General term of sequence 2 An other sequence that arise in context of formula for number of partitions of number natural in parts non greater than 5 is
$81,123,167,229,295,381,473,587, 709, 855,1011,1193,1387,1609,1845,2111,2393,2707,3039,..$
If we try method of finite differences we get following sequences
$ 42 ,44 ,62, 66 ,86 ,92,114, 122,146, 156,   182 ,    194,  222,….$
$ 2 ,   18,    4 ,   20,     6,   22,      8 ,    24,    10,    26,     12,      28,,…$
$16 ,  -14,  16,  -14,  16,    -14,   16,    -14 ,   16 ,   -14,    16,…$
$-30,   30 , -30 ,   30,   -30,   30,    -30,   30,      -30 , 30,…$
Method of finite differences is useless. How to find general term.
 A: Define $\Delta a_n = a_{n+1} - a_n$ (which is the forward difference) and define inductively $\Delta^{(k)}a_n = \Delta^{(k-1)} a_{n+1} - \Delta^{(k-1)} a_n$ (which is the $k^{\text{th}}$ forward difference). It can easily be shown that
$$
\Delta^{(k-1)}a_n = \sum_{i=1}^{n-1} \Delta^{(k)}a_i + \Delta^{(k-1)}a_1
$$
because it is just a telescopic sum. Now consider your sequence ; beginning from the second forward difference (or the third, say fourth if you have a bad-eye), can you guess the other forward differences?
Hint : One sees that 
$$
\Delta^{(2)}a_n = n+1 + 15 \left( \frac{1 + (-1)^n}2 \right)
$$
This gives you the sequence $2, 18, 4, 20, 6, 22, ...$ 
Now you have, using the above identity
$$
\Delta^{(1)}a_n = \sum_{i=1}^{n-1} \Delta^{(2)} a_i + 42 = \sum_{i=1}^{n-1} \left( i+1 + 15 \left( \frac{1 + (-1)^i}2 \right) \right) + 42
$$
You can easily compute this sum if you know elementary identities on summations. (If you can't guess the second forward difference, start with the fourth, and keep using this trick until you get to the second.)
One computes and finds
$$
\Delta a_n = n^2 + 8n + \frac{67}2 - \frac{15}2 \left( \frac{1+(-1)^n}2 \right).
$$
I leave the computation for you as an exercise, and using the identities 
$$
\sum_{i=1}^k i = \frac{k(k+1)}2, \qquad \sum_{i=1}^k i^2 = \frac{k(k+1)(2k+1)}6, \qquad \sum_{i=1}^{n-1} (-1)^i = - \left( \frac{1 + (-1)^n}2 \right)
$$
you can easily find $a_n$, but I don't have the will of computing right now. =)
Hope that helps,
