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A296254
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Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1)^2, where a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
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2
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2, 3, 21, 49, 106, 204, 374, 659, 1133, 1913, 3190, 5272, 8658, 14155, 23069, 37513, 60906, 98780, 160086, 259350, 419965, 679891, 1100481, 1781048, 2882258, 4664090, 7547189, 12212179, 19760329, 31973532, 51734950, 83709638, 135445813, 219156747, 354603929
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OFFSET
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0,1
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COMMENTS
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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). See A296245 for a guide to related sequences.
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LINKS
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FORMULA
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a(n) = H + R, where H = f(n-1)*a(0) + f(n)*a(1) and R = f(n-1)*b(1)^2 + f(n-2)*b(2)^2 + ... + f(2)*b(n-2)^2 + f(1)*b(n-1)^2, where f(n) = A000045(n), the n-th Fibonacci number.
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EXAMPLE
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a(0) = 2, a(1) = 3, b(0) = 1, b(1) = 4;
a(2) = a(0) + a(1) + b(1)^2 = 21;
Complement: (b(n)) = (1, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, ...)
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MATHEMATICA
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a[0] = 2; a[1] = 3; b[0] = 1; b[1] = 4;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-1]^2;
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}] (* A296254 *)
Table[b[n], {n, 0, 20}] (* complement *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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