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
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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jan 16 2011
STATUS
approved