login
A184536
a(n) = floor(1/{(1+n^4)^(1/4)}), where {} = fractional part.
11
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
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
Sequence in context: A322952 A101966 A341423 * A363411 A089574 A077207
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jan 16 2011
STATUS
approved