At least half of the digits of every term (except a(14)) are the same.
Let n > 0:
a(4n) mod 100 = 211;
a(4n+1) mod 1000 = 3325;
a(4n+2) mod 1000000 = 555551;
a(4n+3) mod 100000000 = 77777775;
Proof for a(4n):
If x is divisible by 4 its hereditary representation in base 2 has all summands divisible by 4 and it cannot have the summands 1 and 2.
If we calculate G_1(x) we would end with:
G_1(x) = B_2(x)-1.
Clearly, B_2(x) = 3^a + 3^b + ... is divisible by 3^3 = 27 and that would mean that the representation of B_2(x)-1 would be B_2(x)-1 = X_3 + 2*3^2+2*3+2.
From now on, let X_n be a sum of powers of n (greater than the right term).
We finish proving the statement by calculating G_8(x):
G_2(x) = B_3(X_3 +2*3^2+2*3+2)-1 = X_4 + 2*4^2+2*4+2-1;
G_3(x) = B_4(X_4 +2*4^2+2*4-1)-1 = X_5 + 2*5^2+2*5+1-1;
G_4(x) = B_5(X_5 +2*5^2+2*5)-1 = X_6 + 2*6^2+2*6-1;
G_5(x) = B_6(X_6 +2*6^2+6+5)-1 = X_7 + 2*7^2+7+5-1;
G_6(x) = B_7(X_7 +2*7^2+7+4)-1 = X_8 + 2*8^2+8+4-1;
G_7(x) = B_8(X_8 +2*8^2+8+3)-1 = X_9 + 2*9^2+9+3-1;
G_8(x) = B_9(X_9 +2*9^2+9+2)-1 = X_10 + 2*10^2+10+2-1 = X_10 + 211;
So finally G_8(x) mod 100 = 211.
The other cases can be proved using the same reasoning.
|