

A306805


An irregular fractal sequence: underline a(n) iff [a(n1) + a(n)] is divisible by the leftmost digit of a(n); all underlined terms rebuild the starting sequence.


2



1, 2, 1, 3, 4, 2, 1, 5, 6, 3, 7, 8, 4, 2, 1, 9, 20, 5, 22, 21, 24, 6, 3, 26, 23, 28, 7, 30, 25, 31, 33, 32, 8, 4, 2, 1, 34, 27, 9, 35, 36, 20, 5, 38, 22, 29, 21, 37, 39, 40, 24, 6, 3, 42, 26, 41, 23, 43, 44, 28, 7, 46, 45, 30, 47, 25, 48, 49, 50, 31, 51, 33, 53, 54, 52, 32, 8, 4, 2, 1, 55, 56, 34, 57, 27, 9, 58, 35, 59, 60, 36
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OFFSET

1,2


COMMENTS

The sequence S starts with a(1) = 1 and a(2) = 2. S is extended by duplicating the first term A among the not yet duplicated terms of S, under the condition that the sum [a(n1) + a(n)] is divisible by the leftmost digit of a(n). If this is not the case, we then extend S with the smallest integer X not yet present in S such that the sum [a(n1) + a(n)] is not divisible by the leftmost digit of a(n). This is the lexicographically earliest sequence with this property.


LINKS

JeanMarc Falcoz, Table of n, a(n) for n = 1..10002


EXAMPLE

S starts with a(1) = 1 and a(2) = 2
Can we duplicate a(1) to form a(3)? Yes, as the sum [a(2) + a(3) = 3] is divisible by 1, the leftmost digit of a(3); thus a(3) = 1.
Can we duplicate a(2) to form a(4)? No, as the sum [a(3) + a(4) = 3] is not divisible by 2, the leftmost digit of a(4); we thus extend S with the smallest integer X not yet in S such that the sum [a(3) + X] is not divisible by the leftmost digit of a(4); thus a(4) = 3.
Can we duplicate a(2) to form a(5)? No, as the sum [a(4) + a(5) = 7] is not divisible by 2, the leftmost digit of a(5); we thus extend S with the smallest integer X not yet in S such that the sum [a(4) + X] is not divisible by the leftmost digit of a(5); thus a(5) = 4.
Can we duplicate a(2) to form a(6)? Yes, as the sum [a(5) + a(6) = 6 is divisible by 2, the leftmost digit of a(6); thus a(6) = 2.
Can we duplicate a(3) to form a(7)? Yes, as 1 can always be duplicated; thus a(7) = 1.
Can we duplicate a(4) to form a(8)? No, as the sum [a(7) + a(8) = 5] is not divisible by 2, the leftmost digit of a(8); we thus extend S with the smallest integer X not yet in S such that the sum [a(7) + X] is not divisible by the leftmost digit of a(8); thus a(8) = 5.
Can we duplicate a(4) to form a(9)? No, as the sum [a(8) + a(9) = 8] is not divisible by 3, the leftmost digit of a(9); we thus extend S with the smallest integer X not yet in S such that the sum [a(8) + X] is not divisible by the leftmost digit of a(9); thus a(9) = 6.
Can we duplicate a(4) to form a(10)? Yes, as the sum [a(9) + a(10) = 9] is divisible by 3, the leftmost digit of a(10); thus a(10) = 3.
Etc.


CROSSREFS

Sequence in context: A248514 A123390 A306806 * A162598 A088208 A081878
Adjacent sequences: A306802 A306803 A306804 * A306806 A306807 A306808


KEYWORD

base,nonn


AUTHOR

Eric Angelini and JeanMarc Falcoz, Mar 11 2019


STATUS

approved



