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A296215 Solution of the complementary equation a(n) = a(1)*b(n-2) + a(2)*b(n-3) + ... + a(n-1)*b(0), where a(0) = 1, a(1) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences. 4

%I #5 Dec 09 2017 19:28:23

%S 1,3,6,24,87,321,1176,4314,15822,58032,212847,780672,2863317,10501959,

%T 38518662,141277197,518170812,1900526031,6970672818,25566752964,

%U 93772706622,343935755925,1261473710904,4626782461218,16969926331719,62241612204120,228287277978756

%N Solution of the complementary equation a(n) = a(1)*b(n-2) + a(2)*b(n-3) + ... + a(n-1)*b(0), where a(0) = 1, a(1) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences.

%C The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295862 for a guide to related sequences.

%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.

%e a(0) =1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5

%e a(2) = a(1)*b(0) = 6

%e Complement: (b(n)) = (2, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, ...)

%t a[0] = 1; a[1] = 3; b[0] = 2;

%t a[n_] := a[n] = Sum[a[k]*b[n - k - 1], {k, 1, n - 1}];

%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];

%t u = Table[a[n], {n, 0, 200}]; (* A296215 *)

%t Table[b[n], {n, 0, 20}]

%t N[Table[a[n]/a[n - 1], {n, 1, 200, 10}], 200];

%t RealDigits[Last[t], 10][[1]] (* A296216 *)

%Y Cf. A296000, A296217.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Dec 08 2017

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