A203302 -id:A203302 - cp's OEIS frontend

0, 3, 15, 42, 91, 166, 274, 421, 614, 857, 1157, 1520, 1953, 2460, 3048, 3723, 4492, 5359, 6331, 7414, 8615, 9938, 11390, 12977, 14706, 16581, 18609, 20796, 23149, 25672, 28372, 31255, 34328, 37595, 41063, 44738, 48627, 52734, 57066

Offset: 0

Clark Kimberling, Dec 20 2012

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).
Conjecture:  if d is a positive integer and f(x) = x^(1/d), then the solutions of J(k) form a linearly recurrent sequence.
This conjecture was proved by David Moews; see Problem 21 in "Unsolved Problems and Rewards".  - Clark Kimberling, Feb 06 2013
Guide to related sequences:
f(x) ....... jump sequence ... linear recurrence order
x^(1/2) .... A002378 ......... 3
x^(1/3) .... A219085 ......... 7
x^(2/3) .... A203302 ......... (not linearly recurrent)
x^(1/4) .... A219086 ......... 5
x^(3/4) .... A219087 ......... (not linearly recurrent)
x^(1/5) .... A219088 ......... 21
x^(1/6) .... A219089 ......... 21
x^(1/7) .... A219090 ..........71
x^(1/8) .... A219091 ......... 23
log(x) ..... A219092 ......... (not linearly recurrent)
log_2(x) ... A084188 ......... (not linearly recurrent)

			Let p=1/3.  Then
3^p=1.44... and 4^p=1.58..., so 3 is a jump point.
15^p=2.46... and 16^p=2.51..., so 15 is a jump point.

Clark Kimberling, Table of n, a(n) for n = 0..10000
Clark Kimberling, Unsolved Problems and Rewards, Problem 21.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1,1,-3,3,-1).

Cf. A002378, A203302.

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

a(n)=n^3 + (6*n^2 + 3*n)\4 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = floor((n + 1/2)^3).
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).
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.
a(n) = (n + 1/2)^3 + (2*i^(n*(n-1))+(-1)^n-4)/8, where i=sqrt(-1). - Bruno Berselli, Dec 21 2012

cp's OEIS Frontend

A219085 a(n) = floor((n + 1/2)^3).

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