OFFSET
1,1
COMMENTS
The resistance is measured between two corners of the triangular region.
LINKS
Hugo Pfoertner, Table of n, a(n) for n = 1..50
Hugo Pfoertner, Graph of R(n), bounded or unbounded for n->oo?
EXAMPLE
R(1) = a(1)/A338402(1) = 2/3,
R(2) = a(2)/A338402(2) = 10/9,
R(4) = a(4)/A338402(4) = 206/123.
a(3) = 10: The following network of A045943(3) = 18 one Ohm resistors has a resistance of R(3) = 10/7 Ohm, i.e., the current I driven by the voltage of 1 Volt is 7/10 = A338402(3)/a(3) Ampere.
.
O
__/ \_
/ / \ \
/1/ \1\
/_/ \_\
/ _____ \
O---|__1__|---O
__/ \_ __/ \_
/ / \ \ / / \ \
/1/ \1\ /1/ \1\
/_/ \_\ /_/ \_\
/ _____ \ / _____ \
O---|__1__|---O---|__1__|---O
__/ \__ __/ \__ __/ \_
/ / \ \ / / \ \ / / \ \
/1/ \1\ /1/ \1\ /1/ \1\
/_/ \_\ /_/ \_\ /_/ \_\
/ _____ \ / _____ \ / _____ \
O---|__1__|---O---|__1__|---O---|__1__|---O
| |
| V = 1 Volt |
| | |
-------------------| |-- I=1/R Ampere ---
|
.
With a numbering of the resistors as shown in the following diagram,
.
O
/ \
15 18
/ \
O--14---O
/ \ / \
7 9 13 17
/ \ / \
O-- 6---O--12---O
/ \ / \ / \
2 3 5 8 11 16
/ \ / \ / \
O---1---O---4---O--10---O
|______1 Volt__I=I19____|
.
the currents in Amperes through the 18 resistors, and the current I=I19 through the voltage source of 1 Volt, are [11/30, 1/3, 1/30, 4/15, 2/15, 1/6, 2/15, 2/15, 1/30, 11/30, 1/30, 1/6, 1/30, 1/15, 1/30, 1/3, 2/15, 1/30, 7/10].
PROG
(PARI) a33840_1_2(n)={my(md=3*n*(n+1)/2+1,
T1=matrix(n, n), T2=matrix(n, n), T3=matrix(n, n),
M=matrix(md, md, i, j, 0), U=vector(md),
valid(i, j)=i>0&&i<=n&&j>0&&j<=n&&i>=j, k=0, neq=1);
\\ List of edges
for(i=1, n, for(j=1, i, T1[i, j]=k++; T2[i, j]=k++; T3[i, j]=k++));
\\ In- and outflow of current at all nodes
\\ lower left triangle with inflow of current from source of voltage
M[1, 1]=-1; M[1, 2]=-1; M[1, md]=1;
\\ loops over lower left corners of triangles
for(i=2, n+1, for(j=1, i,
\\ exclude node at top of triangle
if(j<n+1, neq++;
if(valid(i-1, j), M[neq, T1[i-1, j]]=1; M[neq, T3[i-1, j]]=1);
if(valid(i-1, j-1), M[neq, T2[i-1, j-1]]=1; M[neq, T3[i-1, j-1]]=-1);
if ( valid(i, j), M[neq, T1[i, j]]=-1; M[neq, T2[i, j]]=-1);
\\ lower right corner with current through voltage source
if ( i == n+1 && j == 1, M[neq, md] = -1)
)));
\\ sum of voltages around triangles with vertex above base
for( i = 1, n, for( j = 1, i,
neq++; M[neq, T1[i, j]] = 1; M[neq, T2[i, j]] = -1; M[neq, T3[i, j]] = -1 ));
\\ sum of voltages around triangles with vertex below base
for( i = 1, n-1, for( j = 1, i, neq ++;
M[neq, T3[i, j]] = 1; M[neq, T2[i+1, j]] = 1; M[neq, T1[i+1, j+1]] = -1 ));
\\ External voltage applied to lower corners of triangle
neq = neq++; for ( i = 1, n, M[neq, T1[i, 1]] = 1);
\\ Right side of equations; driving voltage 1 Volt
U[neq]=1; 1/matsolve(M, U~)[neq]};
for(n=1, 10, print1(numerator(a33840_1_2(n)), ", "))
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Hugo Pfoertner, Oct 24 2020
STATUS
approved