The pattern to this sequence is that every 10 terms, the sequential differences doubles.
This sequence is a modified version of the sequence I asked about in this post: What's the mathematical formula for the sequence 3 6 12 18 30 42 66 90 ...?
The general formula for 3-6-12-18-30-42-66-90 is $$a_n=(\frac{15}{2}+\frac{3}{2}\cdot (-1)^{n+1})\cdot 2^{\lceil\frac{n-1}{2}\rceil}-6$$
However, the sequence 3-6-9-12-15-18-21-24-27-30-36-42-48-54-60-66-72-78-84-90-102-114-126-138-150 the sequential difference double every 10 terms instead of every 2 terms.
How could we adapt the formula above to support this new series of numbers?
$\endgroup$ 21 Answer
$\begingroup$Start by defining $a_0=0$ and note that $a_{10k}=10\cdot 3 + 10\cdot 6 + 10\cdot 12 + \ldots 10 \cdot 3\cdot 2^{k-1}=10\cdot 3 \cdot (2^k-1)$ and $a_{10k+m}=a_{10k}+m\cdot 3\cdot 2^k$ for $0 \le m \lt 10$ so
write $n=10k+m$ with $0 \le m \lt 10$ and $$a_n=a_{10k+m}=30(2^k-1)+3m2^k$$
$\endgroup$ 3
