|
| |
|
|
A111056
|
|
Minimal set of prime-strings in base 10 for primes of the form 4n+3 in the sense of A071062.
|
|
4
|
|
|
|
3, 7, 11, 19, 59, 251, 491, 499, 691, 991, 2099, 2699, 2999, 4051, 4451, 4651, 5051, 5651, 5851, 6299, 6451, 6551, 6899, 8291, 8699, 8951, 8999, 9551, 9851, 22091, 22291, 66851, 80051, 80651, 84551, 85451, 86851, 88651, 92899, 98299, 98899
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
The basic rule is: if no substring of p matches any smaller prime of the form 4n+3, add p to the list. The basic theorem of minimal sets says that the minimal set is always finite.
The sequence b-file is complete except for the number (2*10^19153 + 691)/9, i.e., the decimal number consisting of 19151 "2"s followed by two "9"s. - Curtis Bright, Jan 23 2015
|
|
|
LINKS
|
J. Shallit, Minimal primes, J. Recreational Mathematics, vol. 30.2, pp. 113-117, 1999-2000.
|
|
|
EXAMPLE
|
a(5) is not 23, even though 23 is the fifth prime of the form 4n+3, since 23 contains a(1)=3 as a substring. Similarly: 31 and 43 contain 3 and 47 contains a(2)=7. Thus a(5)=59.
This sequence contains 2099 since 2, 0, 9, 20, 09, 99, 209, 299, and 099 are not primes of the form 4n+3.
(End)
|
|
|
MAPLE
|
with(StringTools); wc := proc(s) cat("*", Join(convert(s, list), "*"), "*") end; M3:=[]: wcM3:=[]: p:=1: for z from 1 to 1 do for k while p<10^11 do p:=nextprime(p); if k mod 100000 = 0 then print(k, p, evalf((time()-st)/60, 4)) fi; if p mod 4 = 3 then sp:=convert(p, string); if andmap(proc(w) not(WildcardMatch(w, sp)) end, wcM3) then M3:=[op(M3), p]; wcM3:=[op(wcM3), wc(sp)]; print(p) fi fi od od; # Let it run for a couple of days.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
base,fini,nonn,uned
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
