OFFSET
0,1
COMMENTS
The smallest (n+1)-digit prime 5-tuple is given by 10^n + a(n) + D, with either D = {0, 2, 6, 8, 12} or D = {0, 4, 6, 10, 12}. N = 0 is the only case where the last member of the 5-tuple has one digit more than the first member.
Numerical evidence strongly suggests the conjecture that 0 < a(n) < 10^n for all n > 0, but not even the existence of infinitely many prime 5-tuples is proved.
Some further isolated terms, due to Norman Luhn et al., giving the start of the smallest 500, 600, 700, ..., 1200 digit quintuplets of first or second type:
a(499) = min(58195471283341, 69672492141807),
a(599) = min(319491304676641, 12754947401547),
a(699) = min(2254633393747621, 209264286017367),
a(799) = min(2117758391972791, 1299258655252617),
a(899) = min(2365663735968811, 1484244113736867),
a(999) = min(3554007760224751, 3818999670116007),
a(1099) = min(26317044823878361, 15720821612555937),
a(1199) = min(20483870459152351, 7033048489975137).
Terms through a(399) may be determined by taking the minima of those in the linked tables for quintuplets by Norman Luhn et al. - Michael S. Branicky, Jul 24 2021
The first member of the quintuplets of the first type always ends in digit 1 (except for the 5-tuple (5, 7, 11, 13, 17) corresponding to a(0)), for the second type it always ends in digit 7. Therefore all a(n), n > 0, end in a digit 1 or 7, which indicates the type of the 5-tuple, i.e., the set D that has to be added to 10^n + a(n) to get the whole 5-tuple. - M. F. Hasler, Aug 04 2021
LINKS
M. F. Hasler, Table of n, a(n) for n = 0..399 (terms up to a(51) from Michael S. Branicky), Aug 04 2021.
Norman Luhn, Primzahltupel, prime k-tuple: Smallest-n-digit-prime-quintuplets, on mathematikalpha.de, 2020. Tables.
EXAMPLE
a(0) = 4 because {5, 7, 11, 13, 17} is the smallest prime 5-tuple and it starts with the single-digit prime 10^0 + a(0) = 5 = A022006(1).
a(1) = 1 because 10^1 + 1 = 11 = A022006(2) is the 2-digit prime to start a prime 5-tuple {11, 13, 17, 19, 23}, again of the first type.
a(2) = 1 and a(3) = 481 because 10^2 + 1 = 101 = A022006(3) and 10^3 + 481 = 1481 = A022006(4) are the smallest 3-digit, resp. 4-digit, initial members of a prime 5-tuple, both again of the first type.
a(4) = 5727 because 10^4 + 5727 = 15727 = A022007(6) is the smallest 5-digit initial member of a prime 5-tuple, now of the second type.
It appears that for all n > 0, a(n) < 10^n, so that the primes are of the form 10...0XXX where XXX = a(n) and 0...0 stands for a string of zero or more digits 0.
PROG
(PARI) apply( {A343635(n, q=[1..4], i=0)=forprime(p=10^n, , (q[1+i]+12==q[i++]=p) && return(p-12-10^n); i>3 && i=0)}, [0..15]) \\ Shorter but slightly slower (?)
(PARI) apply( {A343635(n, i=ispseudoprime, q)=forprime(p=10^n, , i(p+12) && i(p+6) && (p+6 > q=nextprime(p+2)) && i(q+6) && return(p-10^n))}, [0..15])
(Python)
from sympy import nextprime
def a(n):
p = nextprime(10**n)
q = nextprime(p); r = nextprime(q); s = nextprime(r); t = nextprime(s)
while p < 10**(n+1):
if t - p == 12: return p - 10**n
p, q, r, s, t = q, r, s, t, nextprime(t)
return 0
print([a(n) for n in range(14)]) # Michael S. Branicky, Jul 24 2021
CROSSREFS
KEYWORD
nonn,base,hard
AUTHOR
M. F. Hasler, Jul 17 2021
STATUS
approved