A352065 a(n) is the least prime p that starts a run of 2n+1 consecutive primes whose product is a sum of the same number of (others or same) consecutive primes.

2, 29, 293, 229, 3119, 67, 18121, 59629, 10247, 15391, 5903, 24007, 11783, 39359, 21013, 104917, 38273, 61129, 23663, 2423

Offset

0,1

Links

Table of n, a(n) for n=0..19.

Jean-Marc Rebert, doubleDecomposition

Carlos Rivera, Puzzle 1077. These numbers that are..., The Prime Puzzles and Problems Connection.

Example

a(0) = 2, because 2 = 2, and there is no smaller prime.
a(1) = 29, because 29 * 31 * 37 = 33263 = 11083 + 11087 + 11093, and there is no smaller prime that starts a run of 3 consecutive primes whose product is a sum of 3 consecutive primes.
a(2) = 293, because 293 * 307 * 311 * 313 * 317 = 2775683761181 = 555136752211 + 555136752221 + 555136752227 + 555136752251 + 555136752271, and there is no smaller prime that starts a run of 5 consecutive primes whose product is a sum of 5 consecutive primes.
Let y be the product of the 2n+1 consecutive primes starting with a(n) and let q be the first prime in the sum of 2n+1 consecutive primes. For n = 0..3 we have:
.
  n  2n+1  a(n)                  y  #dgts(y)                 q  #dgts(q)
  -  ----  ----  -----------------  --------  ----------------  --------
  0     1     2                  2         1                 2         1
  1     3    29              33263         5             11083         5
  2     5   293      2775683761181        13      555136752211        12
  3     7   229  52139749485151463        17  7448535640735789        16
.
For more examples, see the "doubleDecomposition" link.

Prog

(Python)
from math import prod
from sympy import prime, nextprime, prevprime
def A352065(n):
    plist = [prime(k) for k in range(1, 2*n+2)]
    pd = prod(plist)
    while True:
        mlist = [nextprime(pd//(2*n+1)-1)]
        for _ in range(n):
            mlist = [prevprime(mlist[0])]+mlist+[nextprime(mlist[-1])]
        if sum(mlist) <= pd:
            while (s := sum(mlist)) <= pd:
                if s == pd:
                    return plist[0]
                mlist = mlist[1:]+[nextprime(mlist[-1])]
        else:
            while (s := sum(mlist)) >= pd:
                if s == pd:
                    return plist[0]
                mlist = [prevprime(mlist[0])]+mlist[:-1]
        pd //= plist[0]
        plist = plist[1:] + [nextprime(plist[-1])]
        pd *= plist[-1] # Chai Wah Wu, Apr 21 2022

Crossrefs

Cf. A203619, A323052.

Sequence in context: A124301 A276196 A077023 * A101750 A160952 A088615

Adjacent sequences: A352062 A352063 A352064 * A352066 A352067 A352068

Keyword

nonn,hard,more

Author

Jean-Marc Rebert, Mar 05 2022

Extensions

a(15)-a(19) from Chai Wah Wu, Apr 21 2022

Status

approved