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A184537
a(n) = floor(1/{(2+n^4)^(1/4)}), where {} = fractional part.
2
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
0,1
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)
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.
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 *)
PROG
(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
CROSSREFS
Sequence in context: A185880 A292243 A105201 * A187809 A092893 A133172
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jan 16 2011
STATUS
approved