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A355565 T(j,k) are the numerators s in the representation R = s/t + (2/Pi)*u/v of the resistance between two nodes separated by the distance vector (j,k) in an infinite square lattice of one-ohm resistors, where T(j,k), j >= 0, 0 <= k <= j, is a triangle read by rows.

%I A355565 #54 Sep 09 2022 14:50:53
%S A355565 0,1,0,2,-1,0,17,-4,1,0,40,-49,6,-1,0,401,-140,97,-8,1,0,1042,-1569,
%T A355565 336,-161,10,-1,0,11073,-4376,4321,-660,241,-12,1,0,29856,-48833,
%U A355565 13342,-9681,1144,-337,14,-1,0,325441,-136488,160929,-33188,18929,-1820,449,-16,1,0
%N A355565 T(j,k) are the numerators s in the representation R = s/t + (2/Pi)*u/v of the resistance between two nodes separated by the distance vector (j,k) in an infinite square lattice of one-ohm resistors, where T(j,k), j >= 0, 0 <= k <= j, is a triangle read by rows.
%C A355565 The recurrence given by Cserti (2000), page 5, (32) is used to calculate the resistance between two arbitrarily spaced nodes in an infinite square lattice whose edges are replaced by one-ohm resistors. The lower triangle, including the diagonal, in Table I of Atkinson and Steenwijk (1999), page 487, is reproduced. The solution to the resistor grid problem shown in the xkcd Web Comic #356 "Nerd Sniping", provided in A211074, is the special case (j,k) = (2,1).
%C A355565 Using the terms of A280079 and A280317 as pairs of grid indices leads to strictly increasing resistances, i.e., R(A280079(m),A280317(m)) > R(A280079(i),A280317(i)) for m > i. This implies that for grid points on the same radius the resistance increases with the circumferential angle between 0 and Pi/4. The further dependence of the resistance along the circumferential angle with a fixed radius results from symmetry. - _Hugo Pfoertner_, Aug 31 2022
%D A355565 See A211074 for more references and links.
%H A355565 Rainer Rosenthal, <a href="/A355565/b355565.txt">Table of n, a(n) for n = 0..135</a>, rows 0..15 of triangle, flattened.
%H A355565 J. Cserti, <a href="http://arxiv.org/abs/cond-mat/9909120">Application of the lattice Green's function for calculating the resistance of infinite networks of resistors</a>, arXiv:cond-mat/9909120 [cond-mat.mes-hall], 1999-2000.
%H A355565 Hugo Pfoertner, <a href="/A355565/a355565_2.txt">Grid points sorted by increasing R values</a>, (2022).
%H A355565 Hugo Pfoertner, <a href="/A355565/a355565.gp.txt">PARI program for inverse problem</a>, (2022). Finds the grid point [x,y] that leads to the best approximation of a given resistance distance R (ohms) between [0,0] and [x,y].
%H A355565 Physics Stack Exchange, <a href="https://physics.stackexchange.com/questions/2072/on-this-infinite-grid-of-resistors-whats-the-equivalent-resistance">On this infinite grid of resistors, what's the equivalent resistance?</a> Answer by user PBS, Apr 21 2018.
%H A355565 Rainer Rosenthal, <a href="/A355565/a355565_1.txt">Maple program</a>
%F A355565 The resistance for the distance vector (j,k) is R(j,k) = T(j,k)/(1+mod(j+k,2)) +(2/Pi)*A355566(j,k)/A355567(j,k), avoiding the use of A131406.
%F A355565 From _Rainer Rosenthal_, Aug 04 2022: (Start)
%F A355565 R(0,0) = 0; R(1,0) = 1/2.
%F A355565 R(n,n) = R(n-1,n-1) + (2/Pi)/(2*n-1) for n >= 1.
%F A355565 R(j,k) = R(k,j) and R(-j,k) = R(j,k).
%F A355565 4*R(j,k) = R(j-1,k) + R(j+1,k) + R(j,k-1) + R(j,k+1) for (j,k) != (0,0).
%F A355565 (End)
%F A355565 T(j+1,0) = A089165(j)/(1 + mod(j,2)) for j >= 0. - _Hugo Pfoertner_, Aug 21 2022
%e A355565 The triangle begins:
%e A355565      0;
%e A355565      1,     0;
%e A355565      2,    -1,   0;
%e A355565     17,    -4,   1,    0;
%e A355565     40,   -49,   6,   -1,  0;
%e A355565    401,  -140,  97,   -8,  1,  0;
%e A355565   1042, -1569, 336, -161, 10, -1, 0
%e A355565 .
%e A355565 The combined triangles used to calculate the resistances are:
%e A355565   \  k      0       |        1        |       2       |      3       |
%e A355565    \    s/t     u/v |    s/t    u/v   |  s/t      u/v |  s/t    u/v  |
%e A355565   j \---------------|-----------------|---------------|--------------|
%e A355565   0 |   0       0   |     .      .    |   .        .  |   .      .   |
%e A355565   1 |   1/2     0   |    0      1     |   .        .  |   .      .   |
%e A355565   2 |   2      -2   |   -1/2    2     |  0        4/3 |   .      .   |
%e A355565   3 |  17/2   -12   |   -4     23/3   |  1/2      2/3 |  0     23/15 |
%e A355565   4 |  40    -184/3 | - 49/2   40     |  6    -118/15 | -1/2   12/5  |
%e A355565   5 | 401/2  -940/3 | -140    3323/15 | 97/2 -1118/15 | -8    499/35 |
%e A355565 .
%e A355565 continued:
%e A355565   \ k     4       |      5       |
%e A355565    \  s/t   u/v   | s/t    u/v   |
%e A355565   j \-------------|--------------|
%e A355565   0 |  .     .    |  .      .    |
%e A355565   1 |  .     .    |  .      .    |
%e A355565   2 |  .     .    |  .      .    |
%e A355565   3 |  .     .    |  .      .    |
%e A355565   4 | 0   176/105 |  .      .    |
%e A355565   5 | 1/2  20/21  | 0    563/315 |
%e A355565 .
%e A355565 E.g., the resistance for a node distance vector (4,1) is R = T(4,1)/A131406(5,2) + (2/Pi)*A355566(4,1)/A355567(4,1) = -49/2 + (2/Pi)*40/1 = 80/Pi - 49/2.
%p A355565 See link.
%t A355565 alphas[beta_] :=
%t A355565 Log[2 - Cos[beta] + Sqrt[3 + Cos[beta]*(Cos[beta] - 4)]];
%t A355565 Rsqu[n_, p_] :=
%t A355565 Simplify[(1/Pi)*
%t A355565    Integrate[(1 - Exp[-Abs[n]*alphas[beta]]*Cos[p*beta])/
%t A355565      Sinh[alphas[beta]], {beta, 0, Pi}]];
%t A355565 Table[Rsqu[n, k], {n, 0, 4}, {k, 0, n}] // TableForm (* _Hugo Pfoertner_, Aug 21 2022, calculates R, after Atkinson and Steenwijk *)
%o A355565 (PARI) R(m,p,x=pi) = {if (m==0 && p==0, return(0)); if (m==1 && p==0, return(1/2)); if (m==1 && p==1, return(2/x)); if(m==p, my(mm=m-1); return(R(mm,mm)*4*mm/(2*mm+1) - R(mm-1,mm-1)*(2*mm-1)/(2*mm+1))); if (p==(m-1), my(mm=m-1); return(2*R(mm,mm) - R(mm,mm-1))); if (p==0, my(mm=m-1); return(4*R(mm,0) - R(mm-1,0) - 2*R(mm,1))); if (p<m && p>0, my(mm=m-1); return(4*R(mm,p) - R(mm-1,p) - R(mm,p+1) - R(mm,p-1)))};
%o A355565 for(j=0,9,for(k=0,j,my(q=pi*R(j,k,pi));print1(numerator(polcoef(q,1,pi)),", "));print())
%Y A355565 A131406 are the corresponding denominators t, with indices shifted by 1.
%Y A355565 A355566 and A355567 are u and v.
%Y A355565 Cf. A025547, A025550, A089165, A211074, A355953, A355955, A356201, A356202.
%Y A355565 Cf. A355585, A355586, A355587, A355588 (same problem for the infinite triangular lattice).
%Y A355565 Cf. A280079, A280317.
%K A355565 tabl,frac,sign
%O A355565 0,4
%A A355565 _Hugo Pfoertner_, Jul 07 2022