A008960 -id:A008960 - cp's OEIS frontend

A070503 a(n) = n^3 mod 41.

0, 1, 8, 27, 23, 2, 11, 15, 20, 32, 16, 19, 6, 24, 38, 13, 37, 34, 10, 12, 5, 36, 29, 31, 7, 4, 28, 3, 17, 35, 22, 25, 9, 21, 26, 30, 39, 18, 14, 33, 40, 0, 1, 8, 27, 23, 2, 11, 15, 20, 32, 16, 19, 6, 24, 38, 13, 37, 34, 10, 12, 5, 36, 29, 31, 7, 4, 28, 3, 17, 35, 22, 25, 9, 21, 26

Offset: 0

N. J. A. Sloane, May 12 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Cf. A008960.

[Modexp(n, 3, 41): n in [0..80]]; // Vincenzo Librandi, Mar 31 2016

Table[Mod[n^3, 41], {n, 0, 100}] (* Vincenzo Librandi, Jun 19 2014 *)
PowerMod[Range[0,100],3,41] (* Harvey P. Dale, Nov 11 2022 *)

a(n)=n^3%41 \\ Charles R Greathouse IV, Apr 06 2016

[power_mod(n,3,41) for n in range(0, 76)] # Zerinvary Lajos, Oct 30 2009

a(n) = a(n-41). - G. C. Greubel, Mar 31 2016

A070597 Duplicate of A070474.

Original entry on oeis.org

0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3, 4, 5, 0, 7, 8, 9, 4, 11, 0, 1, 8, 3

Offset: 0

N. J. A. Sloane, May 13 2002

dead

a(n)=A070474(n) [Proof: n^5-n^3 == 0 (mod 12) is shown explicitly for n=0 to 11, then the induction n->n+12 for the 5th-order polynomial followed by binomial expansion of (n+12)^k concludes that the zero (mod 12) is periodically extended to the other integers.] - R. J. Mathar, Jul 23 2009

Cf. A008960, A070474.

Table[Mod[n^5,12],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 23 2011 *)

[power_mod(n,7,12) for n in range(0, 100)] # Zerinvary Lajos, Oct 28 2009

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

A070697 Duplicate of A070477.

Original entry on oeis.org

0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14, 0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14, 0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14, 0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14, 0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14, 0, 1, 8, 12, 4, 5, 6, 13, 2, 9, 10, 11, 3, 7, 14

Offset: 0

N. J. A. Sloane, May 13 2002

dead

a(n)=A070477(n). [Proof: (n^7-n^3) == 0 (mod 15) is shown for n=0 to 14 explicitly. Then binomial expansion of (n+15)^7 and (n+15)^3 extend this by induction n->n+15 to all other indices.] - R. J. Mathar, Oct 30 2009

Cf. A008960.

[n^3 mod 15: n in [0..80]]; // Vincenzo Librandi, Jun 19 2014

Table[Mod[n^3, 15], {n, 0, 100}] (* Vincenzo Librandi, Jun 19 2014 *)

a(n)=n^7%15 \\ Charles R Greathouse IV, Apr 06 2016

A130875 Absolute difference of final digits of two consecutive cubes.

Original entry on oeis.org

1, 7, 1, 3, 1, 1, 3, 1, 7, 9, 1, 7, 1, 3, 1, 1, 3, 1, 7, 9, 1, 7, 1, 3, 1, 1, 3, 1, 7, 9, 1, 7, 1, 3, 1, 1, 3, 1, 7, 9, 1, 7, 1, 3, 1, 1, 3, 1, 7, 9, 1, 7, 1, 3, 1, 1, 3, 1, 7, 9, 1, 7, 1, 3, 1, 1, 3, 1, 7, 9, 1, 7, 1, 3, 1, 1, 3, 1, 7, 9, 1, 7, 1, 3, 1, 1, 3, 1, 7, 9, 1, 7, 1, 3, 1, 1, 3, 1, 7, 9, 1, 7, 1, 3, 1

Offset: 1

Giovanni Teofilatto, Jul 25 2007

Periodic with period 10.

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

Cf. A008960.

LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 1},{1, 7, 1, 3, 1, 1, 3, 1, 7, 9},105] (* Ray Chandler, Aug 26 2015 *)
Abs[Differences[Mod[#,10]]]&/@Partition[Range[0,120]^3,2,1]//Flatten (* or *) PadRight[{},120,{1,7,1,3,1,1,3,1,7,9}] (* Harvey P. Dale, Oct 27 2019 *)

Vec(-x*(9*x^9+7*x^8+x^7+3*x^6+x^5+x^4+3*x^3+x^2+7*x+1)/((x-1)*(x+1)*(x^4-x^3+x^2-x+1)*(x^4+x^3+x^2+x+1)) + O(x^100)) \\ Colin Barker, Dec 04 2014

a(n) = a(n-10). - Colin Barker, Dec 04 2014

G.f.: -x*(9*x^9+7*x^8+x^7+3*x^6+x^5+x^4+3*x^3+x^2+7*x+1) / ((x-1)*(x+1)*(x^4-x^3+x^2-x+1)*(x^4+x^3+x^2+x+1)). - Colin Barker, Dec 04 2014

cp's OEIS Frontend

A070503 a(n) = n^3 mod 41.

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A070597 Duplicate of A070474.

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A282779 Period of cubes mod n.

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A070697 Duplicate of A070477.

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A130875 Absolute difference of final digits of two consecutive cubes.

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Formula