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
Examples
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.
Links
- Jean-Marc Rebert, doubleDecomposition
- Carlos Rivera, Puzzle 1077. These numbers that are..., The Prime Puzzles and Problems Connection.
Programs
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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
Extensions
a(15)-a(19) from Chai Wah Wu, Apr 21 2022