cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 13 results. Next

A179390 Modulus for Fibonacci-type sequence described by A015134.

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15
Offset: 1

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Author

Will Nicholes, Jul 12 2010

Keywords

Comments

First terms of A015134 are 1, 2, 2 and 4, meaning that there are 1, 2, 2 and 4 Fibonacci-type sequences modulo 1, 2, 3 and 4 respectively. These are:
mod 1: 0
mod 2: 0
mod 2: 0,1,1
mod 3: 0
mod 3: 0,1,1,2,0,2,2,1
mod 4: 0
mod 4: 0,1,1,2,3,1
mod 4: 0,2,2
mod 4: 0,3,3,2,1,3

Crossrefs

A179391 First term in Fibonacci-type sequence described by A015134.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2
Offset: 1

Views

Author

Will Nicholes, Jul 12 2010

Keywords

Comments

First terms of A015134 are 1, 2, 2 and 4, meaning that there are 1, 2, 2 and 4 Fibonacci-type sequences modulo 1, 2, 3 and 4 respectively. These are:
mod 1: 0
mod 2: 0
mod 2: 0,1,1
mod 3: 0
mod 3: 0,1,1,2,0,2,2,1
mod 4: 0
mod 4: 0,1,1,2,3,1
mod 4: 0,2,2
mod 4: 0,3,3,2,1,3

Crossrefs

A179392 Second term in Fibonacci-type sequence described by A015134.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 2, 3, 0, 1, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 4, 6, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 5, 3, 6, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 4, 8, 8, 0, 1, 2, 3, 4, 6, 7, 9, 3, 5, 0, 1, 2, 4, 3, 4, 5, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 8, 3, 8, 13, 6, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 12, 3, 4, 5, 11
Offset: 1

Views

Author

Will Nicholes, Jul 12 2010

Keywords

Comments

First terms of A015134 are 1, 2, 2 and 4, meaning that there are 1, 2, 2 and 4 Fibonacci-type sequences modulo 1, 2, 3 and 4 respectively. These are:
mod 1: 0
mod 2: 0
mod 2: 0,1,1
mod 3: 0
mod 3: 0,1,1,2,0,2,2,1
mod 4: 0
mod 4: 0,1,1,2,3,1
mod 4: 0,2,2
mod 4: 0,3,3,2,1,3

Crossrefs

A179393 Period of the Fibonacci-type sequence described by A015134.

Original entry on oeis.org

1, 1, 3, 1, 8, 1, 6, 3, 6, 1, 20, 4, 1, 24, 8, 3, 1, 16, 16, 16, 1, 12, 6, 12, 3, 6, 12, 12, 1, 24, 24, 8, 24, 1, 60, 20, 3, 12, 4, 1, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 5, 10, 5, 1, 24, 24, 6, 8, 3, 24, 6, 24, 24, 1, 28, 28, 28, 28, 28, 28, 1, 48, 16, 48, 16, 48, 16, 3, 1, 40, 40, 20
Offset: 1

Views

Author

Will Nicholes, Jul 12 2010

Keywords

Comments

First terms of A015134 are 1, 2, 2 and 4, meaning that there are 1, 2, 2 and 4 Fibonacci-type sequences modulo 1, 2, 3 and 4 respectively. These are:
mod 1: 0
mod 2: 0
mod 2: 0,1,1
mod 3: 0
mod 3: 0,1,1,2,0,2,2,1
mod 4: 0
mod 4: 0,1,1,2,3,1
mod 4: 0,2,2
mod 4: 0,3,3,2,1,3
The first sequence for each modulus is the period-1 sequence of 0,0,0... This has the helpful side effect of causing 1 to act as a delimiter between modulus entries: the first 1 indicates the start of modulo-1 sequences, the second 1 indicates the start of modulo-2 sequences, etc.
For each group of sequences (the group start indicated by a 1), the sum of the periods in that group equal the square of the modulus. 1 = 1, (1+3) = 4, (1+8) = 9, (1+6+3+6) = 16, etc.

Crossrefs

A365519 Moebius inversion of A015134.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 4, 3, 2, 13, 4, 6, 3, 8, 8, 8, 9, 21, 8, 24, 13, 11, 16, 10, 6, 9, 12, 62, 8, 33, 16, 24, 24, 24, 36, 18, 63, 24, 32, 42, 24, 21, 48, 24, 33, 69, 64, 21, 10, 32, 24, 26, 27, 144, 48, 40, 62, 61, 32, 62, 99, 72, 32, 48, 24, 33, 96, 88, 24
Offset: 1

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Author

Jay Anderson, Sep 07 2023

Keywords

Examples

			For n=4, the a(4) = 2 cycles counted are
  {0 1 1 2 3 1} repeating
  {0 3 3 2 1 3} repeating
There are 2 other cycles {0} and {0 2 2} mod 4 which are not counted because they are multiples of cycles counted at earlier n (n=1 for {0} and n=2 for {0 1 1}*2 = {0 2 2}).
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Sum[MoebiusMu[d]Part[ResourceFunction["OEISSequence"]["A015134"],n/d],{d,Divisors[n]}]; Array[a,70] (* Stefano Spezia, Sep 27 2023 *)

Formula

a(n) = Sum_{d|n} moebius(d)*A015134(n/d).
A015134(n) = Sum_{d|n} a(d).

A106285 Number of orbits of the 3-step recursion mod n.

Original entry on oeis.org

1, 4, 3, 12, 5, 12, 9, 44, 21, 20, 25, 36, 15, 66, 15, 172, 53, 84, 21, 60, 27, 144, 23, 132, 105, 116, 183, 482, 177, 60, 91, 684, 75, 420, 45, 252, 109, 162, 45, 220, 125, 198, 265, 520, 105, 92, 2259, 516, 359, 420, 159, 884, 2867, 732, 125, 3714, 63, 1408, 59, 180
Offset: 1

Views

Author

T. D. Noe, May 02 2005

Keywords

Comments

Consider the 3-step recursion x(k)=x(k-1)+x(k-2)+x(k-3) mod n. For any of the n^3 initial conditions x(1), x(2) and x(3) in Zn, the recursion has a finite period. Each of these n^3 vectors belongs to exactly one orbit. In general, there are only a few different orbit lengths (A106288) for each n. For instance, the orbits mod 8 have lengths of 1, 2, 4, 8, 16. Interestingly, for n=2^k and n=3^k, the number of orbits appear to be A039301 and A054879, respectively.

Examples

			Orbits for n=2: {(0,0,0)}, {(1,1,1)}, {(0,1,0), (1,0,1)} and {(0,0,1), (0,1,1), (1,1,0), (1,0,0)}
		

Crossrefs

Cf. A015134 (orbits of Fibonacci sequences), A106286 (orbits of 4-step sequences), A106287 (orbits of 5-step sequences), A106288 (number of different orbit lengths), A106307 (n producing a simple orbit structure).

A106286 Number of orbits of the 4-step recursion mod n.

Original entry on oeis.org

1, 4, 6, 28, 3, 24, 10, 220, 91, 12, 130, 240, 343, 40, 168, 1756, 19, 364, 22, 132, 81, 2068, 26, 1968, 253, 1372, 2336, 448, 2557, 672, 16, 14044, 1143, 76, 108, 4612, 1411, 88, 3084, 1860, 11815, 324, 22, 32092, 13213, 104, 50, 15792, 2467, 4012, 168, 17812
Offset: 1

Views

Author

T. D. Noe, May 02 2005

Keywords

Comments

Consider the 4-step recursion x(k)=x(k-1)+x(k-2)+x(k-3)+x(k-4) mod n. For any of the n^4 initial conditions x(1), x(2), x(3) and x(4) in Zn, the recursion has a finite period. Each of these n^4 vectors belongs to exactly one orbit. In general, there are only a few different orbit lengths (A106289) for each n. For instance, the 220 orbits mod 8 have lengths of 1, 5, 10 and 20.

Crossrefs

Cf. A015134 (orbits of Fibonacci sequences), A106285 (orbits of 3-step sequences), A106287 (orbits of 5-step sequences), A106289 (number of different orbit lengths), A106308 (n producing a simple orbit structure).

A106287 Number of orbits of the 5-step recursion mod n.

Original entry on oeis.org

1, 8, 5, 96, 5, 56, 7, 1468, 203, 40, 11, 1312, 13, 56, 25, 23400, 17, 6392, 193, 480, 35, 88, 555, 37180, 2505, 104, 15539, 672, 293, 280, 151, 374292, 55, 136, 35, 199744, 37, 6128, 65, 7340, 41, 392, 1899, 1056, 1015, 6648, 313775, 627280, 14413, 20040, 85
Offset: 1

Views

Author

T. D. Noe, May 02 2005

Keywords

Comments

Consider the 5-step recursion x(k)=x(k-1)+x(k-2)+x(k-3)+x(k-4)+x(k-5) mod n. For any of the n^5 initial conditions x(1), x(2), x(3), x(4) and x(5) in Zn, the recursion has a finite period. Each of these n^5 vectors belongs to exactly one orbit. In general, there are only a few different orbit lengths (A106290). For instance, the 1468 orbits mod 8 have lengths of 1, 2, 3, 6, 12 and 24.

Crossrefs

Cf. A015134 (orbits of Fibonacci sequences), A106285 (orbits of 3-step sequences), A106286 (orbits of 4-step sequences), A106290 (number of different orbit lengths), A106309 (n producing a simple orbit structure).

A088551 Fibonacci winding number: the number of 'mod n' operations in one cycle of the Fibonacci sequence modulo n.

Original entry on oeis.org

1, 3, 2, 8, 11, 7, 4, 11, 28, 3, 9, 12, 23, 19, 9, 16, 11, 7, 28, 5, 12, 23, 9, 48, 40, 35, 19, 4, 59, 12, 19, 15, 16, 39, 9, 36, 6, 27, 28, 19, 19, 43, 11, 59, 23, 15, 9, 55, 148, 35, 38, 52, 35, 6, 21, 31, 16, 26, 57, 28, 12, 21, 43, 68, 51, 67, 14, 19, 119, 32, 7, 72, 112, 99, 5, 33
Offset: 2

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Author

R C Johnson (bob.johnson(AT)dur.ac.uk), Nov 19 2003

Keywords

Comments

If pi(n) is the n-th Pisano number (A001175) then a(n) is usually about pi(n)/2 - and in any case a(n) > pi(n)/4.

Examples

			a(8)=4 because one cycle of the Fibonacci numbers modulo 8 is 0, 1, 1, 2, 3, 5; 0, 5, 5; 2, 7; 1; - including 4 'mod 8' operations, each marked with a semi-colon.
		

Crossrefs

Programs

  • Mathematica
    (* pp = Pisano period = A001175 *) pp[1] = 1;
    pp[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0 && Mod[Fibonacci[k + 1], n] == 1, Return[k]]];
    a[n_] := Sum[Mod[Fibonacci[k], n], {k, 1, pp[n]}]/n;
    Table[a[n], {n, 2, 77}] (* Jean-François Alcover, Sep 05 2017 *)

Formula

n*a(n) = sum{k=1..A001175(n)} fibonacci(k) mod n. [Mircea Merca, Jan 03 2011]

Extensions

More terms from T. D. Noe
Edited by Ray Chandler, Oct 26 2006

A106306 Primes that yield a simple orbit structure in 2-step recursions.

Original entry on oeis.org

2, 3, 7, 13, 17, 23, 37, 41, 43, 47, 53, 61, 67, 73, 83, 89, 97, 103, 107, 109, 113, 127, 137, 149, 157, 163, 167, 173, 193, 197, 223, 227, 233, 241, 257, 263, 269, 277, 281, 283, 293, 307, 313, 317, 337, 347, 353, 367, 373, 383, 389, 397, 401, 409, 421, 433
Offset: 1

Views

Author

T. D. Noe, May 02 2005

Keywords

Comments

Consider the 2-step recursion x(k)=x(k-1)+x(k-2) mod n. For any of the n^2 initial conditions x(1) and x(2) in Zn, the recursion has a finite period. When n is a prime in this sequence, all of the orbits, except the one containing (0,0), have the same length.
Except for 5, this appears to be the complement of A053032, odd primes p with one 0 in Fibonacci numbers mod p. - T. D. Noe, May 03 2005
A prime p is in this sequence if either (1) the polynomial x^2-x-1 mod p has no zeros for x in [0,p-1] (see A086937) or (2) the polynomial has zeros, but none is a root of unity mod p. The first few primes in the second category are 41, 61, 89 and 109. - T. D. Noe, May 12 2005

Crossrefs

Cf. A015134 (orbits of 2-step sequences).
Showing 1-10 of 13 results. Next