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.

A241020 a(n) is the smallest j such that the n-digit number consisting of a 1 in position j and 7's in the other n-1 positions is a prime, or 0 if no such prime exists.

Original entry on oeis.org

1, 0, 1, 2, 0, 0, 6, 0, 1, 2, 0, 2, 1, 0, 3, 0, 0, 5, 2, 0, 6, 4, 0, 7, 4, 0, 12, 0, 0, 19, 8, 0, 26, 5, 0, 0, 33, 0, 6, 11, 0, 1, 23, 0, 18, 34, 0, 15, 0, 0, 1, 22, 0, 1, 50, 0, 32, 15, 0, 15, 25, 0, 21, 10, 0, 29, 47, 0, 0, 11, 0, 56, 14, 0, 2, 0, 0, 54, 3
Offset: 2

Views

Author

Michel Lagneau, Apr 15 2014

Keywords

Comments

Previous name: Let x(1)x(2)... x(n) denote the decimal expansion of a number p having an index j such that x(j) = 1 and x(i) = 7 for i <> j. The sequence lists the smallest index j such that p is prime, or 0 if no such prime exists.
Except 0, the corresponding primes are 17, 0, 1777, 71777, 0, 0, 77777177, 0, 1777777777, 71777777777, 0, 7177777777777, 17777777777777, 0, 7717777777777777, 0, 0, 7777177777777777777, ... .

Crossrefs

Programs

  • Maple
    with(numtheory):nn:=80:T:=array(1..nn):
       for n from 2 to nn do:
         for i from 1 to n do:
         T[i]:=7:
         od:
           ii:=0:
           for j from 1 to n while(ii=0)do:
           T[j]:=1:s:=sum('T[i]*10^(n-i)', 'i'=1..n):
             if type(s,prime)=true
             then
             ii:=1: printf(`%d, `,j):
             else
             T[j]:=7:
             fi:
             od:
              if ii=0
               then
               printf(`%d, `,0):
               else
              fi:
           od:
  • Mathematica
    Flatten[Position[IntegerDigits[#],1]&/@Table[Select[FromDigits/@Permutations[ Join[ {1},PadRight[ {},n,7]]],PrimeQ]/.{}->{0,0},{n,80}][[;;,1]]/.{}->0] (* Harvey P. Dale, Jul 21 2024 *)
  • Python
    from sympy import isprime
    def a(n):
        if (1+7*(n-1))%3 == 0:
            return 0
        base = (10**n-1)//9*7
        for j in range(1, n+1):
            t = base - 6*10**(n-j)
            if isprime(t):
                return j
        return 0
    print([a(n) for n in range(2, 81)]) # Michael S. Branicky, Jun 02 2024

Formula

a(n) = 0 when 7*(n-1) + 1 mod 3 = 0. - Michael S. Branicky, Jun 02 2024

Extensions

Name simplified by Michael S. Branicky, Jun 02 2024