A329549 Numbers 4*k such that 1 is the last integer obtained when 4*k is successively divided by its divisors in increasing order.

8, 24, 40, 56, 64, 120, 144, 280, 320, 448, 704, 720, 832, 1008, 1024, 1152, 2240, 3200, 4928, 5040, 5760, 5824, 6272, 8064, 9152, 10368, 11264, 13312, 17408, 19456, 22400, 23552, 29696, 31744, 32768, 35200, 40320, 41600, 51200, 51840, 64064, 68992, 72576, 81536, 100352, 114048

Offset

1,1

Comments

At sequence A076933, the question is asked: "What is the longest string of ones in this sequence?" As A076933(4*n) is rarely 1, such a string is not very long. The longest starting below 4*10^8 has length 6 and starts at 141. Checking multiples of 4 may help in finding longer such strings.

Terms are also a multiple of 8. Proof: If m = 8*k + 4 then its divisors are 1, 2, 4 (and maybe 3). After dividing by 4 we have a fraction with denominator 2. Before that we did not see 1.

Links

Table of n, a(n) for n=1..46.

Example

The divisors of 8 are 1, 2, 4 and 8. Dividing from left to right gives 8/1 = 8, 8/2 = 4, 4/4 = 1, and then 1/8 isn't an integer so as the last integer we see is 1, 8 is in the sequence.

Crossrefs

Cf. A076933, A240694 (partial products of divisors of n).

Subsequence of A008586 (multiples of 4) and of A008590 (multiples of 8).

Sequence in context: A304475 A316300 A050427 * A031046 A173080 A051062

Adjacent sequences: A329546 A329547 A329548 * A329550 A329551 A329552

Keyword

nonn

Author

David A. Corneth, Nov 16 2019

Status

approved