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Maximal gap between successive elements of Pn with n >= 3. Pn : primes p <= A002110(n) such that q=2*A002110(n)-p is prime. (p, q) is a pair of centered primes at n-th primorial.
(history; published version)
#27 by Charles R Greathouse IV at Thu Jul 30 10:26:15 EDT 2015
STATUS

editing

approved

#26 by Charles R Greathouse IV at Thu Jul 30 10:25:48 EDT 2015
DATA

6, 16, 44, 122, 268, 556, 886, 1446, 1964, 2900

EXTENSIONS

a(12) from Charles R Greathouse IV, Jul 30 2015

STATUS

approved

editing

Discussion
Thu Jul 30
10:26
Charles R Greathouse IV: 139 CPU-hours.
#25 by Charles R Greathouse IV at Tue Jul 28 11:36:02 EDT 2015
STATUS

editing

approved

#24 by Charles R Greathouse IV at Tue Jul 28 11:35:44 EDT 2015
DATA

6, 16, 44, 122, 268, 556, 886, 1446, 1964

EXTENSIONS

a(11) from Charles R Greathouse IV, Jul 28 2015

#23 by Charles R Greathouse IV at Fri Jul 24 11:15:03 EDT 2015
PROG

(PARI) a(n)=my(P=2*prod(i=1, n, prime(i)), r, last=P); forprime(p=2, P/2, if(isprime(P-p), if(last, r=max(r, p-last)); last=p)); r \\ Charles R Greathouse IV, Jul 24 2015

#22 by Charles R Greathouse IV at Fri Jul 24 08:42:06 EDT 2015
NAME

Maximal gap between successive prime elements of Pn with n >= 3. Pn : primes p <= A002110(n) such that q=2*A002110(n)-p is prime. (p ,, q) is a pair of centered primes at n-th primorial.

DATA

6, 16, 44, 122, 268, 556, 886, 1446

PROG

(PARI) lstp(n) = {my(v = []); pn = prod(i=1, n, prime(i)); forprime(p=1, pn-1, if (isprime(2*pn-p), v = concat(v, p))); my(w = vector(#v-1, k, v[k+1] - v[k])); vecmax(w); } \\ Michel Marcus, Jul 24 2015

(PARI) a(n)=my(P=2*prod(i=1, n, prime(i)), r, last); forprime(p=2, P/2, if(isprime(P-p), if(last, r=max(r, p-last)); last=p)); r \\ Charles R Greathouse IV, Jul 24 2015

EXTENSIONS

a(10) from Charles R Greathouse IV, Jul 24 2015

STATUS

proposed

editing

Discussion
Fri Jul 24
09:32
Michel Marcus: Yes Charles P/2
#21 by Michel Marcus at Fri Jul 24 07:32:05 EDT 2015
STATUS

editing

proposed

#20 by Michel Marcus at Fri Jul 24 07:31:34 EDT 2015
EXAMPLE

For n=3, primorial(3)=A002110(3) is 30 and P3 is (7, 13, 17, 19, 23, 29) because (53, 47, 43, 41, 37, 31) are all prime . The maximum gap between two consecutive terms of P3 is 6, obtained for (7,13) or (23,29), so a(3)=6.

Discussion
Fri Jul 24
07:32
Michel Marcus: Ok, fine. Added prog
#19 by Michel Marcus at Fri Jul 24 07:30:55 EDT 2015
PROG

(PARI) lstp(n) = {my(v = []); pn = prod(i=1, n, prime(i)); forprime(p=1, pn-1, if (isprime(2*pn-p), v = concat(v, p))); my(w = vector(#v-1, k, v[k+1] - v[k])); vecmax(w); } \\ Michel Marcus, Jul 24 2015

STATUS

proposed

editing

#18 by Jean-Marc Rebert at Fri Jul 24 07:16:43 EDT 2015
STATUS

editing

proposed