cp's OEIS Frontend

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.

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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.

Original entry on oeis.org

0, 1, 8, -2, 27, -5, 928, -70, 16, 11249, -2671, 123, 46872, -34354, 5992, -438, 1792225, -445535, 28075, -10303, 23152256, -5824226, 1168304, -178754, 38336, 100685835, -25547957, 5343755, -885717, 101355, 3970817992, -338056246, 72962904, -12914726, 1825464, -386166
Offset: 0

Views

Author

Hugo Pfoertner, Jul 09 2022

Keywords

Comments

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.
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.
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)).
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.
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.

Examples

			The triangle begins:
          0;
          1;
          8,        -2;
         27,        -5;
        928,       -70,      16;
      11249,     -2671,     123;
      46872,    -34354,    5992,    -438;
    1792225,   -445535,   28075,  -10303;
   23152256,  -5824226, 1168304, -178754,  38336;
  100685835, -25547957, 5343755, -885717, 101355;
. The combined triangles used to calculate the resistances are:
   \ j                0              |                 1               |
   k\---------- s/t ----------- u/v -|----------- s/t ----------- u/v -|
   0|           0/1             0/ 1 |             .               .   |
   1|           1/3             0/ 1 |             .               .   |
   2|           8/3            -2/ 1 |           -2/3             1/ 1 |
   3|          27/1           -24/ 1 |           -5/1             5/ 1 |
   4|         928/3          -280/ 1 |          -70/1            64/ 1 |
   5|       11249/3         -3400/ 1 |        -2671/3           808/ 1 |
   6|       46872/1       -212538/ 5 |       -34354/3         51929/ 5 |
   7|     1792225/3      -2708944/ 5 |      -445535/3        673429/ 5 |
   8|    23152256/3    -244962336/35 |     -5824226/3      61623224/35 |
   9|   100685835/1   -3195918288/35 |    -25547957/1     810930216/35 |
  10|  3970817992/3  -42013225014/35 |   -338056246/1    2146081719/ 7 |
  11| 52514317745/3 -111125508824/ 7 | -13481564911/3  142641647567/35 |
.
continued
   \ j             2              |               3            |
   k\-------- s/t ---------- u/v -|--------- s/t -------- u/v -|
   4|        16/1          -14/ 1 |           .            .   |
   5|       123/1         -111/ 1 |           .            .   |
   6|      5992/3        -9054/ 5 |       -438/1       1989/5  |
   7|     28075/1      -127303/ 5 |     -10303/3      15576/5  |
   8|   1168304/3    -12361214/35 |    -178754/3    1891328/35 |
   9|   5343755/1   -169618717/35 |    -885717/1   28113999/35 |
  10|  72962904/1  -2315951182/35 |  -12914726/1   81986531/ 7 |
  11| 993810715/1 -31545031729/35 | -184858117/1 5867671888/35 |
.
continued
   \ j           4             |             5           |
   k\------- s/t -------- u/v -|------- s/t ------- u/v -|
   8|    38336/3    -405592/35 |         .           .   |
   9|   101355/1   -3217136/35 |         .           .   |
  10|  1825464/1  -57942922/35 |  -386166/1  12257507/35 |
  11| 28123355/1 -892677136/35 | -3085317/1  97932579/35 |
.
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.

References

  • See A211074 for more references and links (with alternatives).

Links

Crossrefs

A355586 are the corresponding denominators t.
A355587 and A355588 are u and v.
Cf. A307012 (discussion of oblique coordinate system).
Cf. A084768 (when divided by 3 apparently gives the difference between successive values of s/t in column 0).
Cf. A355565, A355566, A355567 (similar problem for the square lattice).

Programs

  • 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};
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
  • PARI
    \\ Alternative method using a recurrence; calculates triangle of s/t
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]};
D(n) = subst(pollegendre(n), 'x, 7);
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};
ST(11)

Formula

T(n,0)/A355586(n,0) = T(n-1,0)/A355586(n-1,0) + A084768(n-1)/3 for n>=1 (conjectured).

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.

Original entry on oeis.org

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
Offset: 0

Views

Author

Hugo Pfoertner, Jul 07 2022

Keywords

Comments

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

Examples

			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
.
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 |
.
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 |
.
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.

References

  • See A211074 for more references and links.

Links

Crossrefs

A131406 are the corresponding denominators t, with indices shifted by 1.
A355566 and A355567 are u and v.
Cf. A025547, A025550, A089165, A211074, A355953, A355955, A356201, A356202.
Cf. A355585, A355586, A355587, A355588 (same problem for the infinite triangular lattice).
Cf. A280079, A280317.

Programs

  • Maple
    See link.
  • Mathematica
    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 *)
  • 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 (p0, 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())

Formula

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.
From Rainer Rosenthal, Aug 04 2022: (Start)
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)
T(j+1,0) = A089165(j)/(1 + mod(j,2)) for j >= 0. - Hugo Pfoertner, Aug 21 2022

A355586 T(j,k) are the denominators t 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.

Original entry on oeis.org

1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 3, 3, 1, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 3, 3, 1, 3, 1, 1, 1, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1, 1, 3, 3, 1, 1, 3, 3, 3, 3, 1
Offset: 0

Views

Author

Hugo Pfoertner, Jul 09 2022

Keywords

Comments

See A355585 for more information.

Examples

			The triangle begins:
  1;
  3;
  3, 3;
  1, 1;
  3, 1, 1;
  3, 3, 1;
  1, 3, 3, 1;
  3, 3, 1, 3;
  3, 3, 3, 3, 3;
  1, 1, 1, 1, 1;

References

  • See A211074 for references and links.

Crossrefs

A355585 are the corresponding numerators.
A355587 and A355588 are u and v.

Programs

A355566 T(j,k) are the numerators u 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.

Original entry on oeis.org

0, 0, 1, -2, 2, 4, -12, 23, 2, 23, -184, 40, -118, 12, 176, -940, 3323, -1118, 499, 20, 563, -24526, 1234, -18412, 13462, -626, 118, 6508, -130424, 721937, -71230, 327143, -1312, 14369, 262, 88069, -4924064, 191776, -6601046, 2395676, -888568, 131972, -300766, 1624, 91072
Offset: 0

Views

Author

Hugo Pfoertner, Jul 07 2022

Keywords

Comments

See A355565 for more information.
On the diagonal we have T(0,0) = 0 and T(n,n) = A350669(n-1) for n > 0. - Rainer Rosenthal, Aug 01 2022

Examples

			The triangle begins:
       0;
       0,    1;
      -2,    2,      4;
     -12,   23,      2,    23;
    -184,   40,   -118,    12,  176;
    -940, 3323,  -1118,   499,   20, 563;
  -24526, 1234, -18412, 13462, -626, 118, 6508;

References

  • See A211074 for references and links.

Links

Crossrefs

A355567 are the corresponding denominators v.
A355565 and A131406 (with changed offset) are s and t.
Cf. A350669.

Programs

  • PARI
    \\ uses function R(m, p, x) given in A355565
for (j=0, 8, for (k=0, j, my(q=(pi/2)*R(j,k)); print1(numerator(polcoef(q,0,pi)),", ")); print())

A355567 T(j,k) are the denominators v 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.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 3, 15, 3, 1, 15, 5, 105, 3, 15, 15, 35, 21, 315, 15, 1, 35, 105, 45, 45, 3465, 15, 105, 21, 315, 7, 693, 231, 45045, 105, 5, 315, 315, 495, 495, 15015, 585, 45045, 7, 315, 45, 3465, 3465, 45045, 45045, 15015, 385, 765765, 315, 35, 3465, 495, 45045, 6435, 15015, 45045, 765765, 9945, 14549535
Offset: 0

Views

Author

Hugo Pfoertner, Jul 07 2022

Keywords

Comments

See A355565 for more information.
On the diagonal we have T(0,0) = 1 and T(n,n) = A350670(n-1) for n > 0. - Rainer Rosenthal, Aug 01 2022

Examples

			The triangle begins:
   1;
   1,  1;
   1,  1,  3;
   1,  3,  3,  15;
   3,  1, 15,   5, 105;
   3, 15, 15,  35,  21, 315;
  15,  1, 35, 105,  45,  45, 3465

References

  • See A211074 for references and links.

Links

Crossrefs

A355566 are the corresponding numerators u.
A355565 and A131406 (with changed offset) are s and t.
Cf. A350670.

Programs

  • PARI
    \\ uses function R(m, p, x) given in A355565
for (j=0, 8, for (k=0, j, my(q=(pi/2)*R(j, k)); print1(denominator(polcoef(q, 0, pi)), ", ")); print())

A355955 a(n) is the least distance of two nodes on the same grid line in an infinite square lattice of one-ohm resistors for which the resistance measured between the two nodes is greater than n ohms.

Original entry on oeis.org

1, 5, 107, 2460, 56922, 1317211, 30481165, 705355254, 16322409116
Offset: 0

Views

Author

Hugo Pfoertner, Jul 23 2022

Keywords

Comments

The terms are obtained by a high-precision evaluation of the integral R(j,k) = (1/Pi) * Integral_{beta=0..Pi} (1 - exp(-abs(j)*alphas(beta))*cos(k*beta)) / sinh(alphas(beta)), with alphas(beta) = log(2 - cos(beta) + sqrt(3 + cos(beta)*(cos(beta) - 4))) such that floor(R(m-1,0)) < floor(R(m,0)). The values of m for which this condition is satisfied are the terms of the sequence. See Atkinson and van Steenwijk (1999, page 491, Appendix B) for a Mathematica implementation of the integral.
a(9) = 377711852375, found by solving R(x) - 9 = 0, using the asymptotic formula provided by Cserti (2000, page 5), R(x) = (log(x) + gamma + log(8)/2)/Pi, needs independent confirmation. gamma is A001620.

Examples

			a(0) = 1: R(1,0) = 1/2 is the first resistance > 0;
a(1) = 5: R(4,0) = 0.953987..., R(5,0) = 1.025804658...;
a(2) = 107: R(106,0) = 1.999103258858..., R(107,0) = 2.002092149977722...;
a(3) = 2460: R(2459,0) = 2.999894481..., R(2460,0) = 3.0000239019301...;
a(4) = 56922: R(56921,0) = 3.99999536602..., R(56922,0) =  4.0000009581... .

Links

Crossrefs

Cf. A355565, A355589 (same problem for triangular lattice).

Programs

  • PARI
    \\ can be used to calculate estimates of terms for n >= 2, using the asymptotic formula. For n <= 8 results identical to those using the exact evaluation of the full integral are produced, but equality for higher terms might not hold, although with extremely remote probability.
a355955_asymp(upto) = {my(c=2.2, Rsqasy(L)=(1/Pi)*(log(L)+Euler+log(8)/2), d, m); for (n=2, upto, d=exp(c*n); d=solve(x=0.5*d, 2.5*d, Rsqasy(x)-n); print1(ceil(d),", "); c=log(d)/n)};
a355955_asymp(8)

A355589 a(n) is the least distance of two nodes on the same grid line in an infinite triangular lattice of one-ohm resistors for which the resistance measured between the two nodes is greater than n ohms.

Original entry on oeis.org

1, 38, 8632, 1991753, 459625866
Offset: 0

Views

Author

Hugo Pfoertner, Jul 23 2022

Keywords

Comments

The terms are obtained by a high-precision evaluation of the integral 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)), such that floor(R(m-1,0)) < floor(R(m,0)). The values of m for which this condition is satisfied are the terms of the sequence. See Atkinson and van Steenwijk (1999, page 491, Appendix B) for a Mathematica implementation of the integral.

Examples

			a(0) = 1: R(1,0) = 1/3 is the first resistance > 0;
a(1) = 38: R(37,0) = 0.9980131561985..., R(38,0) = 1.0029141482654...;
a(2) = 8632: R(8631) = 1.99999787859849..., R(8632) = 2.000019169949784851...;
a(3) = 1991753: R(1991752) = 2.99999998586..., R(1991753) = 3.000000078131...;
a(4) = 459625866: R(459625865)=3.999999999731...; R(459625866)=4.000000000131....
Assuming a fitted asymptotic logarithmic growth of R(x,0) = log(x)/(Pi*sqrt(3)) + 0.334412..., a(5) is approximately 1.06*10^11, but 250 GByte of main memory is not enough for PARI's function intnum to compute the value of the integral for arguments of that size.

Links

  • D. Atkinson and F. J. van Steenwijk, Infinite resistive lattices, Am. J. Phys. 67 (1999), 486-492. (See A211074 for an alternative link.)

Crossrefs

Cf. A355585, A355955 (same problem for square lattice).
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