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A295862
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Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(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.
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32
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1, 3, 9, 18, 34, 60, 104, 175, 291, 479, 784, 1278, 2078, 3373, 5470, 8863, 14354, 23239, 37616, 60879, 98520, 159425, 257972, 417425, 675426, 1092881, 1768338, 2861251, 4629622, 7490908, 12120566, 19611511, 31732115, 51343665, 83075820, 134419526, 217495388
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OFFSET
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0,2
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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). Following is a guide to related sequences:
*****
Complementary equation: a(n) = a(n-1) + a(n-2) + b(n); initial values (a(0), a(1); b(0), b(1), b(2)):
Complementary equation: a(n) = a(n-1) + a(n-2) + b(n) + 1; initial values (a(0), a(1); b(0), b(1), b(2)):
Complementary equation: a(n) = a(n-1) + a(n-2) + b(n) - 1; initial values (a(0), a(1); b(0), b(1), b(2)):
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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(2) + f(n-2)*b(3) + ... + f(2)*b(n-1) + f(1)*b(n), where f(n) = A000045(n), the n-th Fibonacci number.
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EXAMPLE
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a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5, so that
b(3) = 6 (least "new number");
a(2) = a(1) + a(0) + b(2) = 9;
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20, ...)
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MATHEMATICA
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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];
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[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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