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%I #12 Oct 16 2019 11:51:53
%S 1,2,3,4,8,9,10,14,15,16,21,22,26,27,28,32,33,34,38,39,40,44,45,46,51,
%T 52,56,57,58,62,63,64,68,69,70,74,75,76,81,82,86,87,88,92,93,94,98,99,
%U 100,104,105,106,111,112,116,117,118,122,123,124,128,129,130,134,135
%N a(n)=n for n<=3; for n>3, a(n) is next integer sharing common factors with 1 or 2 of previous 3 terms.
%C Does every sequence generated according to this rule, no matter which three initial terms are chosen, eventually fall into the same pattern with respect to the modulus 210 and thus have almost all of its terms in common with A079279?
%C Sequence includes all 3 mod 6 and 4 mod 6 numbers; all 2 mod 6 numbers except those congruent to 20 mod 30; all numbers congruent to 203 and 204 mod 210; and no other numbers except 1.
%F Conjectures from _Colin Barker_, Oct 15 2019: (Start)
%F G.f.: x*(1 + x + x^2 + x^3 + 4*x^4 + x^5 + x^6 + 4*x^7 + x^8 + x^9 + 5*x^10 + x^11 + 4*x^12 + x^13 + 3*x^15) / ((1 - x)^2*(1 + x)*(1 - x + x^2 - x^3 + x^4 - x^5 + x^6)*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)).
%F a(n) = a(n-1) + a(n-14) - a(n-15) for n>16.
%F (End)
%e 5, 6, 7 and 8 have factors in common with 0, 3, 0 and 2 (respectively) of terms a(2) through a(4) (2, 3 and 4); therefore a(5)=8.
%K easy,nonn
%O 1,2
%A _Matthew Vandermast_, Feb 07 2003
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