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A384191 Unique increasing sequence (a(n)) whose complement, (b(n)), satisfies b(n) = a(a(n))+3.

%I A384191 #8 May 28 2025 00:53:17
%S A384191 1,2,3,7,8,9,10,11,12,16,17,18,22,23,24,25,26,27,31,32,33,34,35,36,40,
%T A384191 41,42,46,47,48,49,50,51,55,56,57,61,62,63,64,65,66,70,71,72,73,74,75,
%U A384191 79,80,81,85,86,87,88,89,90,94,95,96,97,98,99,103,104
%N A384191 Unique increasing sequence (a(n)) whose complement, (b(n)), satisfies b(n) = a(a(n))+3.
%C A384191 For k>=1, the Mathematica program gives solutions of the complementary equation b(n)=a(a(n))+k; if k=1, then a=A000201; if k=2, then a=A137708.
%e A384191 a(a(1)) = a(1) = 1, so b(1) = 4.
%e A384191 a(a(2)) = a(2) = 2, so b(2) = 5.
%e A384191 a(a(3)) = a(3) = 3, so b(3) = 6.
%e A384191 a(a(4)) = a(7) = 10, so b(4) = 13.
%t A384191 k = 3; p = k*Array[Floor[#*GoldenRatio] &, 100] ;
%t A384191 seq = ConstantArray[1, Last[p]];
%t A384191 seq[[p]] = k + 1;
%t A384191 u = Join[{1}, Accumulate[seq] + 1]  (* A384191 *)
%t A384191 Complement[Range[Last[#]], #] &[u]  (* A383192 *)
%t A384191 (* _Peter J. C. Moses_, May 12 2025 *)
%Y A384191 Cf. A000201, A137708, A384192 (complement).
%K A384191 nonn
%O A384191 1,2
%A A384191 _Clark Kimberling_, May 21 2025