%I #26 Jan 02 2023 12:30:50

%S 53,135,153,375,553,579,597,753,1135,1153,1575,1755,1795,1975,3375,

%T 3515,3551,3591,3735,3951,5175,5315,5351,5391,5553,5579,5597,5711,

%U 5715,5759,5773,5795,5931,5957,5975,7155,7195,7335,7511,7515,7559,7573,7595,7753,7915,7955,9175,9351,9531,9557,9575,9715,9755

%N Numbers that end in (..., 175, 175, 175, ...) 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 so far, then the sequence is "extended to the left" with the first digit. (It might have been more natural to "extend" with digits 1, i.e., to reduce k to the number of digits if there are less than 4.)

%C Apart from the trivial cycles (0) and (1) and the cycle (175) considered here, the rule also allows for the "constant" cycle (384), cf. A239722, and (128), cf. A240967.

%C In general, such "constant" cycles for k=4 (other than the trivial (0) and (1)) must consist of a 3-digit number (100a+10b+c), a,b,c>0, such that a = (10b+c)/(b*c^2-100). The only solutions are: a=1, b=2, c=8; a=1, b=7, c=5; a=3, b=8, c=4. - _Bob Selcoe_, Aug 04 2014

%H E. Angelini, <a href="http://list.seqfan.eu/oldermail/seqfan/2014-August/013413.html">Multiplication by themselves of the last k digits</a>.

%e For a(1)=53, the sequence is 53,375,315,75,175,175,175,..., since 5*5*5*3 = 375, 3*3*7*5 = 315, 5*3*1*5 = 75, 1*5*7*5 = 175, etc.

%o (PARI) is_A239721(n) = A238984(99, n, 4)==175 \\ Here, the somewhat arbitrary value 99 (number of iterations before checking whether 175 is reached) should be sufficiently large for small n, but might need to be increased for larger starting values n.

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

%K nonn,base

%O 1,1

%A _M. F. Hasler_, Aug 01 2014