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A355565
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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.
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10
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0, 1, 0, 2, -1, 0, 17, -4, 1, 0, 40, -49, 6, -1, 0, 401, -140, 97, -8, 1, 0, 1042, -1569, 336, -161, 10, -1, 0, 11073, -4376, 4321, -660, 241, -12, 1, 0, 29856, -48833, 13342, -9681, 1144, -337, 14, -1, 0, 325441, -136488, 160929, -33188, 18929, -1820, 449, -16, 1, 0
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OFFSET
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0,4
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COMMENTS
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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).
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
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REFERENCES
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See A211074 for more references and links.
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LINKS
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Hugo Pfoertner, PARI program for inverse problem, (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].
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FORMULA
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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.
R(0,0) = 0; R(1,0) = 1/2.
R(n,n) = R(n-1,n-1) + (2/Pi)/(2*n-1) for n >= 1.
R(j,k) = R(k,j) and R(-j,k) = R(j,k).
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).
(End)
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EXAMPLE
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The triangle begins:
0;
1, 0;
2, -1, 0;
17, -4, 1, 0;
40, -49, 6, -1, 0;
401, -140, 97, -8, 1, 0;
1042, -1569, 336, -161, 10, -1, 0
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The combined triangles used to calculate the resistances are:
\ k 0 | 1 | 2 | 3 |
\ s/t u/v | s/t u/v | s/t u/v | s/t u/v |
j \---------------|-----------------|---------------|--------------|
0 | 0 0 | . . | . . | . . |
1 | 1/2 0 | 0 1 | . . | . . |
2 | 2 -2 | -1/2 2 | 0 4/3 | . . |
3 | 17/2 -12 | -4 23/3 | 1/2 2/3 | 0 23/15 |
4 | 40 -184/3 | - 49/2 40 | 6 -118/15 | -1/2 12/5 |
5 | 401/2 -940/3 | -140 3323/15 | 97/2 -1118/15 | -8 499/35 |
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continued:
\ k 4 | 5 |
\ s/t u/v | s/t u/v |
j \-------------|--------------|
0 | . . | . . |
1 | . . | . . |
2 | . . | . . |
3 | . . | . . |
4 | 0 176/105 | . . |
5 | 1/2 20/21 | 0 563/315 |
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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.
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MAPLE
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See link.
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MATHEMATICA
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alphas[beta_] :=
Log[2 - Cos[beta] + Sqrt[3 + Cos[beta]*(Cos[beta] - 4)]];
Rsqu[n_, p_] :=
Simplify[(1/Pi)*
Integrate[(1 - Exp[-Abs[n]*alphas[beta]]*Cos[p*beta])/
Sinh[alphas[beta]], {beta, 0, Pi}]];
Table[Rsqu[n, k], {n, 0, 4}, {k, 0, n}] // TableForm (* Hugo Pfoertner, Aug 21 2022, calculates R, after Atkinson and Steenwijk *)
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PROG
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(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)))};
for(j=0, 9, for(k=0, j, my(q=pi*R(j, k, pi)); print1(numerator(polcoef(q, 1, pi)), ", ")); print())
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CROSSREFS
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A131406 are the corresponding denominators t, with indices shifted by 1.
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KEYWORD
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AUTHOR
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STATUS
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approved
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