Revision History for A295954
(Underlined text is an addition;
strikethrough text is a deletion.)
Showing all changes.
|
| |
|
| A295954
|
|
Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n) + 1, where a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
(history;
published version)
|
| |
|
|
#9 by N. J. A. Sloane at Thu Feb 22 22:40:53 EST 2018
| |
| |
| |
|
|
#8 by N. J. A. Sloane at Thu Feb 22 22:40:51 EST 2018
|
|
| LINKS
|
Clark Kimberling, <a href="/A295954/b295954.txt">Table of n, a(n) for n = 0..99772000</a>
|
|
| STATUS
|
approved
editing
|
| |
| |
|
|
#7 by Susanna Cuyler at Fri Dec 08 16:55:49 EST 2017
| |
| |
| |
|
|
#6 by Jon E. Schoenfield at Fri Dec 08 14:56:01 EST 2017
| |
| |
| |
|
|
#5 by Jon E. Schoenfield at Fri Dec 08 14:55:54 EST 2017
|
|
| COMMENTS
|
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).
|
|
| STATUS
|
proposed
editing
|
| |
|
|
Discussion
|
| Fri Dec 08
| 14:56
| Jon E. Schoenfield: removed 2nd space between sentences
|
| |
| |
|
|
#4 by Clark Kimberling at Fri Dec 08 10:57:22 EST 2017
| |
| |
| |
|
|
#3 by Clark Kimberling at Fri Dec 08 10:20:08 EST 2017
|
|
| FORMULA
|
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.
|
|
| MATHEMATICA
|
While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
|
| |
| |
|
|
#2 by Clark Kimberling at Fri Dec 08 10:08:37 EST 2017
|
|
| NAME
|
allocatedSolution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n) + 1, where a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are forincreasing Clarkcomplementary Kimberlingsequences.
|
|
| DATA
|
2, 4, 12, 23, 43, 75, 128, 214, 354, 582, 951, 1549, 2517, 4084, 6620, 10724, 17365, 28111, 45499, 73635, 119160, 192822, 312010, 504861, 816901, 1321793, 2138726, 3460552, 5599312, 9059899, 14659247, 23719183, 38378468, 62097690, 100476198, 162573929
|
|
| OFFSET
|
0,1
|
|
| COMMENTS
|
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.
|
|
| LINKS
|
Clark Kimberling, <a href="/A295954/b295954.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.
|
|
| FORMULA
|
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.
|
|
| EXAMPLE
|
a(0) = 2, a(1) = 4, b(0) = 1, b(1) = 3, b(2) = 5
b(3) = 6 (least "new number")
a(2) = a(1) + a(0) + b(2) + 1 = 12
Complement: (b(n)) = (1, 3, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, ...)
|
|
| MATHEMATICA
|
a[0] = 2; a[1] = 4; b[0] = 1; 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}]; (* A295954 *)
|
|
| CROSSREFS
|
Cf. A001622, A000045, A295862.
|
|
| KEYWORD
|
allocated
nonn,easy
|
|
| AUTHOR
|
Clark Kimberling, Dec 08 2017
|
|
| STATUS
|
approved
editing
|
| |
| |
|
|
#1 by Clark Kimberling at Thu Nov 30 19:49:14 EST 2017
|
|
| NAME
|
allocated for Clark Kimberling
|
|
| KEYWORD
|
allocated
|
|
| STATUS
|
approved
|
| |
| |
|
|
