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A271979 G_8(n), where G is the Goodstein function defined in A266201.

%I A271979 #44 Sep 25 2020 11:10:12
%S A271979 0,211,3325,555551,77777775,20000000211,30000003325,50000555551,
%T A271979 70077777775,100000000211,100000003325,100000555551,100077777775
%N A271979 G_8(n), where G is the Goodstein function defined in A266201.
%C A271979 At least half of the digits of every term (except a(14)) are the same.
%C A271979 Let n > 0:
%C A271979 a(4n) mod 100 = 211;
%C A271979 a(4n+1) mod 1000 = 3325;
%C A271979 a(4n+2) mod 1000000 = 555551;
%C A271979 a(4n+3) mod 100000000 = 77777775;
%C A271979 Proof for a(4n):
%C A271979 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.
%C A271979 If we calculate G_1(x) we would end with:
%C A271979 G_1(x) = B_2(x)-1.
%C A271979 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.
%C A271979 From now on, let X_n be a sum of powers of n (greater than the right term).
%C A271979 We finish proving the statement by calculating G_8(x):
%C A271979 G_2(x) = B_3(X_3 +2*3^2+2*3+2)-1 = X_4 + 2*4^2+2*4+2-1;
%C A271979 G_3(x) = B_4(X_4 +2*4^2+2*4-1)-1 = X_5 + 2*5^2+2*5+1-1;
%C A271979 G_4(x) = B_5(X_5 +2*5^2+2*5)-1 = X_6 + 2*6^2+2*6-1;
%C A271979 G_5(x) = B_6(X_6 +2*6^2+6+5)-1 = X_7 + 2*7^2+7+5-1;
%C A271979 G_6(x) = B_7(X_7 +2*7^2+7+4)-1 = X_8 + 2*8^2+8+4-1;
%C A271979 G_7(x) = B_8(X_8 +2*8^2+8+3)-1 = X_9 + 2*9^2+9+3-1;
%C A271979 G_8(x) = B_9(X_9 +2*9^2+9+2)-1 = X_10 + 2*10^2+10+2-1 = X_10 + 211;
%C A271979 So finally G_8(x) mod 100 = 211.
%C A271979 The other cases can be proved using the same reasoning.
%C A271979 a(17) = 3.3330...*10^3333, a(18) = 5.555550...*10^555555. - _Pontus von Brömssen_, Sep 25 2020
%H A271979 Pontus von Brömssen, <a href="/A271979/b271979.txt">Table of n, a(n) for n = 3..16</a>
%H A271979 Wikipedia, <a href="https://en.wikipedia.org/wiki/Goodstein&#39;s_theorem">Goodstein's theorem</a>
%e A271979 Calculate G_8(5):
%e A271979 G_1(5) = B_2(5)-1 = B_2(2^2+1)-1 = 27;
%e A271979 G_2(5) = B_3(3^3)-1 = 4^4-1 = 255;
%e A271979 G_3(5) = B_4(3*4^3 + 3*4^2 + 3*4 + 3)-1 = 3*5^3 + 3*5^2 + 3*5 + 3-1 = 467;
%e A271979 G_4(5) = B_5(3*5^3 + 3*5^2 + 3*5 + 2)-1 = 3*6^3 + 3*6^2 + 3*6 + 2-1 = 775;
%e A271979 G_5(5) = B_6(3*6^3 + 3*6^2 + 3*6 + 1)-1 = 3*7^3 + 3*7^2 + 3*7 + 1-1 = 1197;
%e A271979 G_6(5) = B_7(3*7^3 + 3*7^2 + 3*7)-1 = 3*8^3 + 3*8^2 + 3*8-1 = 1751;
%e A271979 G_7(5) = B_8(3*8^3 + 3*8^2 + 2*8 + 7)-1 = 3*9^3 + 3*9^2 + 2*9 + 7-1 = 2454;
%e A271979 G_8(5) = B_9(3*9^3 + 3*9^2 + 2*9 + 6)-1 = 3*10^3 + 3*10^2 + 2*10 + 6-1 = 3325.
%o A271979 (Python)
%o A271979 from sympy.ntheory.factor_ import digits
%o A271979 def bump(n,b):
%o A271979   s=digits(n,b)[1:]
%o A271979   l=len(s)
%o A271979   return sum(s[i]*(b+1)**bump(l-i-1,b) for i in range(l) if s[i])
%o A271979 def A271979(n):
%o A271979   if n==3: return 0
%o A271979   for i in range(2,10):
%o A271979     n=bump(n,i)-1
%o A271979   return n # _Pontus von Brömssen_, Sep 25 2020
%Y A271979 Cf. A056004: G_1(n); A057650: G_2(n); A059934: G_3(n); A059935: G_4(n); A059936: G_5(n); A271977: G_6(n); A271978: G_7(n); this sequence: G_8(n); A271985: G_9(n); A271986: G_10(n); A266201: G_n(n).
%K A271979 nonn
%O A271979 3,2
%A A271979 _Natan Arie Consigli_, Apr 30 2016
%E A271979 Incorrect program and terms removed by _Pontus von Brömssen_, Sep 25 2020