A184537 a(n) = floor(1/{(2+n^4)^(1/4)}), where {} = fractional part.

5, 3, 16, 54, 128, 250, 432, 686, 1024, 1458, 2000, 2662, 3456, 4394, 5488, 6750, 8192, 9826, 11664, 13718, 16000, 18522, 21296, 24334, 27648, 31250, 35152, 39366, 43904, 48778, 54000, 59582, 65536, 71874, 78608, 85750, 93312, 101306, 109744, 118638, 128000, 137842, 148176, 159014, 170368, 182250, 194672, 207646, 221184, 235298, 250000, 265302, 281216, 297754, 314928, 332750, 351232, 370386, 390224, 410758, 432000, 453962, 476656

Offset: 0

Author

Clark Kimberling, Jan 16 2011

Keywords

• nonn
• easy

Comments

Similar to A033431: replacing a(0) by 0 and a(1) by 2 gives A033431, see next comment.

From Bruno Berselli, Feb 04 2011: (Start)

For n >= 1, the value of (n^4+2)^(1/4) is just slightly above n, so the fractional part is (2+n^4)^(1/4)-n. For n > 1, then, 2*n^3 < 1/((2+n^4)^(1/4)-n) < 2*n^3+1.

The proof that 2*n^3*((2+n^4)^(1/4)-n) < 1 follows easily by isolating the quartic root and raising to the 4th power; similarly, 1 < (2*n^3+1)*((2+n^4)^(1/4)-n) needs a sign estimation of a 9th-order polynomial.

In conclusion, a(n)=A033431(n) for n > 1, with g.f. (34*x^2-12*x^3+x^4+x^5+5-17*x) / (x-1)^4. (End)

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A184536. Essentially the same as A033431.

Programs

Magma

I:=[5, 3, 16, 54, 128,250]; [n le 6 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..70]]; // Vincenzo Librandi, Jul 04 2012

Mathematica

f[n_] := Floor[1/FractionalPart[(n^4 + 2)^(1/4)]]; Array[f, 40, 0]
CoefficientList[Series[(34*x^2-12*x^3+x^4+x^5+5-17*x)/(x-1)^4,{x,0,50}],x] (* Vincenzo Librandi, Jul 04 2012 *)
LinearRecurrence[{4,-6,4,-1},{5,3,16,54,128,250},70] (* Harvey P. Dale, Apr 06 2018 *)

Formula

a(n) = floor( 1 / frac((2+n^4)^(1/4)) ).

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 5.