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

1,2

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).

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)

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 *)

Prog

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

Crossrefs

Cf. A184536.

Sequence in context: A147382 A212971 A258440 * A352013 A164303 A129082

Adjacent sequences: A184631 A184632 A184633 * A184635 A184636 A184637

Keyword

nonn

Author

Clark Kimberling, Jan 18 2011

Status

approved