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 Powered Greatest Root (Posted on 2011-02-03)
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 No Rating

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 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             02             1             1             13             1             1             14             1             1             15             1             1             16             1             1             17             1             1             18             8             2             29             8             2             210            8             2             211            8             2             212            8             2             213            8             2             214            8             2             215            8             2             216            8             2             217            8             2             218            8             2             219            8             2             220            27            3             321            27            3             322            27            3             323            27            3             324            27            3             325            27            3             326            27            3             327            27            3             328            27            3             329            27            3             330            27            3             331            27            3             332            27            3             333            27            3             334            27            3             335            27            3             336            27            3             337            27            3             338            64            4             439            64            4             440            64            4             441            64            4             442            64            4             443            64            4             444            64            4             445            64            4             446            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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