login
A329549
Numbers 4*k such that 1 is the last integer obtained when 4*k is successively divided by its divisors in increasing order.
1
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.
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
KEYWORD
nonn
AUTHOR
David A. Corneth, Nov 16 2019
STATUS
approved