

A276633


a(n) = smallest integer not yet in the sequence with no digits in common with a(n1) and a(n2); a(0)=0, a(1)=1.


8



0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 22, 33, 11, 20, 34, 15, 26, 30, 14, 25, 36, 17, 24, 35, 16, 27, 38, 19, 40, 23, 18, 44, 29, 13, 45, 28, 31, 46, 50, 12, 37, 48, 21, 39, 47, 51, 32, 49, 55, 60, 41, 52, 63, 70, 42, 53, 61, 72, 43, 56, 71, 80, 54, 62, 73, 58, 64, 77, 59, 66, 74, 81, 65, 79
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OFFSET

0,3


COMMENTS

The sequence is not a permutation of the positive integers. E.g., 123456789 and 1023456789 (the smallest pandigital number) are not members.
Numbers n such that a(n)=n: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 52, 147, 1619, 6140, ...
The sequence is infinite, since all digits in a(n3) are allowed in a(n).  Robert Israel, Sep 20 2016


LINKS

Zak Seidov and David A. Corneth, Table of n, a(n) for n = 0..19999 (First 2001 terms from Zak Seidov).


EXAMPLE

From David A. Corneth, Sep 22 2016: (Start) Each number can consist of 2^101 sets of distinct digits, i.e., classes. For example, 21132 is in the class {1, 2, 3}. We don't include a number without digits. For this sequence, we can also exclude numbers with only the digit 0. This leaves 1022 classes. We create a list with a place for each class containing the least number from that class not already in the sequence.
To illustrate the algorithm used to create the current bfile, we'll (for brevity) assume we've already calculated all terms for n = 1 to 100 and that we already know which classes will be used to compute the next 10 terms, for n = 101 to 110.
These classes are: {0, 1}, {2, 3}, {5, 9}, {7, 9}, {8, 9}, {0, 1, 6}, {0, 1, 7}, {2, 2, 2} and {2, 2, 4} having the values 110, 223, 95, 97, 89, 106, 107, 222 and 224. a(99) = 104 and a(100) = 88, so from those values we may only choose from {223, 95, 97 and 222}. The least value in the list is 95. Therefore, a(101) = 95. The number for the class is now replaced with the next larger number having digits {5, 9} (=A276769(95)), being 559.
(One may see that in the example I only listed 9 classes. Class {8, 9} occurs twice in the example; a(104) = 89 and a(107) = 98.)
From a list of computed values up to some n, the values for classes may be updated to compute further. E.g., to compute a(20000), one may use the bfile to find the least number not already in the sequence for each class and then proceed from a(19998) and a(19999), etc. (End)


MAPLE

N:= 10^3: # to get all terms before the first > N
for R in combinat:powerset({$0..9}) minus {{}, {$0..9}} do
Lastused[R]:= [];
MR[R]:= Array[0..9];
for i from 1 to nops(R) do MR[R][R[i]]:= i od:
od:
A[0]:= 0: A[1]:= 1:
S:= {0, 1}:
for n from 2 to N do
R:= {$0..9} minus (convert(convert(A[n1], base, 10), set) union convert(convert(A[n2], base, 10), set));
L:= Lastused[R];
x:= 0;
while member(x, S) do
for d from 1 do
if d > nops(L) then
if R[1] = 0 then L:= [op(L), R[2]] else L:= [op(L), R[1]] fi;
break
elif L[d] < R[1] then
L[d]:= R[MR[R][L[d]]+1]; break
else
L[d]:= R[1];
fi
od;
x:= add(L[j]*10^(j1), j=1..nops(L));
od;
A[n]:= x;
S:= S union {x};
Lastused[R] := L;
od:
seq(A[i], i=0..N); # Robert Israel, Sep 20 2016


MATHEMATICA

s={0, 1}; Do[a=s[[2]]; b=s[[1]]; n=2; idab=Union[IntegerDigits[a], IntegerDigits[b]]; While[MemberQ[s, n] Intersection[idab, IntegerDigits[n]]!={}, n++]; AppendTo[s, n], {100}]; s


CROSSREFS

Cf. A067581, A086066, A276512, A276769.
Sequence in context: A116069 A081511 A030283 * A298482 A301801 A308540
Adjacent sequences: A276630 A276631 A276632 * A276634 A276635 A276636


KEYWORD

nonn,base,easy


AUTHOR

Zak Seidov and Eric Angelini, Sep 08 2016


STATUS

approved



