
EXAMPLE

Consider the prime sequence starting from 2: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, ... The first difference sequence is 1, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, ... There exists no prime sequence > 2 which has a first difference of 1.
Therefore we set offset = 2 and begin with the second prime (3, with its first forward difference of 2) and look for the first prime above 3 with a forward difference of 2. That number is 5, so a(2) = 5.
Next, n=3, start now with the third prime p(3)=5, with two forward differences of 2,4. The next prime above 5 which starts out with differences of 2,4 is 11. So a(3) = 11.
Next start with the fourth prime  7, with three forward differences of 4,2,4. The next prime above 7 which starts out with those difference is 13. So a(4) = 13.
a(5) = 101: The differences between the primes 101, 103, 107, 109 & 113 are 2, 4, 2 & 4, the first to match the differences between 11, 13, 17, 19 & 23.


MATHEMATICA

Do[s = Table[Prime[i + 1]  Prime[i], {i, n + 1, 2n}]; p = 0; q = 0; a = s; k = n + 2; While[p = q; q = Prime[k]; a = Drop[a, 1]; a = Append[a, q  p]; s != a, k++ ]; Print[Prime[PrimePi[q]  n]], {n, 1, 8}]


PROG

(PARI) A073615 = n>{d=vector(n1, i, prime(n+i)prime(n)); forprime(pm=prime(n+1), 9e9, for(k=1, #d, isprime(pm+d[k])next(2)); p=pm; for(k=1, #d, (pm+d[k]==p=nextprime(p+1))next(2)); return(pm))} \\ Yields a(8) in 0.1 sec, but is too slow beyond that.  M. F. Hasler, Feb 05 2014
