A184536 a(n) = floor(1/{(1+n^4)^(1/4)}), where {} = fractional part.

5, 32, 108, 256, 500, 864, 1372, 2048, 2916, 4000, 5324, 6912, 8788, 10976, 13500, 16384, 19652, 23328, 27436, 32000, 37044, 42592, 48668, 55296, 62500, 70304, 78732, 87808, 97556, 108000, 119164, 131072, 143748, 157216, 171500, 186624, 202612, 219488, 237276, 256000, 275684, 296352, 318028, 340736, 364500, 389344, 415292, 442368, 470596, 500000, 530604, 562432, 595508, 629856, 665500, 702464, 740772, 780448, 821516, 864000, 907924, 953312, 1000188

Offset

1,1

Comments

For n >= 1, the value of (n^4+1)^(1/4) is just slightly above n, so the fractional part is (n^4+1)^(1/4)-n. For n > 1, then, 4*n^3 < 1/((1+n^4)^(1/4)-n) < 4*n^3+1. [Proof that 4*n^3*((1+n^4)^(1/4)-n) < 1 follows easily by isolating the quartic root and raising to the 4th power; similarly, 1 < (4*n^3+1)*((1+n^4)^(1/4)-n) needs a sign estimation of an 8th-order polynomial.] In conclusion, a(n)=A033430(n) for n > 1. - Bruno Berselli, Jan 30 2011

Links

Table of n, a(n) for n=1..63.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Formula

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

G.f.: x*(x^4-4*x^3+10*x^2+12*x+5)/(x-1)^4. - Colin Barker, Sep 21 2012

Mathematica

p[n_]:=FractionalPart[(n^4+1)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
LinearRecurrence[{4, -6, 4, -1}, {5, 32, 108, 256, 500}, 70] (* Harvey P. Dale, Dec 14 2023 *)

Crossrefs

Cf. A184537, A033430.

Sequence in context: A322952 A101966 A341423 * A363411 A089574 A077207

Adjacent sequences: A184533 A184534 A184535 * A184537 A184538 A184539

Keyword

nonn,easy

Author

Clark Kimberling, Jan 16 2011

Status

approved