OFFSET
1,2
COMMENTS
Also positive integers y in the solutions to x^2 - 5*y^2 + x + 5*y - 2 = 0, the corresponding values of x being A254626.
Also indices of centered pentagonal numbers (A005891) that are also hexagonal numbers (A000384). - Colin Barker, Feb 11 2015
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,18,-18,-1,1).
FORMULA
a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5).
G.f.: x*(1+x-9*x^2-x^3)/((1-x)*(1+4*x-x^2)*(1-4*x-x^2)).
a(n) = (10 - sqrt(5)*(2-sqrt(5))^n - 5*(-2+sqrt(5))^n - 2*sqrt(5)*(-2+sqrt(5))^n + sqrt(5)*(2+sqrt(5))^n + (-2-sqrt(5))^n*(-5+2*sqrt(5)))/20. - Colin Barker, Jun 06 2016
a(2*n+2) = A232970(2*n+1); a(2*n+1) = A110679(2*n). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019
a(n) = (2 +(1+2*(-1)^n)*Fibonacci(3*n) -(-1)^n*Lucas(3*n))/4. - G. C. Greubel, Apr 19 2019
EXAMPLE
2 is in the sequence because the 2nd centered pentagonal number is 6, which is also the 3rd triangular number.
MATHEMATICA
CoefficientList[Series[x (x^3 + 9 x^2 - x - 1)/((x - 1) (x^2 - 4 x - 1) (x^2 + 4 x - 1)), {x, 0, 25}], x] (* Michael De Vlieger, Jun 06 2016 *)
LinearRecurrence[{1, 18, -18, -1, 1}, {1, 2, 11, 28, 189}, 30] (* Harvey P. Dale, Apr 23 2017 *)
PROG
(PARI) Vec(x*(x^3+9*x^2-x-1)/((x-1)*(x^2-4*x-1)*(x^2+4*x-1)) + O(x^30))
(PARI) {a(n) = (2 +(1+3*(-1)^n)*fibonacci(3*n) - 2*(-1)^n*fibonacci(3*n+1))/4}; \\ G. C. Greubel, Apr 19 2019
(Magma) [(2 +(1+2*(-1)^n)*Fibonacci(3*n) -(-1)^n*Lucas(3*n))/4 : n in [1..30]]; // G. C. Greubel, Apr 19 2019
(Sage) [(2 +(1+3*(-1)^n)*fibonacci(3*n) -2*(-1)^n*fibonacci(3*n+1))/4 for n in (1..30)] # G. C. Greubel, Apr 19 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Feb 03 2015
STATUS
approved