|
| |
|
|
A365093
|
|
Smallest k such that A365092(k) = n.
|
|
2
|
|
|
|
2, 3, 4, 5, 7, 10, 11, 20, 22, 23, 43, 46, 47, 92, 94, 139, 188, 235, 283, 461, 517, 659, 941, 1081, 1319, 2027, 2447, 2879, 4139, 5758, 8278, 10343, 13301, 20117, 26179, 30337, 44227, 56281, 61993, 95197, 115009, 135313, 194533, 270626, 366683, 481199, 606743, 811879
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
2,1
|
|
|
COMMENTS
|
a(n) is well-defined for all n >= 2.
Sequence is increasing: write a(n) as the sums and products of 1s by the process described in A365092. Removing any 1 in the expression yields a smaller number.
For m,n >= 2, it is easy to see that if gcd(a(m),a(n)) = 1, then a(m+n) <= a(m)*a(n). This is conjectured to be true for all m,n. In particular, it is conjectured that a(n+2) <= 2*a(n). If a(m+n) <= a(m)*a(n) is true, then by Fekete's subadditive lemma, we have lim_{n->oo} a(n)^(1/n) = inf_{n>=1} a(n)^(1/n) <= a(29)^(1/29) = 2879^(1/29) = 1.3160857758...
Are all terms other than 4, 20, 92 and 188 squarefree?
What are the primes that divide infinitely many terms? In particular, for p = 2 or 47, are there infinitely many terms divisible by p? Is there any term divisible by 3, 5, 7, 11 or 43 other than themselves, 235 and 517?
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
a(9) = 20 since 20 = (1+1)^(1+1)*((1+1)^(1+1)+1) (see A365092) uses nine 1s, and all smaller numbers use fewer than nine 1s.
|
|
|
PROG
|
(PARI) A365093_search(lim) = my(list=[], k=2); for(n=1, oo, if(A365092(n)==k, list=concat(list, n); k++); if(k==lim, return(list))) \\ see A365092 for its program
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,hard
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
