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A184637 a(n) = floor(1/{(n^4+3*n)^(1/4)}), where {}=fractional part. 4
2, 6, 12, 21, 33, 48, 65, 85, 108, 133, 161, 192, 225, 261, 300, 341, 385, 432, 481, 533, 588, 645, 705, 768, 833, 901, 972, 1045, 1121, 1200, 1281, 1365, 1452, 1541, 1633, 1728, 1825, 1925, 2028, 2133, 2241, 2352, 2465, 2581, 2700, 2821, 2945, 3072, 3201, 3333, 3468, 3605, 3745, 3888, 4033, 4181, 4332, 4485, 4641, 4800, 4961, 5125, 5292, 5461, 5633, 5808, 5985, 6165, 6348, 6533, 6721, 6912, 7105, 7301, 7500, 7701, 7905, 8112, 8321, 8533 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

LINKS

G. C. Greubel, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

FORMULA

a(n) = 2*a(n-1) -a(n-2) +a(n-3) -2*a(n-4) +a(n-5) for n>=8.

G.f.: x*(x+1)*(x^5-2*x^4+x^3-2*x^2-2)/((x-1)^3*(x^2+x+1)). - Colin Barker, Oct 07 2012

a(n) = n^2 + floor(n^2/3) with n>2, a(1)=2, a(2)=6. This confirms the g.f. and the recurrence. - Bruno Berselli, Aug 08 2013

MATHEMATICA

p[n_]:=FractionalPart[(n^4+3*n)^(1/4)];

q[n_]:=Floor[1/p[n]];

Table[q[n], {n, 1, 80}]

FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]

Join[{2, 6}, LinearRecurrence[{2, -1, 1, -2, 1}, {12, 21, 33, 48, 65}, 78]] (* Ray Chandler, Aug 02 2015 *)

PROG

(MAGMA) Fp:=func<i | Root(i^4+3*i, 4)-Iroot(i^4+3*i, 4)>; [Floor(1/Fp(n)): n in [1..80]]; // Bruno Berselli, Aug 08 2013

(PARI) x='x+O('x^30); Vec(x*(x+1)*(x^5-2*x^4+x^3-2*x^2-2)/((x-1)^3*(x^2 +x+1))) \\ G. C. Greubel, Apr 04 2018

CROSSREFS

Cf. A184535, A184636.

Sequence in context: A161203 A292768 A352990 * A022450 A011892 A062482

Adjacent sequences:  A184634 A184635 A184636 * A184638 A184639 A184640

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Jan 18 2011

STATUS

approved

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Last modified May 29 05:00 EDT 2022. Contains 354122 sequences. (Running on oeis4.)