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A239721
Numbers that end in (..., 175, 175, 175, ...) under the rule: next term = product of the last four digits in the sequence so far.
5
53, 135, 153, 375, 553, 579, 597, 753, 1135, 1153, 1575, 1755, 1795, 1975, 3375, 3515, 3551, 3591, 3735, 3951, 5175, 5315, 5351, 5391, 5553, 5579, 5597, 5711, 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
OFFSET
1,1
COMMENTS
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.)
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.
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
EXAMPLE
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.
PROG
(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.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
M. F. Hasler, Aug 01 2014
STATUS
approved