A333554 a(0) = 9; thereafter a(n+1) = 3*a(n)^4 + 4*a(n)^3 for n >= 1.

9, 22599, 782534990456559999, 1124958024440103533642098279435875957333450115658804731260154201599999999

Offset

0,1

Comments

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

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).

Sequence has doubly-exponential growth: log log a(n) ~ n. - Charles R Greathouse IV, Mar 27 2020

References

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

Links

Table of n, a(n) for n=0..3.

Example

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

Mathematica

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 *)

Crossrefs

Subsequence of A017377.

Sequence in context: A292562 A140596 A280900 * A055309 A288940 A053931

Adjacent sequences: A333551 A333552 A333553 * A333555 A333556 A333557

Keyword

nonn,base

Author

Bernard Schott, Mar 26 2020

Status

approved