|
|
A022031
|
|
Define the sequence T(a(0),a(1)) by a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n) for n >= 0. This is T(4,17).
|
|
3
|
|
|
4, 17, 72, 304, 1283, 5414, 22845, 96397, 406757, 1716352, 7242319, 30559689, 128949662, 544115986, 2295951781, 9687997993, 40879475731, 172495033261, 727860031657, 3071278144467, 12959565068034, 54684179957837, 230745362360740, 973653116715681, 4108426630946045
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,1
|
|
COMMENTS
|
The empirical g.f. / recurrence agrees with the original definition for at least 2000 terms (and a(2000) ~ 10^1250). - M. F. Hasler, Feb 11 2016
|
|
LINKS
|
|
|
FORMULA
|
Empirical g.f.: -(x^6+x^5+x^4+x^3-x-4) / ((x-1)*(x^6+2*x^5+3*x^4+4*x^3+4*x^2+3*x-1)). - Colin Barker, Sep 18 2015
a(n+1) = ceiling(a(n)^2/a(n-1))-1 for all n > 0. a(n+1)/a(n) ~ 4.219599938... as n -> oo. - M. F. Hasler, Feb 11 2016
|
|
PROG
|
(PARI) a=[4, 17]; for(n=2, 2000, a=concat(a, ceil(a[n]^2/a[n-1])-1)); A022031(n)=a[n+1] \\ M. F. Hasler, Feb 11 2016
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|