A093652 - OEIS

%I #47 Oct 24 2023 03:21:35

%S 1,2,7,15,45,86,239,433,1157,2034,5307,9151,23497,39974,101467,170913,

%T 430089,718946,1796975,2985775,7422437,12272502,30373191,50016721,

%U 123327373,202395986,497484067,814061151,1995542913,3257222726,7965875891,12973832257,31663779857

%N 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.

%C 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.

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

%H Harri Aaltonen, Apr 18 2008, <a href="/A093652/b093652.txt">Table of n, a(n) for n = 1..50</a> [a(49) corrected by _Georg Fischer_, Mar 13 2020]

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,8,0,-18,0,8,0,-1).

%H <a href="/index/Res#resistances">Index to sequences related to resistances</a>.

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

%F From _Colin Barker_, Dec 20 2019: (Start)

%F 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.

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

%F (End)

%p a_list := proc(last) local B, C, k;

%p    B := [1,2,5, 8];

%p    C := [1,2,7,15];

%p    for k from 5 to last do

%p       B := [op(B), 4*B[k-2]-B[k-4]];

%p       C := [op(C), (k*B[k]-B[k-2])/2];

%p    od;

%p C end:

%p a_list(50); # After _Harri Aaltonen_, _Peter Luschny_, Mar 14 2020

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

%Y Cf. A082630.

%K nonn,easy

%O 1,2

%A _Harri Aaltonen_, May 15 2004, Apr 12 2008

%E Edited by _Peter Luschny_, Jun 14 2021