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 A228276 Sum of a(n)+a(n+1) can be written using the digits of {a(n),a(n+1)}; always choose the smallest possible unused positive integer. 5
 1, 10, 2, 19, 72, 100, 3, 20, 4, 30, 5, 40, 6, 50, 7, 60, 8, 70, 9, 12, 79, 18, 13, 68, 113, 198, 21, 91, 32, 181, 130, 170, 131, 82, 120, 80, 101, 89, 109, 23, 69, 27, 14, 28, 54, 17, 24, 58, 117, 124, 90, 11, 200, 15, 36, 26, 37, 126, 93, 41, 273, 52, 163, 29, 63, 107, 123, 96, 71, 230, 73, 241, 83, 152, 64, 182, 31, 92, 119, 78, 102, 99 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS The sequence is a permutation of the natural numbers.  Sketch of proof: (1) all terms are distinct by definition; (2) each term has a successor (with pandigitals as ultimate candidates); (3) an alleged non-occurring number will succeed the first occurred pandigital number. Cf. A245586. - Reinhard Zumkeller, Jul 26 2014 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 E. Angelini, Add neighbours, use their digits, SeqFan list, Nov. 2, 2013 E. Angelini, Add A to B E. Angelini, Add A to B [Cached copy, with permission] EXAMPLE We see that the result of 1+10 uses only digits from the set {1,1,0} (really a multi-set). The same with 10+2 which use some elements of {1,0,2}. Again, 2+19 uses elements of {2,1,9} for its result. 72 is now the smallest integer respecting the constraint (we see that 19+72 is 91 which uses for its transcription only a few elements of {1,9,7,2}. PROG (PARI) {subseq(a, b, j)=!for(i=1, #a, while(j<#b, a[i]==b[j++]&&next(2)); return)} {u=0; a=1; for(n=1, 99, print1(a", "); u+=1<

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Last modified January 20 08:18 EST 2020. Contains 331081 sequences. (Running on oeis4.)