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These are the analogs of the powers of 2 in carryless arithmetic mod 10.

Let G = {2,4,8,6}. Let o be defined as XoY = least significant digit in XY. Then (G,o) is an Abelian group wherein 2 is a generator (also see the first comment under A001148). - K.V.Iyer, Mar 12 2010

This is also the decimal expansion of 227/1818. - Kritsada Moomuang, Dec 21 2021

Periodic with period 4: [1, 2, 4, 3]. - Washington Bomfim, Nov 23 2010

a(n) will always be a divisor of Phi(100) = 40.

This sequence is periodic with a period of 100 because n^i mod 100 = (n + 100)^i mod 100.

For the odd numbers n ending in {1, 3, 7, 9} which are coprime to 10, we can expect the powers of n mod 100 to loop back to 1, with the value of n^a(n) mod 100 = 1, but for the other numbers n that are not coprime to 10, they do not loop back to 1.

For the even numbers n ending in {2, 4, 6, 8}, the value of n^a(n) mod 100 will be equal to 76.

For the numbers n that are congruent to 5 (mod 20), the value of n^i mod 100 will be equal to 25, for all i >= 2.

For the numbers n that are congruent to 15 (mod 20), the value of n^a(n) mod 100 will be equal to 25, with n^i mod 100 = 25 for the even values of i and n^i mod 100 = 75 for the odd values of i, for all i >= 2.

For the numbers n ending in 0, i.e., that are congruent to 0 (mod 10), the value of n^i mod 100 will be equal to 0, for all i >= 2.

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.

a(n) will always be a divisor of Phi(1000000) = 400000.

This sequence is periodic with a period of 1000000 because n^i mod 1000000 = (n + 1000000)^i mod 1000000.

For the odd numbers n ending in {1, 3, 7, 9} which are coprime to 10, we can expect the powers of n mod 1000000 to loop back to 1, with the value of n^a(n) mod 1000000 = 1, but for the other numbers n that are not coprime to 10, they do not loop back to 1.

For the even numbers n ending in {2, 4, 6, 8}, n^a(n) mod 1000000 = 109376.

For the numbers n ending in 5, n^(16*i) mod 1000000 = 890625, for all i >= 1.

For the numbers n ending in 0, n^i mod 1000000 = 0, for all i >= 6.

a(n) will always be a divisor of Phi(1000) = 400.

This sequence is periodic with a period of 1000 because n^i mod 1000 = (n + 1000)^i mod 1000.

For the odd numbers n ending in {1, 3, 7, 9} which are coprime to 10, we can expect the powers of n mod 1000 to loop back to 1, with the value of n^a(n) mod 1000 = 1, but for the other numbers n that are not coprime to 10, they do not loop back to 1.

For the even numbers n ending in {2, 4, 6, 8}, n^a(n) mod 1000 = 376.

For the numbers n ending in 5, n^(2*i) mod 1000 = 625, for all i >= 2.

For the numbers n ending in 0, n^i mod 1000 = 0, for all i >= 3.

a(n) will always be a divisor of Phi(10000) = 4000.

This sequence is periodic with a period of 10000 because n^i mod 10000 = (n + 10000)^i mod 10000.

For the odd numbers n ending in {1, 3, 7, 9} which are coprime to 10, we can expect the powers of n mod 10000 to loop back to 1, with the value of n^a(n) mod 10000 = 1, but for the other numbers n that are not coprime to 10, they do not loop back to 1.

For the even numbers n ending in {2, 4, 6, 8}, n^a(n) mod 10000 = 9376.

For the numbers n ending in 5, n^(4*i) mod 10000 = 625, for all i >= 1.

For the numbers n ending in 0, n^i mod 10000 = 0, for all i >= 4.

a(n) will always be a divisor of Phi(100000) = 40000.

This sequence is periodic with a period of 100000 because n^i mod 100000 = (n + 100000)^i mod 100000.

For the odd numbers n ending in {1, 3, 7, 9} which are coprime to 10, we can expect the powers of n mod 100000 to loop back to 1, with the value of n^a(n) mod 100000 = 1, but for the other numbers n that are not coprime to 10, they do not loop back to 1.

For the even numbers n ending in {2, 4, 6, 8}, n^a(n) mod 100000 = 9376.

For the numbers n ending in 5, n^(8*i) mod 100000 = 90625, for all i >= 1.

For the numbers n ending in 0, n^i mod 100000 = 0, for all i >= 5.