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While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];

Cf. A001622, A296245, A296491, A296492A298171, A298172.

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allocated for Clark KimberlingSolution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n) + 2*n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 3, 13, 28, 56, 102, 179, 305, 511, 846, 1391, 2274, 3705, 6022, 9773, 15844, 25669, 41568, 67295, 108924, 176283, 285274, 461627, 746974, 1208678, 1955732, 3164493, 5120311, 8284893, 13405296, 21690284, 35095678, 56786063, 91881845, 148668015, 240549970

0,2

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622). See A296245 for a guide to related sequences.

Clark Kimberling, <a href="/A296776/b296776.txt">Table of n, a(n) for n = 0..1000</a>

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

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

a(2) = a(0) + a(1) + b(2) + 4 = 13

Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, ...)

a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4; b[2] = 5;

a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n] + 2 n;

j = 1; While[j < 16, k = a[j] - j - 1;

While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];

u = Table[a[n], {n, 0, k}]; (* A296776 *)

Table[b[n], {n, 0, 20}] (* complement *)

Cf. A001622, A296245, A296491, A296492.

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nonn,easy

Clark Kimberling, Jan 06 2018

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