A167176 -id:A167176 - cp's OEIS frontend

A008960 Final digit of cubes: n^3 mod 10.

0, 1, 8, 7, 4, 5, 6, 3, 2, 9, 0, 1, 8, 7, 4, 5, 6, 3, 2, 9, 0, 1, 8, 7, 4, 5, 6, 3, 2, 9, 0, 1, 8, 7, 4, 5, 6, 3, 2, 9, 0, 1, 8, 7, 4, 5, 6, 3, 2, 9, 0, 1, 8, 7, 4, 5, 6, 3, 2, 9, 0, 1, 8, 7, 4, 5, 6, 3, 2, 9, 0, 1, 8, 7, 4, 5, 6, 3, 2, 9, 0, 1, 8, 7, 4, 5

Offset: 0

N. J. A. Sloane

Decimal expansion of 208284810/1111111111. - Alexander R. Povolotsky, Mar 08 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Index entries for sequences related to final digits of numbers
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,1).

Cf. A167176.

Cf. A010879, A008959, A070514. - Doug Bell, Jun 15 2015

[n^3 mod 10: n in [0..80]]; // Vincenzo Librandi, Mar 26 2013

Table[Mod[n^3,10],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 23 2011 *)
LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 1},{0, 1, 8, 7, 4, 5, 6, 3, 2, 9},81] (* Ray Chandler, Aug 26 2015 *)

a(n)=n^3%10 \\ Charles R Greathouse IV, Mar 08 2013

concat(0, Vec(x*(1+8*x+7*x^2+4*x^3+5*x^4+6*x^5+3*x^6+2*x^7+9*x^8) / ((1-x)*(1+x)*(1-x+x^2-x^3+x^4)*(1+x+x^2+x^3+x^4)) + O(x^100))) \\ Colin Barker, Nov 30 2015

[power_mod(n,3,10 ) for n in range(0, 81)] # Zerinvary Lajos, Oct 29 2009

Periodic with period 10. - Franklin T. Adams-Watters, Mar 13 2006
a(n) = 4.5 -cos(Pi*n/5) +(1/2*(-(5-5^(1/2))^(1/2) +(5+5^(1/2))^(1/2))*2^(1/2))*sin(Pi*n/5) -cos(2*Pi*n/5) +(-1/10*(-(5-5^(1/2))^(1/2)+3*(5+5^(1/2))^(1/2))*2^(1/2))*sin(2*Pi*n/5) -cos(3*Pi*n/5) +(-1/2*((5-5^(1/2))^(1/2) +(5+5^(1/2))^(1/2))*2^(1/2))*sin(3*Pi*n/5) -cos(4*Pi*n/5) +( -1/10*(3*(5-5^(1/2))^(1/2) +(5 +5^(1/2))^(1/2))*2^(1/2))*sin(4*Pi*n/5) -0.5*(-1)^n. - Richard Choulet, Dec 12 2008
a(n) = n^k mod 10; for k > 0 where k mod 4 = 3. - Doug Bell, Jun 15 2015
G.f.: x*(1+8*x+7*x^2+4*x^3+5*x^4+6*x^5+3*x^6+2*x^7+9*x^8) / ((1-x)*(1+x)*(1-x+x^2-x^3+x^4)*(1+x+x^2+x^3+x^4)). - Colin Barker, Nov 30 2015

A070474 a(n) = n^3 mod 12, n^5 mod 12.

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 12 2002

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

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, 1).

Cf. A167176.

[Modexp(n, 3, 12 ): n in [0..100]]; // Vincenzo Librandi, Mar 27 2016

Table[Mod[n^3,12],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 23 2011 *)
PowerMod[Range[0,100],3,12] (* Harvey P. Dale, Oct 29 2014 *)

a(n)=n^3%12 \\ Charles R Greathouse IV, Apr 06 2016

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

From R. J. Mathar, Jul 23 2009: (Start)
a(n) = a(n-12).
G.f.: -x*(1 + 8*x + 3*x^2 + 4*x^3 + 5*x^4 + 7*x^6 + 8*x^7 + 9*x^8 + 4*x^9 + 11*x^10)/ ((x-1) *(1+x+x ^2) *(1+x) *(1-x+x^2) *(1+x^2) *(x^4-x^2+1)). (End)

A224484 Numbers which are the sum of two positive cubes and divisible by 3.

Original entry on oeis.org

9, 54, 72, 126, 189, 243, 351, 432, 468, 513, 576, 756, 855, 945, 1008, 1125, 1332, 1395, 1458, 1512, 1674, 1755, 1944, 2205, 2322, 2331, 2457, 2709, 2745, 2808, 3087, 3402, 3456, 3528, 3591, 3744, 4104, 4221, 4608, 4914, 4941

Offset: 1

Vincenzo Librandi, May 10 2013

If 12*h-27 is a square then some values of 3*h are in this sequence. It is easy to verify that h is of the form 3*m^2-3*m+3, and therefore 9*(m^2-m+1) = (2-m)^3+(m+1)^3.

All entries are multiples of 9. [Proof: the cubes mod 3 are A010872. So the two cubes are either of the form (3i)^3 and (3j)^3 or (3i+1)^3 and (3j+2)^3. The same 3-periodic pattern is seen in the cubes modulo 9, A167176.] - R. J. Mathar, Aug 24 2016

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A224485 (divisible by k=5), A101421 (k=7), A101852 (k=11), A094447 (k=13), A099178 (k=17), A102619 (k=19), A101806 (k=23), A224483 (k=29), A102658 (k=31), A102618 (k=37).

upto[n_] := Block[{t}, Union@ Reap[ Do[If[Mod[t = x^3 + y^3, 3] == 0, Sow@ t], {x, n^(1/3)}, {y, Min[x, (n - x^3)^(1/3)]}] ][[2, 1]]]; upto[5000] (* Giovanni Resta, Jun 12 2020 *)
Module[{nn=20},Select[Union[Total/@Tuples[Range[nn]^3,2]],Mod[#,3]==0 && #Harvey P. Dale, Mar 06 2022 *)

A070473 a(n) = n^3 mod 11.

Original entry on oeis.org

0, 1, 8, 5, 9, 4, 7, 2, 6, 3, 10, 0, 1, 8, 5, 9, 4, 7, 2, 6, 3, 10, 0, 1, 8, 5, 9, 4, 7, 2, 6, 3, 10, 0, 1, 8, 5, 9, 4, 7, 2, 6, 3, 10, 0, 1, 8, 5, 9, 4, 7, 2, 6, 3, 10, 0, 1, 8, 5, 9, 4, 7, 2, 6, 3, 10, 0, 1, 8, 5, 9, 4, 7, 2, 6, 3, 10, 0, 1, 8, 5, 9, 4, 7, 2, 6, 3, 10, 0, 1, 8, 5, 9, 4, 7, 2, 6, 3, 10

Offset: 0

N. J. A. Sloane, May 12 2002

Vincenzo Librandi, 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, 1).

Cf. A008960, A070474, A167176.

[n^3 mod 11: n in [0..80] ]; // Vincenzo Librandi, Jun 01 2011

Table[Mod[n^3,11],{n,0,200}] (* Vladimir Joseph Stephan Orlovsky, Apr 23 2011 *)
PowerMod[Range[0,120],3,11] (* or *) PadRight[{},120,{0,1,8,5,9,4,7,2,6,3,10}] (* Harvey P. Dale, May 04 2019 *)

a(n)=n^3%11 \\ Charles R Greathouse IV, Apr 06 2016

[power_mod(n,3,11 ) for n in range(0, 99)] # Zerinvary Lajos, Oct 29 2009

a(n) = a(n-11). - G. C. Greubel, Mar 26 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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A008960 Final digit of cubes: n^3 mod 10.

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A070474 a(n) = n^3 mod 12, n^5 mod 12.

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A224484 Numbers which are the sum of two positive cubes and divisible by 3.

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A070473 a(n) = n^3 mod 11.

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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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