cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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).

Original entry on oeis.org

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
Offset: 1

Views

Author

Eric Angelini and Hans Havermann, Mar 07 2020

Keywords

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

Examples

			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.
		

Crossrefs

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

Programs

  • 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

Extensions

Edited by M. F. Hasler, Mar 18 2020