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Numbers that end in (..., 384, 384, 384, ...) under the rule: next term = product of the last four digits in the sequence so far.
5

%I #13 Apr 18 2018 02:39:18

%S 12,13,14,17,19,21,24,28,38,42,61,62,68,69,112,113,114,117,119,121,

%T 122,123,126,128,129,131,133,134,136,137,142,143,144,148,162,163,173,

%U 177,179,182,184,192,197,199,211,212,217,218,221,224,228,231,242,244,249,267,268,271,274,278,281,282,284,287,294,299,311,312,321

%N Numbers that end in (..., 384, 384, 384, ...) under the rule: next term = product of the last four digits in the sequence so far.

%C Additional rule: If there are fewer than k=4 digits in the sequence so far, then it is "extended to the left" with the first digit (i.e., that digit is repeated as often as necessary).

%C Apart from the trivial cycles (0) (cf. A239616) and (1) and the cycle (384) considered here, the rule also allows the "constant" cycle (175), cf. A239721.

%o (PARI) is_A239722(n) = A238984(99,n,4)==384 \\ The "99" here should be large enough to reach the cycle (384), for small initial values n. It might be necessary to increase this value in other cases.

%Y Cf. A238984, A239419, A239616, A239721.

%K nonn,base

%O 1,1

%A _M. F. Hasler_, Aug 01 2014