A109718 -id:A109718 - cp's OEIS frontend

A204689 a(n) = n^n (mod 4).

1, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0, 3, 0, 1, 0

Offset: 0

José María Grau Ribas, Jan 18 2012

Apart from a(0), the same as A109718. [Joerg Arndt, Sep 17 2013]

Periodic for n>0 with period 4 = A174824(4): repeat [1, 0, 3, 0].

Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Cf. A000007, A010684, A109718, A174824.

[1] cat &cat [[1, 0, 3, 0]^^30]; // Wesley Ivan Hurt, Jun 15 2016

A204689:=n->n^n mod 4: seq(A204689(n), n=0..150); # Wesley Ivan Hurt, Jun 15 2016

Mathematica
```
Table[PowerMod[n, n, 4], {n,0,140}]
```

From Bruno Berselli, Jan 18 2012: (Start)
G.f.: (1+x+3x^3-x^4)/(1-x^4).
a(n) = (1-(-1)^n)*(2+i^(n+1))/2 with i=sqrt(-1), a(0)=1.
a(n) = A109718(n) for n>0. (End)
a(2k) = A000007(k), a(2k+1) = A010684(k). - Wesley Ivan Hurt, Jun 15 2016

A282779 Period of cubes mod n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 16, 17, 6, 19, 20, 21, 22, 23, 24, 25, 26, 9, 28, 29, 30, 31, 32, 33, 34, 35, 12, 37, 38, 39, 40, 41, 42, 43, 44, 15, 46, 47, 48, 49, 50, 51, 52, 53, 18, 55, 56, 57, 58, 59, 60, 61, 62, 21, 64, 65, 66, 67, 68, 69, 70, 71, 24, 73, 74, 75, 76, 77, 78, 79, 80, 27

Offset: 1

Ilya Gutkovskiy, Feb 21 2017

The length of the period of A000035 (n=2), A010872 (n=3), A109718 (n=4), A070471 (n=5), A010875 (n=6), A070472 (n=7), A109753 (n=8), A167176 (n=9), A008960 (n = 10), etc. (see also comment in A000578 from R. J. Mathar).
Conjecture: let a_p(n) be the length of the period of the sequence k^p mod n where p is a prime, then a_p(n) = n/p if n == 0 (mod p^2) else a_p(n) = n.
For example: sequence k^7 mod 98 gives 1, 30, 31, 18, 19, 48, 49, 50, 79, 80, 67, 68, 97, 0, 1, 30, 31, 18, 19, 48, 49, 50, 79, 80, 67, 68, 97, 0, ... (period 14), 7 is a prime, 98 == 0 (mod 7^2) and 98/7 = 14.

			a(9) = 3 because reading 1, 8, 27, 64, 125, 216, 343, 512, ... modulo 9 gives 1, 8, 0, 1, 8, 0, 1, 8, 0, ... with period length 3.

Ray Chandler, Table of n, a(n) for n = 1..10000
Ilya Gutkovskiy, Extended graphical example
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, -1).

Cf. A000035, A000578, A008960, A010872, A010875, A046530, A070471, A070472, A109718, A109753, A167176, A186646.

a[1] = 1; a[n_] := For[k = 1, True, k++, If[Mod[k^3, n] == 0 && Mod[(k + 1)^3 , n] == 1, Return[k]]]; Table[a[n], {n, 1, 81}]

Apparently: a(n) = 2*a(n-9) - a(n-18).

Empirical g.f.: x*(1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5 + 7*x^6 + 8*x^7 + 3*x^8 + 8*x^9 + 7*x^10 + 6*x^11 + 5*x^12 + 4*x^13 + 3*x^14 + 2*x^15 + x^16) / ((1 - x)^2*(1 + x + x^2)^2*(1 + x^3 + x^6)^2). - Colin Barker, Feb 21 2017

A337596 Largest m such that k^n (mod m) is always either 0, +1, or -1.

Original entry on oeis.org

3, 5, 9, 16, 11, 13, 4, 32, 27, 25, 23, 16, 4, 29, 31, 64, 4, 37, 4, 41, 49, 23, 47, 32, 11, 53, 81, 29, 59, 61, 4, 128, 67, 8, 71, 73, 4, 8, 79, 41, 83, 49, 4, 89, 31, 47, 4, 97, 4, 125, 103, 53, 107, 109, 121, 113, 9, 59, 4, 61, 4, 8, 127, 256, 131, 67, 4, 137

Offset: 1

Elliott Line, Sep 02 2020

nonn

For a given n, for all k, k^n mod a(n) will always be either 0, 1 or a(n)-1. This will not be true for numbers larger than a(n).

It appears that a(m) = 4 for m in A045979. - Michel Marcus, Sep 04 2020

			For n = 5 all fifth powers of natural numbers: 1,32,243,1024, etc. are either a multiple of 11, or 1 greater or 1 less than a multiple of 11. There is no greater number than 11 for which all fifth powers are at most 1 different from a multiple. So a(5) = 11.

Cf. A010872, A070430, A167176, A000035, A070596, A070636, A109718.

Cf. residues: A096008 (for n=2), A096087 (for n=3).

More terms from Michel Marcus, Sep 04 2020

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A204689 a(n) = n^n (mod 4).

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A282779 Period of cubes mod n.

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A337596 Largest m such that k^n (mod m) is always either 0, +1, or -1.

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