b(n-2). These sequences modulo n cycle. The number of different cycles is A015134(n).
This sequence divided by n^2 is the average cycle length per different starting pairs modulo n, see A233248.
If n is in A064414, then a(n)/n^2 is the average distance between two neighboring multiples of n.
If n is in A064414, then a(n)/2n^2 is the average distance to the next zero over all starting pairs of remainders.

			For n=4 there are four possible cycles: A trivial cycle of length 1: 0; two cycles of length 6: 0,1,1,2,3,1; and a cycle of length 3: 0,2,2. Hence, a(4)=1+9+36+36=82.

B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614, 2014 and J. Int. Seq. 17 (2014) # 14.8.5

Cf. A233248, A064414.

cl[i_, j_, n_] := (step = 1; first = i; second = j;
  next = Mod[first + second, n];
  While[second != i || next != j, step++; first = second;
   second = next; next = Mod[first + second, n]]; step)
Table[Total[
  Flatten[Table[cl[i, j, n], {i, 0, n - 1}, {j, 0, n - 1}]]], {n, 50}]

A266295 2-free tetranacci sequence beginning 1,3,5,7.

Original entry on oeis.org

1, 3, 5, 7, 1, 1, 7, 1, 5, 7, 5, 9, 13, 17, 11, 25, 33, 43, 7, 27, 55, 33, 61, 11, 5, 55, 33, 13, 53, 77, 11, 77, 109, 137, 167, 245, 329, 439, 295, 327, 695, 439, 439, 475, 1, 677, 199, 169, 523, 49, 235, 61, 217, 281, 397, 239, 567, 371, 787

Offset: 1

Jeremy F. Alm, Dec 28 2015

nonn

For n>4, a(n) = (a(n-1) + a(n-2) + a(n-3) + a(n-4)) / 2^d, where 2^d is the largest power of 2 dividing a(n-1) + a(n-2) + a(n-3) + a(n-4). In other words, sum the previous four terms, then divide by two until the result is odd.

Alm, Herald, Miller, and Sexton, 2-Free Tetranacci Sequences, unpublished.

Jeremy F. Alm, Table of n, a(n) for n = 1..10000

Cf. A000265, A232666, A233248, A233526.

nxt[{a_, b_, c_, d_}] := {b, c, d, (a + b + c + d)/2^IntegerExponent[ a + b + c + d, 2]}; NestList[nxt,{1,3,5,7},60][[All,1]] (* Harvey P. Dale, Nov 09 2020 *)

lista(nn) = {print1(x = 1, ", "); print1(y = 3, ", "); print1(z = 5, ", "); print1(t = 7, ", "); for (n=5, nn, tt = (x+y+z+t); tt /= 2^valuation(tt, 2); print1(tt, ", "); x=y; y=z; z=t; t=tt;);} \\ Michel Marcus, Dec 29 2015

### CREATES A b-FILE ###
def main():
    name = "b266295.txt"
    file = open(name, 'w')
    file.write('1' + ' ' + '1\n')
    file.write('2' + ' ' + '3\n')
    file.write('3' + ' ' + '5\n')
    file.write('4' + ' ' + '7\n')
    a, b, c, d = 1, 3, 5, 7
    for i in range(5,10001):
        x=a+b+c+d
        while x%2==0:
            x /= 2
        a, b, c, d = b, c, d, x
        file.write(str(i) + ' ' + str(int(d)) + '\n')
    file.close()
main()

a(n) = (a(n-1) + a(n-2) + a(n-3) + a(n-4)) / 2^d, where 2^d is the largest power of 2 dividing a(n-1) + a(n-2) + a(n-3) + a(n-4).

a(n) = A000265(a(n-1) + a(n-2) + a(n-3) + a(n-4)). - Michel Marcus, Dec 29 2015

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A233246 Sum of squares of cycle lengths for different cycles in Fibonacci-like sequences modulo n.

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A266295 2-free tetranacci sequence beginning 1,3,5,7.

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