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A355585 T(j,k) are the numerators s in the representation R = s/t + (2*sqrt(3)/Pi)*u/v of the resistance between two nodes separated by the distance (j,k) in an infinite triangular lattice of one-ohm resistors, where T(j,k), j >= 0, 0 <= k <= floor(j/2) is an irregular triangle read by rows.

%I A355585 #38 Sep 19 2022 13:59:56
%S A355585 0,1,8,-2,27,-5,928,-70,16,11249,-2671,123,46872,-34354,5992,-438,
%T A355585 1792225,-445535,28075,-10303,23152256,-5824226,1168304,-178754,38336,
%U A355585 100685835,-25547957,5343755,-885717,101355,3970817992,-338056246,72962904,-12914726,1825464,-386166
%N A355585 T(j,k) are the numerators s in the representation R = s/t + (2*sqrt(3)/Pi)*u/v of the resistance between two nodes separated by the distance (j,k) in an infinite triangular lattice of one-ohm resistors, where T(j,k), j >= 0, 0 <= k <= floor(j/2) is an irregular triangle read by rows.
%C A355585 The distance vector (j,k) is defined in an oblique coordinate system with an angle of 120 degrees between the axes, see e.g. A307012.
%C A355585 Atkinson and Steenwijk (1999) (see links in A211074) provided a generalization of the method used to calculate the resistance between two arbitrary nodes in an infinite square lattice of one-ohm resistors to infinite triangular lattices. Similar to the square lattice, the integral describing the resistance distance between nodes can exactly be represented by an expression of the form given in the name of this sequence with integer coefficients. Atkinson and Steenwijk, page 489, provided results for j <= 3 found by evaluation of the integral (17) (given below) and application of Mathematica's "Simplify" function.
%C A355585 R(j,k) = (1/Pi) * Integral_{y=0..Pi/2} (1 - exp(-|j-k|*x)*cos((j+k)*y)) / (sinh(x)*cos(y)) dy, with x = arccosh(2/cos(y)-cos(y)).
%C A355585 It would be useful to know whether, since the publication cited, a recurrence analogous to that known for the square lattice (used in A355565) for determining the coefficients has also been found for the triangular lattice.
%C A355585 The results in this sequence were found by systematic parameter variation of u and v and continued fraction expansion of the difference from the exact value of the integral for the resistance distance to determine s/t.
%D A355585 See A211074 for more references and links (with alternatives).
%H A355585 D. Atkinson and F. J. van Steenwijk, <a href="http://dx.doi.org/10.1119/1.19311">Infinite resistive lattices</a>, Am. J. Phys. 67 (1999), 486-492.
%H A355585 R. J. Mathar, <a href="http://vixra.org/abs/2208.0111">Recurrence for the Atkinson-Steenwijk Integrals for Resistors in the Infinite Triangular Lattice</a>, viXra:2208.0111 (2022).
%H A355585 Hugo Pfoertner, <a href="/A355585/a355585.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].
%F A355585 T(n,0)/A355586(n,0) = T(n-1,0)/A355586(n-1,0) + A084768(n-1)/3 for n>=1 (conjectured).
%e A355585 The triangle begins:
%e A355585           0;
%e A355585           1;
%e A355585           8,        -2;
%e A355585          27,        -5;
%e A355585         928,       -70,      16;
%e A355585       11249,     -2671,     123;
%e A355585       46872,    -34354,    5992,    -438;
%e A355585     1792225,   -445535,   28075,  -10303;
%e A355585    23152256,  -5824226, 1168304, -178754,  38336;
%e A355585   100685835, -25547957, 5343755, -885717, 101355;
%e A355585 . The combined triangles used to calculate the resistances are:
%e A355585    \ j                0              |                 1               |
%e A355585    k\---------- s/t ----------- u/v -|----------- s/t ----------- u/v -|
%e A355585    0|           0/1             0/ 1 |             .               .   |
%e A355585    1|           1/3             0/ 1 |             .               .   |
%e A355585    2|           8/3            -2/ 1 |           -2/3             1/ 1 |
%e A355585    3|          27/1           -24/ 1 |           -5/1             5/ 1 |
%e A355585    4|         928/3          -280/ 1 |          -70/1            64/ 1 |
%e A355585    5|       11249/3         -3400/ 1 |        -2671/3           808/ 1 |
%e A355585    6|       46872/1       -212538/ 5 |       -34354/3         51929/ 5 |
%e A355585    7|     1792225/3      -2708944/ 5 |      -445535/3        673429/ 5 |
%e A355585    8|    23152256/3    -244962336/35 |     -5824226/3      61623224/35 |
%e A355585    9|   100685835/1   -3195918288/35 |    -25547957/1     810930216/35 |
%e A355585   10|  3970817992/3  -42013225014/35 |   -338056246/1    2146081719/ 7 |
%e A355585   11| 52514317745/3 -111125508824/ 7 | -13481564911/3  142641647567/35 |
%e A355585 .
%e A355585 continued
%e A355585    \ j             2              |               3            |
%e A355585    k\-------- s/t ---------- u/v -|--------- s/t -------- u/v -|
%e A355585    4|        16/1          -14/ 1 |           .            .   |
%e A355585    5|       123/1         -111/ 1 |           .            .   |
%e A355585    6|      5992/3        -9054/ 5 |       -438/1       1989/5  |
%e A355585    7|     28075/1      -127303/ 5 |     -10303/3      15576/5  |
%e A355585    8|   1168304/3    -12361214/35 |    -178754/3    1891328/35 |
%e A355585    9|   5343755/1   -169618717/35 |    -885717/1   28113999/35 |
%e A355585   10|  72962904/1  -2315951182/35 |  -12914726/1   81986531/ 7 |
%e A355585   11| 993810715/1 -31545031729/35 | -184858117/1 5867671888/35 |
%e A355585 .
%e A355585 continued
%e A355585    \ j           4             |             5           |
%e A355585    k\------- s/t -------- u/v -|------- s/t ------- u/v -|
%e A355585    8|    38336/3    -405592/35 |         .           .   |
%e A355585    9|   101355/1   -3217136/35 |         .           .   |
%e A355585   10|  1825464/1  -57942922/35 |  -386166/1  12257507/35 |
%e A355585   11| 28123355/1 -892677136/35 | -3085317/1  97932579/35 |
%e A355585 .
%e A355585 Using the terms for (j,k) = (10,5) with {s, t, u, v} = {-386166, 1, 12257507, 35} the resistance is R = T(10,5)/A355586(10,5) + (2*sqrt(3)/Pi) * A355587(10,5)/A355588(10,5) = -386166/1 + (2*sqrt(3)/Pi)*12257507/35 = 0.731139136228538824636... . This equals the integral for the resistance distance R(j,k) after substitution of j=10 and k=5.
%o A355585 (PARI) Rtri(n,p)={my(alphat(beta)=acosh(2/cos(beta)-cos(beta))); intnum (beta=0, Pi/2, (1 - exp (-abs(n-p) * alphat(beta))*cos((n+p)*beta)) / (cos(beta)*sinh(alphat(beta)))) / Pi};
%o A355585 searchr (target, maxn=1000000, maxd=10, maxrat=1000, minn=0, mind=1) = {my (Rcons=2*sqrt(3)/Pi, delta=oo); for (d=mind, maxd, my(PP=Rcons/d); for (nn=minn, maxn, foreach ([-nn,nn], n, my (P=PP*n, T=target-P, Q = bestappr(T,maxrat), D=abs(target-P-Q)); if(D<delta, delta=D; if (delta < 10^(-getlocalprec()*5\6), return ([numerator(Q), denominator(Q),n,d]) )))))};
%o A355585 \\ j=6 takes a while; prints [j,k][s,t,u,v]
%o A355585 for (j=0,6, for (k=0,j\2, my(R=Rtri(j,k)); S=searchr(R,500000); print([j,k],S)));
%o A355585 (PARI) \\ Alternative method using a recurrence; calculates triangle of s/t
%o A355585 jk(j,k) = {my(jj=j,kk=k); if(k<1,jj=j-k+1;kk=2-k); my(km=(jj+1)/2); if(kk>km, kk=2*km-kk); [jj,kk]};
%o A355585 D(n) = subst(pollegendre(n), 'x, 7);
%o A355585 ST(nend) = {my(nmax=nend+1, N=matrix(nmax,(nmax+1)\2)); for (n=2, nmax, N[n,1]=(1/3) * sum(k=0,n-2,D(k))); for (n=3, nmax, N[n,2] = (1/2)*(6*N[n-1,1] - 2*N[jk(n-1,2)[1],jk(n-1,2)[2]] - N[n-2,1] - N[n,1])); for (n=5, nmax, for (m=3, (n+1)\2, N[n,m] = 6*N[jk(n-1,m-1)[1],jk(n-1,m-1)[2]] - N[jk(n-1,m)[1],jk(n-1,m)[2]] - N[jk(n-2,m-1)[1],jk(n-2,m-1)[2]] - N[jk(n-2,m-2)[1],jk(n-2,m-2)[2]] - N[jk(n-1,m-2)[1],jk(n-1,m-2)[2]] - N[jk(n,m-1)[1],jk(n,m-1)[2]] )); N};
%o A355585 ST(11)
%Y A355585 A355586 are the corresponding denominators t.
%Y A355585 A355587 and A355588 are u and v.
%Y A355585 Cf. A307012 (discussion of oblique coordinate system).
%Y A355585 Cf. A084768 (when divided by 3 apparently gives the difference between successive values of s/t in column 0).
%Y A355585 Cf. A355565, A355566, A355567 (similar problem for the square lattice).
%K A355585 tabf,frac,sign
%O A355585 0,3
%A A355585 _Hugo Pfoertner_, Jul 09 2022