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

Author

Clark Kimberling, Jan 16 2011

Keywords

• nonn
• easy

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

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

Crossrefs

Cf. A184537. Essentially the same as A033430.

Programs

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 *)

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