来自在线整数百科全书的问候语!http://oeis.org/ Search: id:a296258 Showing 1-1 of 1 %I A296258 %S A296258 1,3,8,27,60,123,232,436,768,1325,2237,3731,6164,10120,16540,26949, %T A296258 43813,71123,115336,186900,302720,490149,793445,1284219,2078340, %U A296258 3363343,5442524,8806767,14250252,23058043,37309384,60368583,97679192,158049071,255729632 %N A296258 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2)^2, where a(0) = 1, a(1) = 3, b(0) = 2, and (a(n)) and (b(n)) are increasing complementary sequences. %C A296258 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=黄金比(A00 1622)。有关相关序列的指南见A29 6245。n,a(n)n=0…1000的表%H A29 6258 Clark Kimberling,互补方程J. Int. Seq。19(2007),1-13.0%F A29 6258A(n)=H+R,其中H=f(n-1)*a(0)+f(n)*a(1)和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. %e A296258 a(0) = 1, a(1) = 3, b(0) = 2; %e A296258 a(2) = a(0) + a(1) + b(0)^2 = 8; %e A296258 Complement: (b(n)) = (2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, ...) %t A296258 a[0] = 1; a[1] = 3; b[0] = 2; %t A296258 a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-2]^2; %t A296258 j = 1; While[j < 6 , k = a[j] - j - 1; %t A296258 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; %t A296258 Table[a[n], {n, 0, k}] (* A296258 *) %t A296258 Table[b[n], {n, 0, 20}] (* complement *) %Y A296258 Cf. A001622, A296245. %K A296258 nonn,easy %O A296258 0,2 %A A296258 _Clark Kimberling_, Dec 11 2017 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE