login
A184633
Floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.
1
1, 4, 12, 28, 55, 96, 152, 227, 324, 444, 591, 768, 976, 1219, 1500, 1820, 2183, 2592, 3048, 3555, 4116, 4732, 5407, 6144, 6944, 7811, 8748, 9756, 10839, 12000, 13240, 14563, 15972, 17468, 19055, 20736, 22512, 24387, 26364, 28444, 30631, 32928, 35336, 37859, 40500, 43260, 46143, 49152, 52288, 55555, 58956, 62492, 66167, 69984, 73944, 78051, 82308, 86716, 91279, 96000, 100880, 105923, 111132, 116508, 122055, 127776, 133672, 139747, 146004, 152444, 159071, 165888, 172896, 180099, 187500
OFFSET
1,2
LINKS
Edward Jiang and Ray Chandler, Table of n, a(n) for n = 1..10000 (first 1000 terms from Edward Jiang)
FORMULA
a(n) = floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.
It appears that a(n) = 3a(n-1)-3a(n-2)+2a(n-3)-3a(n-4)+3a(n-5)-a(n-6) for n>=9.
Empirical g.f.: x*(x+1)*(x^6-3*x^5+3*x^4-x^3+3*x^2+1) / ((x-1)^4*(x^2+x+1)). - Colin Barker, Jun 13 2015
MATHEMATICA
p[n_]:=FractionalPart[(n^4+9)^(1/4)]; q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1, 4}, LinearRecurrence[{3, -3, 2, -3, 3, -1}, {12, 28, 55, 96, 152, 227}, 73]] (* Ray Chandler, Aug 02 2015 *)
PROG
(PARI) a(n)=my(t=(9+n^4)^(1/4)); 1\(t-t\1) \\ Charles R Greathouse IV, Sep 12 2014
CROSSREFS
Cf. A184536.
Sequence in context: A109629 A112087 A166019 * A006000 A161216 A085622
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 18 2011
STATUS
approved