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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.
3
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
1,2
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]
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)
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 *)
CROSSREFS
Cf. A082630.
Sequence in context: A065506 A330454 A121165 * A200862 A096690 A279286
KEYWORD
nonn,easy
AUTHOR
Harri Aaltonen, May 15 2004, Apr 12 2008
EXTENSIONS
Edited by Peter Luschny, Jun 14 2021
STATUS
approved