A239721 Numbers that end in (..., 175, 175, 175, ...) under the rule: next term = product of the last four digits in the sequence so far.

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

Links

Table of n, a(n) for n=1..53.

E. Angelini, Multiplication by themselves of the last k digits.

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

Cf. A238984, A239419, A239616, A239722.

Sequence in context: A323590 A044304 A044685 * A142259 A161611 A251076

Adjacent sequences: A239718 A239719 A239720 * A239722 A239723 A239724

Keyword

nonn,base

Author

M. F. Hasler, Aug 01 2014

Status

approved