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A111057
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Minimal set in the sense of A071062 of prime-strings in base 12 for primes of the form 4n+1.
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0
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5, 13, 37, 73, 97, 109, 313, 337, 373, 409, 421, 577, 601, 661, 709, 1009, 1033, 1093, 1129, 1489, 1609, 1669, 3457, 7537, 12721, 13729, 17401, 17569, 19009, 19141, 20593, 20641, 165877, 208501, 221173, 225781, 226201, 226357, 228793, 246817, 246937, 248821, 1097113, 2695813, 2735269, 2736997, 2737129, 32555521, 388177921
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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J. Shallit, Minimal primes, J. Recreational Mathematics, vol. 30.2, pp. 113-117, 1999-2000.
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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base,fini,full,nonn,uned
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AUTHOR
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STATUS
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approved
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