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A276633
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a(n) = smallest integer not yet in the sequence with no digits in common with a(n-1) and a(n-2); a(0)=0, a(1)=1.
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15
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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
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0,3
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COMMENTS
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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(n-3) are allowed in a(n). - Robert Israel, Sep 20 2016
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LINKS
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EXAMPLE
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Each number can consist of 2^10-1 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 b-file, 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 b-file to find the least number not already in the sequence for each class and then proceed from a(19998) and a(19999), etc. (End)
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MAPLE
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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[n-1], base, 10), set) union convert(convert(A[n-2], 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^(j-1), j=1..nops(L));
od;
A[n]:= x;
S:= S union {x};
Lastused[R] := L;
od:
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MATHEMATICA
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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
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PROG
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(Python)
from itertools import count, islice, product as P
def only(s, D=1): # numbers with >= D digits only from s
yield from (int("".join(p)) for d in count(D) for p in P(s, repeat=d))
def agen(): # generator of terms
aset, an1, an, minan = {0, 1}, 0, 1, 2
yield from [0, 1]
while True:
an1, an, s = an, minan, set(str(an) + str(an1))
use = "".join(c for c in "0123456789" if c not in s)
for an in only(use, D=len(str(minan))):
if an not in aset: break
aset.add(an)
yield an
while minan in aset: minan += 1
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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