%I #22 Oct 31 2020 03:47:32
%S 2,10,10,206,3326,43118,150806,11591578,436494606,1008712015454,
%T 382034633808890,13187511533010430,2111825680430510462,
%U 171204772756285452656378,89579048665281690355286,1013412795315891086553473628734,20023655015717377508089133638478,24678955315461926144059519221489609194
%N a(n) is the numerator of the resistance R(n) = a(n)/A338402(n) of a triangular network of 3*n*(n+1)/2 one Ohm resistors in a hexagonal lattice arrangement.
%C The resistance is measured between two corners of the triangular region.
%H Hugo Pfoertner, <a href="/A338401/b338401.txt">Table of n, a(n) for n = 1..50</a>
%H Hugo Pfoertner, <a href="https://oeis.org/plot2a?graph=1&name1=A338401&name2=A338402&tform1=untransformed&tform2=untransformed&shift=0&radiop1=ratio&drawpoints=true&drawlines=true">Graph of R(n)</a>, bounded or unbounded for n->oo?
%e R(1) = a(1)/A338402(1) = 2/3,
%e R(2) = a(2)/A338402(2) = 10/9,
%e R(4) = a(4)/A338402(4) = 206/123.
%e 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.
%e .
%e O
%e __/ \_
%e / / \ \
%e /1/ \1\
%e /_/ \_\
%e / _____ \
%e O---|__1__|---O
%e __/ \_ __/ \_
%e / / \ \ / / \ \
%e /1/ \1\ /1/ \1\
%e /_/ \_\ /_/ \_\
%e / _____ \ / _____ \
%e O---|__1__|---O---|__1__|---O
%e __/ \__ __/ \__ __/ \_
%e / / \ \ / / \ \ / / \ \
%e /1/ \1\ /1/ \1\ /1/ \1\
%e /_/ \_\ /_/ \_\ /_/ \_\
%e / _____ \ / _____ \ / _____ \
%e O---|__1__|---O---|__1__|---O---|__1__|---O
%e | |
%e | V = 1 Volt |
%e | | |
%e -------------------| |-- I=1/R Ampere ---
%e |
%e .
%e With a numbering of the resistors as shown in the following diagram,
%e .
%e O
%e / \
%e 15 18
%e / \
%e O--14---O
%e / \ / \
%e 7 9 13 17
%e / \ / \
%e O-- 6---O--12---O
%e / \ / \ / \
%e 2 3 5 8 11 16
%e / \ / \ / \
%e O---1---O---4---O--10---O
%e |______1 Volt__I=I19____|
%e .
%e 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].
%o (PARI) a33840_1_2(n)={my(md=3*n*(n+1)/2+1,
%o T1=matrix(n,n),T2=matrix(n,n),T3=matrix(n,n),
%o M=matrix(md,md,i,j,0),U=vector(md),
%o valid(i,j)=i>0&&i<=n&&j>0&&j<=n&&i>=j,k=0,neq=1);
%o \\ List of edges
%o for(i=1,n,for(j=1,i,T1[i,j]=k++;T2[i,j]=k++;T3[i,j]=k++));
%o \\ In- and outflow of current at all nodes
%o \\ lower left triangle with inflow of current from source of voltage
%o M[1,1]=-1;M[1,2]=-1;M[1,md]=1;
%o \\ loops over lower left corners of triangles
%o for(i=2,n+1,for(j=1,i,
%o \\ exclude node at top of triangle
%o if(j<n+1,neq++;
%o if(valid(i-1,j),M[neq,T1[i-1,j]]=1;M[neq,T3[i-1,j]]=1);
%o if(valid(i-1,j-1),M[neq,T2[i-1,j-1]]=1; M[neq,T3[i-1,j-1]]=-1);
%o if ( valid(i,j),M[neq,T1[i,j]]=-1;M[neq,T2[i,j]]=-1);
%o \\ lower right corner with current through voltage source
%o if ( i == n+1 && j == 1, M[neq,md] = -1)
%o )));
%o \\ sum of voltages around triangles with vertex above base
%o for( i = 1, n, for( j = 1, i,
%o neq++; M[neq,T1[i,j]] = 1; M[neq,T2[i,j]] = -1; M[neq,T3[i,j]] = -1 ));
%o \\ sum of voltages around triangles with vertex below base
%o for( i = 1, n-1, for( j = 1, i, neq ++;
%o M[neq,T3[i,j]] = 1; M[neq,T2[i+1,j]] = 1; M[neq,T1[i+1,j+1]] = -1 ));
%o \\ External voltage applied to lower corners of triangle
%o neq = neq++; for ( i = 1, n, M[neq,T1[i,1]] = 1);
%o \\ Right side of equations; driving voltage 1 Volt
%o U[neq]=1;1/matsolve(M,U~)[neq]};
%o for(n=1,10,print1(numerator(a33840_1_2(n)),", "))
%Y Cf. A000217, A045943, A048211, A174283, A338402.
%K nonn,frac
%O 1,1
%A _Hugo Pfoertner_, Oct 24 2020