Revision History for A295955
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Showing entries 11-13
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| A295955
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Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n) + 1, where a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
(history;
published version)
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#3 by Clark Kimberling at Fri Dec 08 10:19:39 EST 2017
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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 nth Fibonacci number.
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| MATHEMATICA
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While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
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#2 by Clark Kimberling at Fri Dec 08 10:10:46 EST 2017
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| NAME
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allocatedSolution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n) + 1, where a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 1, b(2) = 5, and (a(n)) and (b(n)) are forincreasing Clarkcomplementary Kimberlingsequences.
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| DATA
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3, 4, 13, 24, 45, 78, 133, 222, 367, 602, 984, 1602, 2603, 4223, 6845, 11088, 17954, 29064, 47041, 76129, 123196, 199352, 322576, 521957, 844563, 1366551, 2211146, 3577730, 5788910, 9366675, 15155621, 24522333, 39677992, 64200364, 103878396, 168078801
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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 A295862 for a guide to related sequences.
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| LINKS
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Clark Kimberling, <a href="/A295955/b295955.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.
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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 nth Fibonacci number.
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| EXAMPLE
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a(0) = 3, a(1) = 4, b(0) = 1, b(1) = 2, b(2) = 5
b(3) = 6 (least "new number")
a(2) = a(1) + a(0) + b(2) + 1 = 13
Complement: (b(n)) = (1, 2, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, ...)
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| MATHEMATICA
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a[0] = 3; a[1] = 4; b[0] = 1; b[1] = 2; 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}]; (* A295955 *)
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| CROSSREFS
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Cf. A001622, A000045, A295862.
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| KEYWORD
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allocated
nonn,easy
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| AUTHOR
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Clark Kimberling, Dec 08 2017
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| STATUS
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approved
editing
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#1 by Clark Kimberling at Thu Nov 30 19:49:14 EST 2017
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| NAME
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allocated for Clark Kimberling
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| KEYWORD
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allocated
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| STATUS
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approved
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