A269028 a(n) = 40*a(n - 1) - a(n - 2) for n>1, a(0) = 1, a(1) = 1.

1, 1, 39, 1559, 62321, 2491281, 99588919, 3981065479, 159143030241, 6361740144161, 254310462736199, 10166056769303799, 406387960309415761, 16245352355607326641, 649407706263983649879, 25960062898203738668519, 1037753108221885563090881

Offset: 0

Author

Ilya Gutkovskiy, Feb 18 2016

Keywords

• nonn
• easy

Comments

In general, the ordinary generating function for the recurrence relation b(n) = k*b(n - 1) - b(n - 2) with n>1 and b(0)=1, b(1)=1, is (1 - (k - 1)*x)/(1 - k*x +x^2). This recurrence gives the closed form b(n) = (2^( -n - 1)*((k - 2)*(k - sqrt(k^2 - 4))^n + sqrt(k^2 - 4)*(k - sqrt(k^2 - 4))^n - (k - 2)*(sqrt(k^2 - 4) + k)^n + sqrt(k^2 - 4)*(sqrt(k^2 - 4) + k)^n))/sqrt(k^2 - 4).

Links

• Index entries for linear recurrences with constant coefficients, signature (40,-1).

Crossrefs

Cf. A001519, A001835, A001653, A049685, A070997, A070998, A072256, A078922, A160682, A007805, A075839, A157014, A159664, A159668, A157877, A238379, A097315.

Programs

Magma

[n le 2 select 1 else 40*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 19 2016

Mathematica

Table[Cosh[n Log[20 + Sqrt[399]]] - Sqrt[19/21] Sinh[n Log[20 + Sqrt[399]]], {n, 0, 17}]
Table[(2^(-n - 2) (38 (40 - 2 Sqrt[399])^n + 2 Sqrt[399] (40 - 2 Sqrt[399])^n - 38 (40 + 2 Sqrt[399])^n + 2 Sqrt[399] (40 + 2 Sqrt[399])^n))/Sqrt[399], {n, 0, 17}]
LinearRecurrence[{40, -1}, {1, 1}, 17]

Formula

G.f.: (1 - 39*x)/(1 - 40*x + x^2).

a(n) = cosh(n*log(20 + sqrt(399))) - sqrt(19/21)*sinh(n*log(20 + sqrt(399))).

a(n) = (2^(-n - 2)*(38*(40 - 2*sqrt(399))^n + 2*sqrt(399)*(40 - 2*sqrt(399))^n - 38*(40 + 2*sqrt(399))^n + 2*sqrt(399)*(40 + 2*sqrt(399))^n))/sqrt(399).

Sum_{n>=0} 1/a(n) = 2.0262989201139499769986...