A333905 Lexicographically earliest sequence of distinct positive integers such that a(n) divides the concatenation of a(n+1) to a(n+2).

1, 2, 3, 4, 5, 6, 10, 8, 20, 16, 40, 32, 80, 64, 160, 128, 320, 256, 640, 512, 1280, 1024, 2560, 2048, 5120, 4096, 10240, 8192, 20480, 16384, 40960, 32768, 81920, 65536, 163840, 131072, 327680, 262144, 655360, 524288, 1310720, 1048576, 2621440, 2097152, 5242880, 4194304, 10485760, 8388608, 20971520, 16777216, 41943040

Offset: 1

Author

Eric Angelini, Apr 09 2020

Keywords

• base
• nonn

Examples

			a(1) = 1 divides 23 (and 23 is a(2) = 2 concatenated to a(3) = 3);
a(2) = 2 divides 34 (and 34 is a(3) = 3 concatenated to a(4) = 4);
a(3) = 3 divides 45 (and 45 is a(4) = 4 concatenated to a(5) = 5);
a(4) = 4 divides 56 (and 56 is a(5) = 5 concatenated to a(6) = 6);
a(5) = 5 divides 610 (and 610 is a(6) = 6 concatenated to a(7) = 10);
a(6) = 6 divides 108 (and 108 is a(7) = 10 concatenated to a(8) = 8);
From a(7) = 10 on, the pattern of the sequence is regular.
		

Crossrefs

Cf. A085946 (a(1) = 1, a(2) = 2 and a(n) = smallest number not included earlier that divides the concatenation a(n-2), a(n-1)).

Formula

Conjectures from Colin Barker, Apr 09 2020: (Start)

G.f.: x*(1 + 2*x + x^2 - x^4 - 2*x^5 - 4*x^7) / (1 - 2*x^2).

a(n) = 2*a(n-2) for n>6.

(End)

Conjecture: a(n) = 2^((n-7)/2)*(5 + 2*sqrt(2) + (2*sqrt(2) - 5)*(-1)^n) for n > 6. - Stefano Spezia, Oct 23 2021