A184634 a(n) = floor(1/{(10+n^4)^(1/4)}), where {}=fractional part.

1, 3, 11, 25, 50, 86, 137, 204, 291, 400, 532, 691, 878, 1097, 1350, 1638, 1965, 2332, 2743, 3200, 3704, 4259, 4866, 5529, 6250, 7030, 7873, 8780, 9755, 10800, 11916, 13107, 14374, 15721, 17150, 18662, 20261, 21948, 23727, 25600, 27568, 29635, 31802, 34073, 36450, 38934, 41529, 44236, 47059, 50000, 53060, 56243, 59550, 62985, 66550, 70246, 74077, 78044, 82151, 86400, 90792, 95331, 100018, 104857, 109850, 114998, 120305, 125772, 131403, 137200

Offset: 1

Author

Clark Kimberling, Jan 18 2011

Keywords

• nonn

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (3, -3, 1, 0, 1, -3, 3, -1).

Crossrefs

Cf. A184536.

Programs

Maple

f:= proc(n) local k, t;
  t:= n mod 5;
  k:= (n-t)/5;
  [50*k^3, 50*k^3 + 30*k^2 + 6*k,
50*k^3 + 60*k^2 + 24*k + 3,
50*k^3 + 90*k^2 + 54*k + 10,
50*k^3 + 120*k^2 + 96*k + 25][t+1]
end proc:
f(1):= 1: f(3):= 11:
map(f, [$1..100]); # Robert Israel, Feb 25 2019

Mathematica

p[n_]:=FractionalPart[(n^4+10)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1,3,11},LinearRecurrence[{3,-3,1,0,1,-3,3,-1},{25,50,86,137,204,291,400,532},67]] (* Ray Chandler, Aug 02 2015 *)

PARI

a(n)=1\frac(sqrtn(n^4+10,4)) \\ Charles R Greathouse IV, Feb 07 2016

Formula

a(n)=floor(1/{(10+n^4)^(1/4)}), where {}=fractional part.

It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3)+a(n-5)-3a(n-6)+3a(n-7)-a(n-8).

From Robert Israel, Feb 25 2019: (Start)

2*n^3/5 + 3/(2*n) > 1/{(10+n^4)^(1/4)} > 2*n^3/5 for all n.

From this we can show that a(5*k) = 50*k^3 for k >= 1,

a(5*k+1) = 50*k^3 + 30*k^2 + 6*k for k >= 1,

a(5*k+2) = 50*k^3 + 60*k^2 + 24*k + 3,

a(5*k+3) = 50*k^3 + 90*k^2 + 54*k + 10 for k >= 1,

a(5*k+4) = 50*k^3 + 120*k^2 + 96*k + 25.

This implies the conjectured recurrence for n >= 12. (End)