OFFSET
1,1
COMMENTS
Maple worksheet available upon request. Here is the minimal set of primes of the form 4n+1 in base 12, where X is ten and E is eleven. 5, 11, 31, 61, 81, 91, 221, 241, 271, 2X1, 2E1, 401, 421, 471, 4E1, 701, 721, 771, 7X1, X41, E21, E71, 2001, 4441, 7441, 7E41, X0X1, X201, E001, E0E1, EE01, EE41, 7EEE1, X07E1, X7EE1, XX7E1, XXXX1, XXEE1, E04X1, EXX01, EXXX1, EEEE1, 44XXX1, XX00E1, XEXXE1, XEEXE1, XEEEX1, XXX0001, XX000001. Note that the last prime in the set is the same as the last prime in the minimal set of all primes. See A110600. I am checking certain ranges past this last prime but flow-charting the possibilities leads me to believe I have found the full sequence. The minimal set of prime strings in base 12 for primes of the form 4n+3 is [3, 7, E] since every 4n+3 prime greater than 3 ends in either 7 or E.
LINKS
J. Shallit, Minimal primes, J. Recreational Mathematics, vol. 30.2, pp. 113-117, 1999-2000.
EXAMPLE
a(11)=421="2E1" since the pattern "*2*E*1*" does not occur in any previously found prime of the form 4n+1. Assuming all previous members of the list have been similarly recursively constructed, then "401" (577 in base 10) is the next prime in the list. The basic rule is: if no substring of p matches any previously found prime, add p to the list. The basic theorem of minimal sets says that this process will terminate, that is, the minimal set is always finite.
CROSSREFS
KEYWORD
base,fini,full,nonn,uned
AUTHOR
Walter Kehowski, Oct 06 2005
STATUS
approved