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A333085
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Sequence of primes in which each term a(n) = 10*i + d gives the position i and value d of a digit in the concatenation of all terms (see comments).
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1
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11, 41, 61, 83, 113, 101, 151, 181, 233, 223, 263, 293, 353, 383, 419, 401, 479, 467, 541, 1009, 599, 631, 661, 691, 727, 751, 787, 797, 809, 877, 907, 919, 967, 991, 9001, 1031, 1063, 1151, 1171, 1187, 1201, 1237, 1303, 1321, 1361, 1373, 1453, 1481, 1597, 1601
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OFFSET
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1,1
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COMMENTS
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a(n) = p says "In position 'floor(p/10)' is a digit 'p mod 10'."
Each term must be the smallest possible prime not used earlier.
a(1447) = 19173719153, a(3868) = 371379371929.
No further record value up to n = 10^4. Some earlier record values: a(19) = 541, a(20) = 1009, a(35) = 9001, a(110) = 10007, ..., a(142) = 30011, ..., a(278) = 70001, ..., a(474) = 90001, a(523) = 101009, a(657) = 191339, a(902) = 300007, ..., a(1386) = 300221. - M. F. Hasler, Mar 19 2020
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LINKS
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EXAMPLE
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a(1) = 11 says "In position 1 is a 1" - which is compatible with the term itself. Since any term must have at least two digits, this is certainly the smallest possibility.
a(2) = 41 says "In position 4 is a 1" - which is indeed the last digit of a(2). There is no smaller solution: the term cannot refer to the 2nd nor the 3rd digit of the sequence, since neither 21 nor 33 is prime.
a(3) = 61 says "In position 6 is a 1"; again, there's no smaller solution.
a(4) = 83 says "In position 8 is a 3", and this is again the smallest solution.
a(5) = 113 says "In position 11 is a 3": again the last digit of a(5) itself, and there is no smaller solution.
a(6) = 101 says: "In position 10 is a 1." (This term wasn't possible earlier, but at this position it is.)
a(19) = 541 says "In position 54 there is a 1", which is not yet there: position 54 is the first digit of a(20). So a(20) must start with a digit 1, and the smallest solution is a(20) = 1009, predicting a digit 9 in position 100.
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PROG
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(PARI) A333085_vec(n, d=[], U=[], F=[], k)={vector(n, i, forprime(p=11, , setsearch(U, p)&& next; k=divrem(p, 10); k[1] > #d + logint(k[1], 10)+1 || k[2] == if( k[1]<=#d, d[ k[1]], digits( k[1] )[ k[1]-#d ]) || next; for(i=1, #F, F[i][1] > #d + logint(p, 10)+1 && break; F[i][2] == digits(p)[ F[i][1]-#d ] || next(2)); d=concat(d, digits(p)); break); while(#F && F[1][1]<=#d, F=F[^1]); k[1]>#d && F=setunion(F, [k]); U=setunion(U, primes([k[1], k[1]+1]*10)); [10, 1]*k)} \\ For n > 500, use the much faster code given in LINKS. - M. F. Hasler, Mar 18 2020
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CROSSREFS
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Cf. A264646 (n concatenated with the n-th digit of S).
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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