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A299402
Lexicographic first sequence of positive integers such that a(n)*a(n+1) has a digit 2, and no term occurs twice.
15
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
1,2
COMMENTS
A permutation of the positive integers.
LINKS
EXAMPLE
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.
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
PROG
(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}
CROSSREFS
Cf. A299403, A298974, ..., A298979 (analog with digit 3, ..., 9).
Cf. A299957, A299969, ..., A299988 (analog with addition instead of multiplication, and different digits).
Sequence in context: A333520 A003571 A068457 * A286451 A102510 A206225
KEYWORD
nonn,base
AUTHOR
M. F. Hasler, Feb 22 2018
STATUS
approved