How to determine this sequence 1 21 467

Question

1 21 467 5216

Answer 1

Note that the question is "how to determine" the sequence, rather than simply find the function which controls it. So below is a method I've used to try to find what controls the sequence.

If the sequence is 1, 21, 467, 5216, and we can take rough logs (base 10) of these numbers we get:

0, 1.32, 2.67, 3.72 which implies that the sequence could be logarithmic, because the ratios are roughly arithmetic with a common difference about 1.3 which is roughly the log of 20. This would generate the sequence 1, 20, 400, 8000. 8000 is much greater than 5216, so let's see what happens if we use 19 instead of 20: 1, 19, 361, 6859. The last term is still larger than we would like but the second term is too small.

Now try something else. Suppose the sequence is of the form a_nxⁿn+b_n for each term, where integer n starts at 0, and x, a_n and b_n are constants then:

b₀=1, and the other terms are a₁x+b₁=21, a₂x²+b₂=467, a₃x³+b₃=5216.

Let's try some possible values for x (some figures are approximate) :

x=15:

15a₁+b₁=21; 225a₂+b₂=467; 3375a₃+b₃=5216, a₁=(21-b₁)/15=1.4-b₁/15; a₂=(467-b₂)/225=2.076-b₂/225; a₃=(5216-b₃)/3375=1.545-b₃/3375;

x=16:

16a₁+b₁=21; 256a₂+b₂=467; 4096a₃+b₃=5216; a₁=(21-b₁)/16=1.313-b₁/16; a₂=(467-b₂)/256=1.824-b₂/256; a₃=(5216-b₃)/4096=1.545-b₃/4096;

x=17:

17a₁+b₁=21; 289a₂+b₂=467; 4913a₃+b₃=5216; a₁=(21-b₁)/17=1.235-b₁/17; a₂=(467-b₂)/289=1.616-b₂/289; a₃=(5216-b₃)/4913=1.062-b₃/4913;

x=18:

18a₁+b₁=21;324a₂+b₂=467; 4913a₃+b₃=5832; a₁=(21-b₁)/18=1.167-b₁/18; a₂=(467-b₂)/324=1.441-b₂/324; a₃=(5216-b₃)/5832=0.894-b₃/5832.

From these examples we hope to get a common value for a_n, or a pattern in terms of n for a_n, and a pattern in terms of n for b_n. Unfortunately no such patterns emerge. 

It's always possible to find a formula for the nth term of a series when there are so few terms. The result is likely to be a formula with ungainly coefficients, but nevertheless it illustrates a method.

Let T(n)=a₀+a₁n+a₂n²+a₃n³, where a_n values are constants.

T(0)=a₀=1;

T(1)=a₀+a₁+a₂+a₃=21, a₁+a₂+a₃=20;

T(2)=a₀+2a₁+4a₂+8a₃=467, 2a₁+4a₂+8a₃=466⇒a₁+2a₂+4a₃=233;

T(3)= 3a₁+9a₂+27a₃=5215.

Now we have a system of 3 equations and 3 unknowns.

(1) T(2)-T(1)=a₂+3a₃=213

(2) T(3)-3T(1)=6a₂+24a₃=5155

(3) (2)-6(1)=6a₃=3877, a₃=3877/6⇒a₂=-3451/2⇒a₁=3298/3.

So T(n)=1+3298n/3-3451n²/2+3877n³/6. This produces the series: 1, 21, 467, 5216, 18145, 43131, ...

However, it's unlikely to be the right solution.

Comments

Hello @Rod, using that formula, If you don't mind, can you give me more the next 30 or 40 numbers in the sequence because I want to confirm something, I really don't know how to calculate using the formula that you have given thank you very much for your help...

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The series continues:

..., 84051, 144782, 229201, 341185, 484611, 663356, 881297, 1142311, 1450275, 1809066, 2222561, 2694637, 3229171, 3830040, 4501121, 5246291, 6069427, 6974406, 7965105, 9045401, 10219171, 11490292, 12862641, 14340095, 15926531, 17625826, 19441857, 21378501, 23439635, 25629136, 27950881, 30408747, 33006611, 35748350, 38637841, ...

As I said, the formula I gave may be one of many, and mine could easily be the wrong one.