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A360146
Integers d such that the longest possible arithmetic progression (AP) of primes with common difference d has exactly 10 elements.
6
210, 420, 630, 840, 1050, 1260, 1470, 1680, 1890, 2100, 2520, 2730, 2940, 3150, 3360, 3570, 3780, 3990, 4200, 4410, 4830, 5040, 5250, 5460, 5670, 5880, 6090, 6300, 6510, 6720, 7140, 7350, 7560, 7770, 7980, 8190, 8400, 8610, 8820, 9030, 9450, 9660, 9870, 10080, 10290, 10500, 10710, 10920
OFFSET
1,1
COMMENTS
The 10 elements are not necessarily consecutive primes.
All the terms are positive multiples of 210 = 7# but are not multiples of 11 and also must not belong to A206045, where the first term is 1536160080; indeed, terms d' in A206045 correspond to the longest possible APs of primes that have exactly 11 elements with these common differences d'.
A342309(d) gives the first element of the smallest AP with 10 elements whose common difference is a(n) = d.
FORMULA
m is a term iff A123556(m) = 10.
EXAMPLE
d = 210 is a term because the longest possible APs of primes with common difference d = 210 all have 10 elements. The first such AP is (199, 409, 619, 829, 1039, 1249, 1459, 1669, 1879, 2089), then 2299 = 11*209.
d = 420 is another term because the longest possible APs of primes with common difference d = 420 all have 10 elements; the first such APs start with 52879, 3544939, ... The smallest one is (52879, 53299, 53719, 54139, 54559, 54979, 55399, 55819, 56239, 56659), then 57079 = 11*5189.
PROG
(PARI) A053669(n) = forprime(p=2, , if(n%p, return(p)));
f(n) = my(p=A053669(n)); for (i=1, p-1, if (!isprime(p+i*n), return(p-1))); p; \\ A123556
isok(n) = f(n) == 10; \\ Michel Marcus, Mar 10 2023
CROSSREFS
Common differences for longest possible APs of primes with exactly k elements: A007921 (k=1), A359408 (k=2), A206037 (k=3), A359409 (k=4), A206039 (k=5), A359410 (k=6), A206041 (k=7), this sequence (k=10), A206045 (k=11).
Sequence in context: A258359 A325991 A264664 * A147571 A306508 A358988
KEYWORD
nonn
AUTHOR
Bernard Schott, Mar 09 2023
STATUS
approved