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Clark Kimberling, <a href="/A295957/b295957.txt">Table of n, a(n) for n = 0..99772000</a>

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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).

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

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

1, 4, 11, 22, 41, 72, 123, 206, 342, 562, 919, 1497, 2433, 3948, 6400, 10368, 16789, 27179, 43992, 71196, 115214, 186437, 301679, 488145, 789854, 1278030, 2067916, 3345979, 5413929, 8759943, 14173908, 22933888, 37107834, 60041761, 97149635, 157191437

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 A295862 for a guide to related sequences.

Clark Kimberling, <a href="/A295957/b295957.txt">Table of n, a(n) for n = 0..9977</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) = 4, b(0) = 2, b(1) = 3, b(2) = 5

b(3) = 6 (least "new number")

a(2) = a(1) + a(0) + b(2) + 1 = 11

Complement: (b(n)) = (2, 3, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, ...)

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

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

j = 1; While[j < 6, k = a[j] - j - 1;

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

Table[a[n], {n, 0, k}]; (* A295957 *)

Cf. A001622, A000045, A295862.

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Clark Kimberling, Dec 08 2017

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