A339386 Erroneous version of A338487.
1, 5, 37, 226, 1460, 9235
Offset: 5
Keywords
Links
- Joel Karnofsky, Solution of problem from Technology Review's Puzzle Corner Oct 3, 2003, Feb 23 2004.
This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(n) counts all resistances that can be obtained with fewer than n resistors as well as with exactly n resistors. Without a resistor the resistance is infinite, i.e., a(0) = 1. One 1-ohm resistor adds resistance 1, so a(1) = 2. Two resistors in parallel give 1/2 ohm, while in series they give 2 ohms. So a(2) is the number of elements in the set {infinity, 1, 1/2, 2}, i.e., a(2) = 4. - _Rainer Rosenthal_, Feb 07 2021
(* See link. *)
a(6) = 44 because the resistances 11/13 and 13/11 (in units of resistor value) are representable in addition to the A051389(6)=42 resistances that can be achieved by only serial and parallel configurations with exactly 6 resistors and not by a network with fewer than 6 resistors.
Example 1: Five equal unit resistors. Each arm of the bridge has one unit resistor, leading to an equivalent resistance of 1; so the set is {1} and its order is 1. Example 2: Six equal unit resistors. Four arms have one unit resistor each and the fifth arm has two unit resistors. Two resistors in the same arm, when combined in series and parallel result in 2 and 1/2 respectively (corresponding to 2: {1/2, 2} in A048211). The set {1/2, 2}, in the diagonal results in {1}. Set {1/2, 2} in any of the four arms results in {11/13, 13/11}. Consequently, with six equal resistors, we have the set {11/13, 1, 13/11}, whose order is 3. Union of the previous terms is {1} and the union with these three is again {11/13, 1, 13/11}. So the terms for five and six resistors are 1 and 3 respectively.
The a(3) = 3 CDE-descendants of A-Z with 3 edges are
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A A A
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o o - o o - o
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Z Z Z
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DCC DD DE
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\\ See A339065 for G. InvEulerT(v)={my(p=log(1+x*Ser(v))); dirdiv(vector(#v,n,polcoef(p,n)), vector(#v,n,1/n))} seq(n)={my(A=O(x*x^n), g=G(2*n, x+A,[]), gr=G(2*n, x+A,[1])/g, u=InvEulerT(Vec(-1+G(2*n, x+A,[1,1])/(g*gr^2))), t=InvEulerT(Vec(-1+G(2*n, x+A,[2])/(g*subst(gr,x,x^2)))), v=vector(n)); for(n=1, #v, v[n]=(u[n]+t[n]-if(n%2==0,u[n/2]-v[n/2]))/2); v} \\ Andrew Howroyd, Nov 20 2020
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a(6) = 4, because the last of these 5 networks (Fig. 5) is not 2-connected: when the middle vertex is removed, then A and Z are part of two separated subgraphs.
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A A A A A
// \ / \ d \ / \ /|
// \ /___\ / \ / \ / |
o-----o o --- o o-----o o--o--o o--o--o
\ / \ / \ / \ / | /
\ / \ / \ / \ / |/
Z Z Z Z Z
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Fig. 1 Fig. 2 Fig. 3 Fig. 4 Fig. 5
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Figures 1 to 4 correspond to N1, N2, N4 and N5 in the example section of A338487.
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