A333554 - OEIS

%I #24 May 04 2020 00:23:34

%S 9,22599,782534990456559999,

%T 1124958024440103533642098279435875957333450115658804731260154201599999999

%N a(0) = 9; thereafter a(n+1) = 3*a(n)^4 + 4*a(n)^3 for n >= 1.

%C Proposition: a(n) ends with exactly 2^n 9's.

%C The number of digits in a(n) is respectively 1, 5, 18, 73, 289, 1156, 4622, ... At each step until a(6), the number of digits of a(n+1) is nearly 4 times the number of digits of a(n), and exactly so from a(4) to a(5).

%C Sequence has doubly-exponential growth: log log a(n) ~ n. - _Charles R Greathouse IV_, Mar 27 2020

%D Eric Billault, Walter Damin, Robert Ferréol et al., MPSI - Classes Prépas, Khôlles de Maths, Ellipses, 2012, exercice 4.21 pages 79 and 91.

%e a(2) = 3*22599^4 + 4*22599^3 = 782534990456559999 ends with 2^2 digits 9's.

%t a[0] = 9; a[n_] := a[n] = 3*a[n-1]^4 + 4*a[n-1]^3; Array[a, 4, 0] (* _Amiram Eldar_, Mar 26 2020 *)

%Y Subsequence of A017377.

%K nonn,base

%O 0,1

%A _Bernard Schott_, Mar 26 2020