A276633 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.
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
Offset: 0
Examples
From _David A. Corneth_, Sep 22 2016: (Start)
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)
Links
- Zak Seidov and David A. Corneth, Table of n, a(n) for n = 0..19999 (First 2001 terms from Zak Seidov).
Programs
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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[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: 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 -
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 print(list(islice(agen(), 75))) # Michael S. Branicky, Jun 30 2022
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