A299402 Lexicographic first sequence of positive integers such that a(n)*a(n+1) has a digit 2, and no term occurs twice.

1, 2, 6, 4, 3, 7, 16, 8, 9, 14, 13, 17, 12, 10, 20, 11, 19, 15, 18, 24, 5, 25, 21, 22, 26, 27, 23, 34, 28, 29, 32, 31, 33, 37, 35, 36, 42, 30, 40, 38, 39, 48, 43, 44, 46, 45, 47, 41, 49, 50, 51, 52, 53, 54, 55, 59, 58, 56, 57, 60, 62, 61, 66, 64, 63, 67, 69, 68, 65, 80, 74, 71, 72, 73, 77, 76, 70, 75, 79, 78, 84, 83, 81, 82, 86, 87, 95, 96, 92

Offset: 1

Author

M. F. Hasler, Feb 22 2018

Keywords

• nonn
• base

Comments

A permutation of the positive integers.

Examples

			a(1) = 1 is the least positive integer, it has no other requirement to satisfy.
a(2) = 2 is the least positive integer > a(1) = 1, and a(2)*a(1) = 2 has a digit 2.
a(3) = 6 is the least positive integer > a(2) = 2 such that a(3)*a(2) (= 12) has a digit 2: The smaller choices 3, 4 or 5 do not satisfy this.
a(4) = 4 is the least positive integer > a(2) = 2 such that a(4)*a(3) (= 24) has a digit 2: The smaller choice 3 yields 3*6 = 18 and does not satisfy this.
Now, the least available positive integer a(5) = 3 is such that 3*4 = 12, which has again a digit 2. And so on.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A299403, A298974, ..., A298979 (analog with digit 3, ..., 9).

Cf. A299957, A299969, ..., A299988 (analog with addition instead of multiplication, and different digits).

Programs

Maple

N:= 100: # to get a(1)..a(n) where a(n+1) > N
S:= [$2..N]: nS:= N-1:
R:= 1: x:= 1; found:= true;
while found do
  found:= false;
  for i from 1 to nS do
    if member(2, convert(S[i]*x,base,10)) then
       found:= true;
       x:= S[i];
       R:= R,x;
       S:= subsop(i=NULL,S);
       nS:= nS-1;
       break
    fi
  od
od:
R; # Robert Israel, Feb 12 2023

PARI

a(n,f=1,d=2,a=1,u=[a])={for(n=2,n,f&&if(f==1,print1(a","),write(f,n-1," "a));for(k=u[1]+1,oo,setsearch(u,k)&&next;setsearch(Set(digits(a*k)),d)&&(a=k)&&break);u=setunion(u,[a]);while(#u>1&&u[2]==u[1]+1,u=u[^1]));a}