(In reply to
Where to find the sequence by Charlie)
For part a, we can use, from Sloane,
The number of terms less than or equal to n is the Sum_{i = 0…floor(log(3, n))}, floor(log(2, n/3^i) + 1).  Robert G. Wilson v, Aug 17 2012
For part b, we can first get an approximation via
An asymptotic formula for a(n) is roughly : a(n)= 1/sqrt(6) * exp(sqrt(2*log(2)*log(3)*n)).  Benoit Cloitre, Nov 20 2001
also from the Sloane site.
The functions are written as:
FUNCTION term (n)
approx = INT(EXP(SQR(2 * LOG(2) * LOG(3) * n)) / SQR(6))
nval = termsLE(approx)
WHILE nval >= n
approx = approx  1
nval = termsLE(approx)
WEND
WHILE nval < n
approx = approx + 1
nval = termsLE(approx)
WEND
term = approx
END FUNCTION
FUNCTION termsLE (n)
IF n = 0 THEN termsLE = 0: EXIT FUNCTION
fudgefactor = .00000000000001#
top = INT(LOG(n) / LOG(3) + fudgefactor)
sum = 0
FOR i = 0 TO top
sum = sum + INT(LOG(n / 3 ^ i) / LOG(2) + 1 + fudgefactor)
NEXT
termsLE = sum
END FUNCTION
The fudge factor is placed there as we are dividing logarithms to get base2 or base3 logs, and that inherently imposes the possibility of a slight error, which, if on the low side, could result in taking the INT(), i.e., the floor function, on a number such as 11.99999999998, say, giving an 11 rather than the proper 12.
In finding the nth term, we start with the asymptotic approximation and lower the number until the number of terms less than or equal to the given number is too low. Then we raise one by one until the proper number is achieved.
Using a driver to test:
DEFDBL AZ
FOR i = 1 TO 55
PRINT USING "## #####"; i; term(i)
NEXT
we get, in agreement with the list found in Sloane:
1 1
2 2
3 3
4 4
5 6
6 8
7 9
8 12
9 16
10 18
11 24
12 27
13 32
14 36
15 48
16 54
17 64
18 72
19 81
20 96
21 108
22 128
23 144
24 162
25 192
26 216
27 243
28 256
29 288
30 324
31 384
32 432
33 486
34 512
35 576
36 648
37 729
38 768
39 864
40 972
41 1024
42 1152
43 1296
44 1458
45 1536
46 1728
47 1944
48 2048
49 2187
50 2304
51 2592
52 2916
53 3072
54 3456
55 3888
This tests both functions, as the term(n) function invokes the termsLE(n) function in order to home in on the actual answer from the asymptotic approximation.

Posted by Charlie
on 20130909 13:03:14 