A184637 a(n) = floor(1/{(n^4+3*n)^(1/4)}), where {}=fractional part.
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
Offset: 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).
Programs
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Magma
Fp:=func; [Floor(1/Fp(n)): n in [1..80]]; // Bruno Berselli, Aug 08 2013
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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 *) -
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
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