Problems & Puzzles: Puzzles

Puzzle 71.- Consecutive primes and Cunningham chains

A Cunningham chain of 1st order and length L is a sequence of L numbers: p, 2p + 1, 4p + 3, 8p + 7, … 2^(L - 1)p + 2^(L - 1) - 1, such that all the numbers are prime.

A Cunningham chain of 2nd order and length L is a sequence of L numbers: p, 2p - 1, 4p - 3, 8p - 7, … 2^(L - 1)p - (2^(L - 1) - 1), such that all the numbers are prime.

Below you can find my little search for the least sets of K consecutive primes for Cunningham chains of length 2 <= L <= 4, first and second order.

1st Order L=2 L=3 L=4
K  p such that p & 2p + 1 are primes p such that p, 2p + 1, &  4p+3 are primes p such that p, 2p + 1, 4p +3 & 8p +7 are primes
2 2, 3 1889, 1901 34139879, 34139909
3 2, 3, 5 66961409, 66961439, 66961451 ?
4 1433849,  1433891,  1433903,  1433909 ? ?
5 9816899,  9816923,  9816941,  9816953,  9816959 ? ?
6 445480319,  445480331,  445480361,  445480391,  445480421,  445480439  (This is my A047984) ? ?
7 ? ? ?

 

2nd Order L=2 L=3 L=4
K  p such that p & 2p - 1 are primes P such that p, 2p - 1, &  4p - 3 are primes P such that p, 2p - 1, 4p - 3 & 8p - 7 are primes
2 31, 37 25609   25621 66954961        66954991
3 3169,  3181,  3187 26923669        26923681        26923711 ?
4 63199,  63211,  63241,  63247 ? ?
5 17742877,  17742889,  17742919, 17742931,  17742937 ? ?
 

6

 

 86257279,  86257387,  86257411,  86257417,  86257459,  86257489 ? ?
7 ? ? ?

Questions: Can you extend and/or complete the table?

Jud McCranie (17/10/99) sent the following comments: " I worked on puzzle 71, but didn't find anything new.  I tried L=2 - first order and second order, p<2^32.  I didn't find any longer sequences.  I verified your results for L=2 in both cases, except that for 2nd order 2 & 3 are the smallest solutions instead of 31 & 37"

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Giovanni Resta wrote (Nov. 2004):

(Consecutive primes and Cunningham chains) 1st order:
L=2 k=7 (298098924131, 298098924143, 298098924173, 298098924209,
298098924251, 298098924341, 298098924443)
L=3 k=4 (58308965339, 58308965369, 58308965381, 58308965411)
L=5 k=2 (5133129899, 5133129929)
(no other new solutions for L<=5 and p<=321,839,692,501)

2nd order:
L=5 k=2 (5412418021, 5412418051)
(no other new solutions for L<=5 and p<=305,882,832,161)
 

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On Set 8, 2026 Diego González Gauss wrote:

I have a formal certificate for this if useful.
2632484059949, 2632484060009, 2632484060099.
These are three consecutive ordinary primes. Each starts a first-kind
Cunningham chain of four primes. The three chains are:
2632484059949, 5264968119899, 10529936239799, 21059872479599;
2632484060009, 5264968120019, 10529936240039, 21059872480079;
2632484060099, 5264968120199, 10529936240399, 21059872480799.
The starting value is minimal, using the published finite prime-gap result
of Richard P. Brent, Mathematics of Computation 35 (1980), pp. 1435-1436:
ordinary-prime gaps are at most 652 for starting primes <=4.444*10^12.
Above 5, every four-prime-chain starter is 29 modulo 30. Therefore any earlier
qualifying consecutive-prime triple has two positive gaps in {30,60,...,630}.
Of the 441 ordered shapes, 54 are impossible modulo 7. Every remaining shape
has been completely searched through the displayed starting prime, using two
independent affine-sieve implementations. The complete survivor lists agree
literally. Exact arithmetic certificates exclude every retained tuple except
this triple. Starts at most 20000 are separately checked by exact sieving.
The attached package contains the source programs, candidate lists, full
coverage records, recursive Lucas primality certificates and proper-divisor
consecutivity certificates. The historical prime-gap computation was not
rerun here; no previous Cunningham minimum is assumed.
The chains have forward lengths exactly four from the displayed starters.
The third starter has a prime predecessor, so backward maximality is not
claimed and is not required by the puzzle's stated definition.

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