What is the next number in the sequence: $24, 30, 33 , 39 , 51,…$

What is the next number in the sequence:

$24, 30, 33 , 39 , 51,...$

Here all numbers are divisible by $3$, difference between numbers are $6,3,6,12,...$ but I can't find common relationship in the sequence.

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Let me introduce you to this site. – David Mitra Jun 21 '13 at 16:29
@DavidMitra Only three sequences found with this? Tsk, tsk .. – Hagen von Eitzen Jun 21 '13 at 16:36

Looks to me like each term is the preceding term, plus the preceding term's sum of digits. The sequence would then be: $24, 30, 33, 39, 51, 57, 69, 84, 96,...$

Note that if a number is divisible by 3, then the sum of its digits is also divisible by 3, resulting in all of the numbers in the sequence being divisible by 3. Because of this, there are probably many other ways to define the sequence, but I find the one I gave to be fairly natural.

Edit: Just noticed @DavidMitra's earlier comment which links to an OEIS page which includes the answer I gave, but I'll leave my answer here as reference.

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24 or 45 or 1000 or 286575! Perhaps anything will work. A theorem states that given any values, a polynomial exists which takes all those values. So this question has a pretty basic flaw in it!

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Why would you only consider polynomials? – Eckhard Jun 21 '13 at 16:50
@Eckhard If you consider other functions than polynomials, then the answer would be anything and something more... The point is that polynomials already produce everything... – N. S. Jun 21 '13 at 17:05
@user83376 Actually anything works. For example, is $P(X)$ is a polynomial (or any function) such that $P(1)=24, P(2)=30, P(3)=33, P(4)=39, P(5)=51, P(6)=0$ then $f(x)=P(x) + \frac{1}{6!}(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)\cdot [anything]$ produces the desired outcome. So the answer is obviously RED, since we can replace anything by red... – N. S. Jun 21 '13 at 17:08

I see two sequences interwoven: $a_n=\{24,33,51,...\}$ and $b_n=\{30,39,...\}$

I'd continue them as $$\begin{eqnarray} a_n &=& \{24,24+9,33+18,51+27,78+36,...\}\\ & = & 24+ 9 \cdot\{0,1,3,6,10,...\} \\ & = & 24 + 9 \cdot \binom{n}{2} \\ & = & \{24,33,51,78,114,...\} \end{eqnarray}$$ and $$\begin{eqnarray} b_n &=&\{30,30+9,39+18,57+27,...\} \\ &=& 30+9\cdot \{0,1,3,6,...\} \\ & =& 30 + 9 \cdot \binom{n}{2} \\ & = & \{30,39,57,84,120,...\} \end{eqnarray}$$

The indexes should then be adapted to reproduce the intervowen sequence: $$c_n = \{24,30,33,39,51,57,78,84,114,120...\}$$

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I see five sequences intervowen. – Did Jun 21 '13 at 17:23
@did : och mann... ;-) – Gottfried Helms Jun 21 '13 at 17:29
Naja, wirklich. :-) – Did Jun 21 '13 at 18:10

I think the sequence may be a 4th degree sequence.

24 30 33 39 51
+6 +3 +6 +12
-3 +3 +6
+6 +3
-3
In a 4th degree sequence, the difference of the difference of the difference of the difference is constant, so the next number would be 75:

24 30 33 39 51 69
+6 +3 +6 +12 +18
-3 +3 +6 +6
+6 +3 +0
-3 -3

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 24,30,33,39,51,....?
1st term =  24 and 2+4=6
now 24+6 =  30
2nd term =  30 and 3+0=3
now 30+3 =  33
3rd term =  33 and 3+3=6
now 33+6 =  39
4th term =  39 and 3+9=12
now 39+12 = 51
5th term =  51 and 5+1=6
now 51+6 =  57
6th term =  57

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