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Numbers that end in (..., 128, 128, 128, ...) under the rule: next term = product of the last four digits in the sequence so far.
1

%I #11 Apr 18 2018 02:39:50

%S 248,418,681,733,871,1288,1448,1828,2188,2248,2428,2614,2622,2641,

%T 2818,3414,3422,3441,3773,4148,4228,4314,4322,4341,4418,4891,4981,

%U 6214,6222,6241,6681,6861,7373,7733,7881,8128,8218,8491,8661,8781,8871,8941,9481,9841

%N Numbers that end in (..., 128, 128, 128, ...) 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) (entered for initial values 1 or 11 or 111 or 1111+x*10^5) and the cycle (128) considered here, the rule also allows the "constant" cycles (175) and (384), cf. A239721-A239722. These seem to be the only cycles allowed for this rule.

%C See A238984 for further information, including a link to Eric Angelini's SeqFan post motivating this sequence.

%o (PARI) is(n) = A238984(99,n,4)==128 \\ The "99" here should be large enough to reach the cycle (128) for small initial values n. It might be necessary to increase this value in other cases.

%Y Cf. A238984, A239419, A239616, A239721, A239722.

%K nonn,base

%O 1,1

%A _M. F. Hasler_ and _Bob Selcoe_, Aug 05 2014