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Powered Greatest Root (Posted on 2011-02-03) Difficulty: 3 of 5
The sequence {An, n ≥ 0} is defined by:

A0 = 0, and:

An+1 = [3√(An + n)]3

Determine an explicit formula representing An in terms of n.

Hence, or otherwise, determine all n such that An = n

Note: [x] represents the greatest integer ≤ x

No Solution Yet Submitted by K Sengupta    
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Solution solution via research and quad formula, with a fudge for n=1 | Comment 3 of 4 |

DEFDBL A-Z
CLS
a = 0: n = 0
FOR n = 0 TO 4500
  newa = INT((a + n) ^ (1 / 3)) ^ 3
  IF a <> newa THEN PRINT n + 1, newa, INT((a + n) ^ (1 / 3))
  a = newa
NEXT

lists the values when they change from one n to the next:

 2             1             1
 8             8             2
 20            27            3
 38            64            4
 62            125           5
 92            216           6
 128           343           7
 170           512           8
 218           729           9
 272           1000          10
 332           1331          11
 398           1728          12
 470           2197          13
 548           2744          14
 632           3375          15
 722           4096          16
 818           4913          17
 920           5832          18
 1028          6859          19
 1142          8000          20
 1262          9261          21
 1388          10648         22
 1520          12167         23
 1658          13824         24
 1802          15625         25
 1952          17576         26
 2108          19683         27
 2270          21952         28
 2438          24389         29
 2613          27000         30
 2793          29791         31
 2978          32768         32
 3170          35937         33
 3368          39304         34
 3572          42875         35
 3782          46656         36
 3998          50653         37
 4220          54872         38
 4448          59319         39

The column on the right is the cube root of A(n).

The sequence 2, 8, 20, 38, ..., which are the values of n at which A(n) changes, are found as Sloane's A077588, the Maximum number of regions the plane is divided into by n triangles.

The formula for that is given as 3*n^2 - 3*n + 2, but I'll change the letter to m, to avoid confusion with the puzzle's n: 3*m^2 - 3*m + 2.  For a given n in the puzzle problem we want to know the m value that leads to a number no larger than that, so solve 3*m^2 - 3*m + 2 = n. So m = [(3 + sqrt(9 + 12*(n-2)) / 6].

Comparing,

DEFDBL A-Z
CLS
a = 0: n = 0
FOR n = 0 TO 45
  newa = INT((a + n) ^ (1 / 3)) ^ 3
  PRINT n + 1, newa, INT((a + n) ^ (1 / 3)), INT((3 + SQR(ABS(9 + 12 * (n + 1 - 2)))) / 6)
  a = newa
NEXT

finds

1             0             0             0
2             1             1             1
3             1             1             1
4             1             1             1
5             1             1             1
6             1             1             1
7             1             1             1
8             8             2             2
9             8             2             2
10            8             2             2
11            8             2             2
12            8             2             2
13            8             2             2
14            8             2             2
15            8             2             2
16            8             2             2
17            8             2             2
18            8             2             2
19            8             2             2
20            27            3             3
21            27            3             3
22            27            3             3
23            27            3             3
24            27            3             3
25            27            3             3
26            27            3             3
27            27            3             3
28            27            3             3
29            27            3             3
30            27            3             3
31            27            3             3
32            27            3             3
33            27            3             3
34            27            3             3
35            27            3             3
36            27            3             3
37            27            3             3
38            64            4             4
39            64            4             4
40            64            4             4
41            64            4             4
42            64            4             4
43            64            4             4
44            64            4             4
45            64            4             4
46            64            4             4

so it would seem that, cubing the given formula, the answer would be

(INT((3 + SQR(ABS(9 + 12 * (n - 2)))) / 6)) ^ 3

or, using the bracket terminology

[(3 + sqrt(9 + 12 * (n  - 2))) / 6] ^ 3

for all values of n greater than 1, as the formula needed fiddling, with an absolute value function for n=1. (Note also that the transition to the next n was done late enough in the program so that N+1 had to be used to represent it).


  Posted by Charlie on 2011-02-03 14:57:31
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