Sequences and Patterns

This pattern could also be represented by:

a(n) = (1/2)(n-1)^3 - (3/2)(n-1)^2 + 4(n-1) + 2

in which case a(41) = 29766

In general there are an infinite number of ways of representing these simple series!

The equation to this sequence is 3N-1 

So then you just plugin 41 and you get 122 

Hmmmm.... Guest 1,       '14' doesn't fit the  3n-1  solution....

Observe the pattern: 2,5,8,14,... if the pattern continues, what is the 41st number?

$\begin{array}{r|r} \hline & \text{Primes of the form 4n-1 } & \\ & p & \frac{3\cdot p - 1}{4} \\ \hline 1 & 3 & a_1 = \frac{3\cdot 3 - 1}{4} =2 \\ 2 & 7 & a_2 = \frac{3\cdot 7 - 1}{4} =5 \\ 3 & 11 & a_3 = \frac{3\cdot 11 - 1}{4} = 8 \\ 4 & 19 & a_4 = \frac{3\cdot 19 - 1}{4} = 14 \\ 5 & 23 & a_5 = \frac{3\cdot 23 - 1}{4} = 17 \\ 6 & 31 & 23 \\ 7 & 43 & 32 \\ 8 & 47 & 35 \\ 9 & 59 & 44 \\ 10 & 67 & 50 \\ 11 & 71 & 53\\ 12 & 79 & 59 \\ 13 & 83 & 62 \\ 14 & 103 & 77 \\ 15 & 107 & 80 \\ 16 & 127 & 95 \\ 17 & 131 & 98\\ 18 & 139 & 104 \\ 19 & 151 & 113\\ 20 & 163 & 122 \\ 21 & 167 & 125 \\ 22 & 179 & 134 \\ 23 & 191 & 143 \\ 24 & 199 & 149 \\ 25 & 211 & 158 \\ 26 & 223 & 167 \\ 27 & 227 & 170 \\ 28 & 239 & 179\\ 29 & 251 & 188\\ 30 & 263 & 197 \\ 31 & 271 & 203 \\ 32 & 283 & 212 \\ 33 & 307 & 230 \\ 34 & 311 & 233 \\ 35 & 331 & 248 \\ 36 & 347 & 260 \\ 37 & 359 & 269 \\ 38 & 367 & 275 \\ 39 & 379 & 284 \\ 40 & 383 & 287 \\ 41 & 419 & a_{41} = \frac{3\cdot 419 - 1}{4} = 314 \\ 42 & 431 & 323 \\ 43 & 439 & 329 \\ 44 & 443 & 332 \\ 45 & 463 & 347 \\ \cdots & \cdots & \cdots \\ \hline \end{array}$