

A295754


Solution of the complementary equation a(n) = a(n1) + a(n3) + a(n4) + b(n4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.


4



1, 2, 3, 4, 12, 23, 37, 61, 105, 175, 284, 463, 757, 1231, 1994, 3231, 5237, 8481, 13726, 22215, 35955, 58186, 94152, 152348, 246516, 398882, 645411, 1044305, 1689734, 2734059, 4423808, 7157881, 11581709, 18739612, 30321339, 49060968, 79382329, 128443321
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OFFSET

0,2


COMMENTS

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.
a(n)/a(n1) > (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growthrate of the Fibonacci numbers (A000045).


LINKS

Table of n, a(n) for n=0..37.
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 113.


EXAMPLE

a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, so that
b(4) = 9 (least "new number")
a(4) = a(3) + a(1) + a(0) + b(0) = 12
Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 13, 14, 15, ...)


MATHEMATICA

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
a[n_] := a[n] = a[n  1] + a[n  3] + a[n  4] + b[n  4];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n  1}]]];
z = 36; Table[a[n], {n, 0, z}] (* A295754 *)
Table[b[n], {n, 0, 20}] (*complement *)


CROSSREFS

Cf. A001622, A000045, A293411, A295755.
Sequence in context: A117555 A243392 A239943 * A059810 A037397 A168047
Adjacent sequences: A295751 A295752 A295753 * A295755 A295756 A295757


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Nov 30 2017


STATUS

approved



