A093652 Let a(1) = 1, a(2) = 2, a(3) = 7, a(4) = 15 and for n >= 5 set a(n) = (n*b(n) - b(n-2)) / 2, where b(n) = 4*b(n-2) - b(n-4) for n >= 5 and b(1) = 1, b(2) = 2, b(3) = 5, b(4) = 8.

1, 2, 7, 15, 45, 86, 239, 433, 1157, 2034, 5307, 9151, 23497, 39974, 101467, 170913, 430089, 718946, 1796975, 2985775, 7422437, 12272502, 30373191, 50016721, 123327373, 202395986, 497484067, 814061151, 1995542913, 3257222726, 7965875891, 12973832257, 31663779857

Offset: 1

Author

Harri Aaltonen, May 15 2004, Apr 12 2008

Keywords

• nonn
• easy

Comments

a(n)/b(n) gives the ohm value of a ladder of unit resistors measured from opposite corners. The ladder is best described as a line of n squares, where every segment has a resistance of 1 ohm.

1/(n - 2*a(n)/b(n)) approaches 2 + sqrt(3) as n increases.

Links

• Harri Aaltonen, Apr 18 2008, Table of n, a(n) for n = 1..50 [a(49) corrected by _Georg Fischer_, Mar 13 2020]
• Index entries for linear recurrences with constant coefficients, signature (0,8,0,-18,0,8,0,-1).
• Index to sequences related to resistances.

Crossrefs

Cf. A082630.

Programs

Maple

a_list := proc(last) local B, C, k;
   B := [1,2,5, 8];
   C := [1,2,7,15];
   for k from 5 to last do
      B := [op(B), 4*B[k-2]-B[k-4]];
      C := [op(C), (k*B[k]-B[k-2])/2];
   od;
C end:
a_list(50); # After Harri Aaltonen, Peter Luschny, Mar 14 2020

Mathematica

LinearRecurrence[{0, 8, 0, -18, 0, 8, 0, -1}, {1, 2, 7, 15, 45, 86, 239, 433}, 50] (* Jean-François Alcover, Oct 24 2023 *)

Formula

Conjecture: b(n) = A082630(n). If true, we can write a(n) = (n*A082630(n) - A082630(n-2)) / 2.

From Colin Barker, Dec 20 2019: (Start)

G.f.: x*(1 + 2*x - x^2 - x^3 + 7*x^4 + 2*x^5 - 3*x^6 - x^7) / (1 - 4*x^2 + x^4)^2.

a(n) = 8*a(n-2) - 18*a(n-4) + 8*a(n-6) - a(n-8) for n>8.

(End)

Extensions

Edited by Peter Luschny, Jun 14 2021