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 A333554 a(0) = 9; thereafter a(n+1) = 3*a(n)^4 + 4*a(n)^3 for n >= 1. 0
 9, 22599, 782534990456559999, 1124958024440103533642098279435875957333450115658804731260154201599999999 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified August 11 06:25 EDT 2020. Contains 336422 sequences. (Running on oeis4.)