

A322466


Lexicographically first sequence of distinct terms such that the a(n)th digit after a(n) shares a(n)'s parity.


1



1, 3, 4, 2, 5, 6, 8, 9, 7, 10, 20, 12, 11, 13, 14, 15, 16, 18, 17, 19, 30, 22, 21, 31, 24, 23, 26, 32, 25, 33, 27, 28, 34, 35, 29, 36, 38, 37, 39, 50, 40, 51, 41, 42, 52, 44, 53, 43, 45, 54, 46, 55, 47, 48, 49, 56, 57, 58, 59, 60, 62, 61, 63, 70, 64, 65, 66, 72, 67, 71, 68, 73, 69, 74, 75, 76, 78, 80, 77, 79, 81
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OFFSET

1,2


COMMENTS

This sequence is conjectured to be a permutation of the positive integers.


LINKS



EXAMPLE

The sequence starts with 1,3,4,2,5,6,8,9,7,10,20,12,...
a(1) = 1 forces the next digit to be odd;
Thus a(2) = 3 as this 3 is the smallest available integer not leading to a contradiction; this 3 forces the 3rd digit after 3 to be odd;
Could a(3) be equal to 2 (instead of 4, here)? No, because this would lead to a contradiction as we've just seen that the 3rd digit after 3 must be odd;
Thus a(3) = 4; this 4 forces the 4th digit after 4 to be even;
a(4) = 2 as this 2 is the smallest available integer not leading to a contradiction; this 2 forces the 2nd digit after 2 to be even;
a(5) = 5 as the first digit of a(5) must be odd and 5 is the smallest available integer not leading to a contradiction; this 5 forces the 5th digit after 5 to be odd;
a(6) = 6 as the first digit of a(6) must be even and 6 is the smallest available integer not leading to a contradiction; this 6 forces the 6th digit after 6 to be even;
a(7) = 8 as the first digit of a(7) must be even and 8 is the smallest available integer not leading to a contradiction; this 8 forces the 8th digit after 8 to be even;
Could a(8) be equal to 7 (instead of 9, here)? No, because this would lead to a contradiction as we've just seen that the 8th digit after 8 must be even;
Thus a(8) = 9 as this 9 is the smallest available integer not leading to a contradiction; this 9 forces the 9th digit after 9 to be odd;
a(9) = 7 as this 7 is the smallest available integer not leading to a contradiction; this 7 forces the 7th digit after 7 to be odd;
a(10) = 10 as the smallest available integer starting with an even digit and not leading to a contradiction is 10; this 10 forces the 10th digit after 10 to be even;
a(11) = 20 as the smallest available integer starting with an even digit and not leading to a contradiction is 20; this 20 forces the 10th digit after 20 to be even;
a(12) = 12 as the smallest available integer ending with an even digit and not leading to a contradiction is 12; this 12 forces the 12th digit after 12 to be even;
etc.


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



