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A296257
Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2)^2, where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.
4
1, 2, 12, 30, 67, 133, 249, 446, 776, 1322, 2219, 3710, 6125, 10060, 16441, 26790, 43555, 70706, 114661, 185808, 300953, 487290, 788819, 1276734, 2066229, 3343692, 5410705, 8755238, 14166904, 22923166, 37091159, 60015481, 97107865, 157124642, 254233876
OFFSET
0,2
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 A296245 for a guide to related sequences.
LINKS
Clark Kimberling, Complementary equations, 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(0)^2 + f(n-2)*b(1)^2 + ... + f(2)*b(n-3)^2 + f(1)*b(n-2)^2, where f(n) = A000045(n), the n-th Fibonacci number.
EXAMPLE
a(0) = 1, a(1) = 2, b(0) = 3;
a(2) = a(0) + a(1) + b(0)^2 = 12;
Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, ...)
MATHEMATICA
a[0] = 1; a[1] = 2; b[0] = 3;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-2]^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}] (* A296257 *)
Table[b[n], {n, 0, 20}] (* complement *)
CROSSREFS
Sequence in context: A249055 A127118 A259127 * A375623 A301774 A286230
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Dec 11 2017
STATUS
approved