A220468 Number of cyclotomic cosets of n mod 10.

1, 10, 2, 4, 3, 2, 5, 4, 2, 6, 1, 10, 2, 4, 3, 2, 5, 4, 2, 6, 1, 10, 2, 4, 3, 2, 5, 4, 2, 6, 1, 10, 2, 4, 3, 2, 5, 4, 2, 6, 1, 10, 2, 4, 3, 2, 5, 4, 2, 6, 1, 10, 2, 4, 3, 2, 5, 4, 2, 6, 1, 10, 2, 4, 3, 2, 5, 4, 2, 6, 1, 10, 2, 4, 3, 2, 5, 4, 2, 6, 1, 10, 2, 4, 3, 2, 5, 4, 2, 6

Offset: 0

Author

V. Raman, Jan 03 2013

Keywords

• nonn
• base
• less
• easy

Comments

In other words, number of different cycles mod 10 obtained by repeatedly multiplying by n, with different starting elements. There are four different cycles for n = 3: {1, 3, 9, 7}, {2, 6, 8, 4}, {5}, {0}. By starting with a random number below 10, the numbers obtained by repeatedly multiplying by 3 and then taking modulo 10 repeat through one of these four different cycles.

There are two different cycles for n = 2: {2, 4, 8, 6}, {0}. Note that this does not cover all the positive integers less than 10. The elements 1, 3, 7, 9 belong to the cycle {2, 4, 8, 6} since on starting with them, the numbers obtained by the process of repeatedly multiplying by 2 and then taking modulo 10 repeat through the elements of this cycle. Likewise the element 5 belongs to the cycle {0}. For n coprime to 10, the different cycles will cover all the elements less than 10 and will form different equivalence classes by themselves. For other values of n, the cycles will not cover all the elements less than 10.

This sequence is periodic with period of 10, since x^i == (x + 10)^i mod 10.

Examples

			The following are the different cycles obtained by repeatedly multiplying by n, and then taking mod 10, with different starting elements.
n = 0: {0}.
n = 1: {0}, {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}, {9}.
n = 2: {2, 4, 8, 6}, {0}.
n = 3: {1, 3, 9, 7}, {2, 6, 8, 4}, {5}, {0}.
n = 4: {4, 6}, {2, 8}, {0}.
n = 5: {5}, {0}.
n = 6: {2}, {4}, {6}, {8}, {0}.
n = 7: {1, 7, 9, 3}, {2, 4, 8, 6}, {5}, {0}.
n = 8: {8, 4, 2, 6}, {0}.
n = 9: {1, 9}, {3, 7}, {2, 8}, {4, 6}, {5}, {0}.
		

Links

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

Crossrefs

Cf. A173635, A220022-A220026.

Programs

Mathematica

iter[n_] := Table[ FixedPoint[ Union[#, Mod[n*#, 10]] &, {m}], {m, 0, 9}]; classes[n_] := iter[n] //. {a___List, b_List, c___List, d_List, e___List} /; Intersection[b, d] != {} :> {a, Union[b, d], c, e}; a[n_] := Length[classes[n]]; Table[a[n], {n, 0, 89}] (* Jean-François Alcover, Jan 08 2013 *)

PARI

k=10; j=1; for(i=0, 100, m=0; n=vector(k, X, -1); for(l=0, k-1, if(n[((l*i^j)%k)+1]>=0, n[l+1]=n[((l*i^j)%k)+1]; continue, n[l+1]=m; p=l; for(o=1, eulerphi(k), p=(p*i)%k; if(n[p+1]>-1, break); n[p+1]=m); m++)); print1(m", "))

PARI

a(n)=[1, 10, 2, 4, 3, 2, 5, 4, 2, 6][n%10+1] \\ Charles R Greathouse IV, Jan 08 2013

Formula

G.f.: -(6*x^9+2*x^8+4*x^7+5*x^6+2*x^5+3*x^4+4*x^3+2*x^2+10*x+1) / (x^10-1). - Colin Barker, Apr 13 2013