

A333085


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).


1



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

1,1


COMMENTS

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


LINKS



EXAMPLE

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.


PROG

(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


CROSSREFS

Cf. A264646 (n concatenated with the nth digit of S).


KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



