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A295862 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. 32

%I #14 Aug 14 2018 21:07:22

%S 1,3,9,18,34,60,104,175,291,479,784,1278,2078,3373,5470,8863,14354,

%T 23239,37616,60879,98520,159425,257972,417425,675426,1092881,1768338,

%U 2861251,4629622,7490908,12120566,19611511,31732115,51343665,83075820,134419526,217495388

%N 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.

%C 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:

%C *****

%C Complementary equation: a(n) = a(n-1) + a(n-2) + b(n); initial values (a(0), a(1); b(0), b(1), b(2)):

%C A295862: (1,3; 2,4,5)

%C A295947: (2,4; 1,3,5)

%C A295948: (3,4; 1,2,5)

%C A295949: (1,2; 3,4,5)

%C A295950: (1,4; 2,3,5)

%C A295951: (2,3; 1,4,5)

%C A295952: (1,5; 2,3,4)

%C 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)):

%C A295953: (1,3; 2,4,5)

%C A295954: (2,4; 1,3,5)

%C A295955: (3,4; 1,2,5)

%C A295956: (1,2; 3,4,5)

%C A295957: (1,4; 2,3,5)

%C A295958: (2,3; 1,4,5)

%C A295959: (1,5; 2,3,4)

%C 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)):

%C A295860: (1,3; 2,4,5)

%C A295961: (2,4; 1,3,5)

%C A295962: (3,4; 1,2,5)

%C A295963: (1,2; 3,4,5)

%C A295964: (1,4; 2,3,5)

%C A295965: (2,3; 1,4,5)

%C A295966: (1,5; 2,3,4)

%H Clark Kimberling, <a href="/A295862/b295862.txt">Table of n, a(n) for n = 0..3000</a>

%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.

%F 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.

%e a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5, so that

%e b(3) = 6 (least "new number");

%e a(2) = a(1) + a(0) + b(2) = 9;

%e Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20, ...)

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

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

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

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

%t Table[a[n], {n, 0, k}] (*A295862*)

%t Table[b[n], {n, 0, 20}] (*complement*)

%Y Cf. A001622, A000045, A295947.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Dec 08 2017

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