A219085 - OEIS

%I #32 Feb 14 2024 10:46:26

%S 0,3,15,42,91,166,274,421,614,857,1157,1520,1953,2460,3048,3723,4492,

%T 5359,6331,7414,8615,9938,11390,12977,14706,16581,18609,20796,23149,

%U 25672,28372,31255,34328,37595,41063,44738,48627,52734,57066

%N a(n) = floor((n + 1/2)^3).

%C a(n) is the number k such that {k^p} < 1/2 < {(k+1)^p}, where p = 1/3 and { } = fractional part.  In general, suppose that f is a continuous strictly increasing downward concave function, with f(1)>=0 and f(k)+1/2 not an integer.  Let J(k) denote the inequality {f(k)} < 1/2 < {f(k+1)}, where {}= fractional part; equivalently, [{f(k)} + 1/2] = 0 and [{f(k+1)} + 1/2] = 1, where [ ] = floor.  Thus J(k) holds if the integer nearest f(k+1) exceeds the integer nearest f(k), so that k can be regarded as a "jump point for f".  The solutions of J(k) are the numbers [g(n)+1/2)] for n >= 0, where g = (inverse of f).

%C Conjecture:  if d is a positive integer and f(x) = x^(1/d), then the solutions of J(k) form a linearly recurrent sequence.

%C This conjecture was proved by David Moews; see Problem 21 in "Unsolved Problems and Rewards".  - _Clark Kimberling_, Feb 06 2013

%C Guide to related sequences:

%C f(x) ....... jump sequence ... linear recurrence order

%C x^(1/2) .... A002378 ......... 3

%C x^(1/3) .... A219085 ......... 7

%C x^(2/3) .... A203302 ......... (not linearly recurrent)

%C x^(1/4) .... A219086 ......... 5

%C x^(3/4) .... A219087 ......... (not linearly recurrent)

%C x^(1/5) .... A219088 ......... 21

%C x^(1/6) .... A219089 ......... 21

%C x^(1/7) .... A219090 ..........71

%C x^(1/8) .... A219091 ......... 23

%C log(x) ..... A219092 ......... (not linearly recurrent)

%C log_2(x) ... A084188 ......... (not linearly recurrent)

%H Clark Kimberling, <a href="/A219085/b219085.txt">Table of n, a(n) for n = 0..10000</a>

%H Clark Kimberling, <a href="http://faculty.evansville.edu/ck6/integer/unsolved.html">Unsolved Problems and Rewards, Problem 21</a>.

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1,1,-3,3,-1).

%F a(n) = floor((n + 1/2)^3).

%F a(n) = 3*a(n-1) -3*a(n-2) +a(n-3) +a(n-4) -3*a(n-5) +3*a(n-6) -a(n-7).

%F G.f.: (3*x +6*x^2 +6*x^3 +7*x^4 +x^5 +x^6)/(u*v), where u = (1 - x)^4, v = 1 + x + x^2 + x^3.

%F a(n) = (n + 1/2)^3 + (2*i^(n*(n-1))+(-1)^n-4)/8, where i=sqrt(-1). - _Bruno Berselli_, Dec 21 2012

%e Let p=1/3.  Then

%e 3^p=1.44... and 4^p=1.58..., so 3 is a jump point.

%e 15^p=2.46... and 16^p=2.51..., so 15 is a jump point.

%t Table[Floor[(n + 1/2)^3], {n, 0, 100}]

%o (PARI) a(n)=n^3 + (6*n^2 + 3*n)\4 \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A002378, A203302.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Dec 20 2012