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%I #11 Jun 25 2018 17:37:55
%S 2,3,11,33,104,323,1007,3136,9769,30431,94791,295274,919773,2865082,
%T 8924690,27800290,86597525,269750118,840267961,2617423311,8153238141,
%U 25397226311,79112015761,246432856920,767635009499,2391172651130,7448470401642,23201884354901
%N Solution of the complementary equation a(n) = a(0)*b(n-1) + a(1)*b(n-2) + ... + a(n-1)*b(0), where a(0) = 2, a(1) = 3, b(0) = 1, 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. a(n)/a(n-1) -> 3.114986447390302... (as in A296006). See A296000 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) = 2, a(1) = 3, b(0) = 1, b(1) = 4, so that
%e a(2) = a(0)*b(1) + a(1)*b(0) = 11
%e Complement: (b(n)) = (1, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, ...)
%t mex[list_] := NestWhile[# + 1 &, 1, MemberQ[list, #] &];
%t a[0] = 2; a[1] = 3; b[0] = 1; a[n_] := a[n] = Sum[a[k]*b[n - k - 1], {k, 0, n - 1}];
%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t Table[a[n], {n, 0, 100}]; (* A296005 *)
%t t = N[Table[a[n]/a[n - 1], {n, 1, 500, 100}], 200]
%t Take[RealDigits[Last[t], 10][[1]], 100] (* A296006 *)
%Y Cf. A296000, A296006.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Dec 07 2017
%E Conjectured g.f. removed by _Alois P. Heinz_, Jun 25 2018
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