OFFSET
1,2
COMMENTS
From Ant King, Oct 18 2011: (Start)
lim(n->oo, u(2n+1)/u(2n)) = 1/2(2207+987*sqrt(5)),
lim(n->oo, u(2n)/u(2n-1)) = 1/2(47+21*sqrt(5)). (End)
From Raphie Frank, Nov 30 2012: (Start)
Where L_n is a Lucas number and F_n is Fibonacci number:
lim(n->oo, u(2n+1)/u(2n)) = 1/2(L_16+F_16*sqrt(5)),
lim(n->oo, u(2n)/u(2n-1)) = 1/2(L_8+F_8*sqrt(5)),
a(n) = L_1*a(n-1) + L_24*a(n-2) - L_24*a(n-3)- L_1*a(n-4) + L_1*a(n-5). (End)
LINKS
Colin Barker, Table of n, a(n) for n = 1..399
J. C. Su, On some properties of two simultaneous polygonal sequences, JIS 10 (2007) 07.10.4, example 4.4
Eric Weisstein's World of Mathematics, Heptagonal Triangular Number
Index entries for linear recurrences with constant coefficients, signature (1,103682,-103682,-1,1).
FORMULA
The two bisections satisfy the same recurrence relation: a(n+2)=103682*a(n+1)-a(n)+18144 or a(n+1)=51841*a(n)+9072+2898*(320*a(n)^2+112*a(n)+9)^0.5. The g.f. satisfies f(z)=(z+55*z^2+18088*z^3+18088*z^4+55*z^5+z^6)/((1-z^2)*(1-103682*z^2+z^4))=1*z+55*z^2+121771*z^3+... - Richard Choulet, Sep 20 2007
From Ant King, Oct 18 2011: (Start)
a(n) = a(n-1)+103682a(n-2)-103682a(n-3)-a(n-4)+a(n-5).
a(n) = 1/80*((3-sqrt(5)*(-1)^n)*(2+sqrt(5))^(4n-2)+(3+sqrt(5)*(-1)^n)*(2-sqrt(5))^(4n-2)-14).
a(n) = floor(1/80*(3-sqrt(5)*(-1)^n)*(2+sqrt(5))^(4n-2)).
G.f.: x(1+54*x+18034*x^2+54*x^3+x^4)/((1-x)(1-322*x+x^2)(1+322*x+x^2)). (End)
MATHEMATICA
LinearRecurrence[{1, 103682, -103682, -1, 1}, {1, 55, 121771, 5720653, 12625478965}, 12] (* Ant King, Oct 18 2011 *)
PROG
(PARI) a(n)=((3-sqrt(5)*(-1)^n)*(2+sqrt(5))^(4*n-2)+(3+sqrt(5)*(-1)^n)*(2-sqrt(5))^(4*n-2)-14)\/80 \\ Charles R Greathouse IV, Oct 18 2011
(PARI) Vec(-x*(x^4+54*x^3+18034*x^2+54*x+1)/((x-1)*(x^2-322*x+1)*(x^2+322*x+1)) + O(x^20)) \\ Colin Barker, Jun 23 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved