

A219085


Floor((n + 1/2)^3).


9



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
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OFFSET

0,2


COMMENTS

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)


LINKS

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


FORMULA

a(n) = floor((n + 1/2)^3).
a(n) = 3*a(n1) 3*a(n2) +a(n3) +a(n4) 3*a(n5) +3*a(n6) a(n7).
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*(n1))+(1)^n4)/8, where i=sqrt(1).  Bruno Berselli, Dec 21 2012


EXAMPLE

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.


MATHEMATICA

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


PROG

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


CROSSREFS

Cf. A002378, A203302.
Sequence in context: A012222 A069267 A059270 * A093627 A192060 A316642
Adjacent sequences: A219082 A219083 A219084 * A219086 A219087 A219088


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Dec 20 2012


STATUS

approved



