OFFSET
2,1
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(n-1, 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
CROSSREFS
KEYWORD
hard,more,nonn
AUTHOR
Amarnath Murthy, Aug 06 2002
EXTENSIONS
Edited and extended by Robert G. Wilson v, Aug 09 2002
Edited by N. J. A. Sloane, Feb 05 2014 following suggestions from Don Reble, T. D. Noe and Harvey P. Dale.
a(10) from Ray Chandler, Feb 05 2014
a(12)-a(14) from Don Reble, Feb 06 2014
STATUS
approved