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.

Showing 1-2 of 2 results.

A338158 Triangle read by rows: T(n,k) is the coefficient of x^k in the ZZ polynomial of the hexagonal graphene flake O(3,3,n).

Original entry on oeis.org

20, 30, 12, 1, 175, 450, 425, 180, 33, 2, 980, 3308, 4458, 3065, 1140, 225, 22, 1, 4116, 16468, 27293, 24262, 12521, 3796, 653, 58, 2, 14112, 63522, 120848, 126518, 79506, 30681, 7132, 933, 58, 1, 41580, 204180, 429030, 503664, 361690, 163380, 45885, 7588, 648, 20
Offset: 1

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Author

Henryk A. Witek, Oct 14 2020

Keywords

Comments

The maximum k for which T(n,k) is nonzero, denoted as Cl(n), is usually referred to as the Clar number of O(3,3,n); one has: Cl(1)=3, Cl(2)=5, Cl(3)=7, Cl(4)=8, and Cl(n)=9 for n>4.
T(n,k) denotes the number of Clar covers of order k in the hexagonal graphene flake O(3,3,n).
The Kekulé number of O(3,3,n) is given by T(n, 0).
ZZ polynomials of hexagonal graphene flakes O(3,3,n) with n=1..10 are listed in Eq.(36) of Chou, Li and Witek.
ZZ polynomials of hexagonal graphene flakes O(3,3,n) with any n can be obtained from Eq.(13) of Witek, Langner, Mos and Chou.
ZZ polynomials of hexagonal graphene flakes O(3,3,n) can be also computed using ZZDecomposer (see links below), a graphical program to compute ZZ polynomials of general benzenoids.

Examples

			Triangle begins:
       k=0    k=1    k=2    k=3    k=4    k=5   k=6  k=7 k=8 k=9
n=1:    20     30     12      1
n=2:   175    450    425    180     33      2
n=3:   980   3308   4458   3065   1140    225    22    1
n=4:  4116  16468  27293  24262  12521   3796   653   58   2
n=5: 14112  63522 120848 126518  79506  30681  7132  933  58   1
n=6: 41580 204180 429030 503664 361690 163380 45885 7588 648  20
   ...
Row n=4 corresponds to the polynomial 4116 + 16468*x + 27293*x^2 + 24262*x^3 + 12521*x^4 + 3796*x^5 + 653*x^6 + 58*x^7 + 2*x^8.
		

Crossrefs

Column k=0 is A047819.
Other representation of ZZ polynomials of O(3,3,n) is given by A338217.

Programs

  • Maple
    (n,k)->add(binomial(i+k,k)*(binomial(9,i+k)*binomial(n,i+k)+(10*binomial(7,i+k-2)-binomial(6,i+k-2))*binomial(n+1,i+k)+(20*binomial(5,i+k-4)+binomial(3,i+k-3)-binomial(3,i+k-5))*binomial(n+2,i+k)+(10*binomial(3,i+k-6)+binomial(2,i+k-5)+binomial(3,i+k-5))*binomial(n+3,i+k)+binomial(2,i+k-7)*binomial(n+4,i+k)),i = 0..9)

Formula

T(n,k) = Sum_{l=0..9} C(k+l,k) * (C(9,k+l)*C(n,k+l) + (10*C(7,k+l-2) - C(6,k+l-2)) * C(n+1,k+l) + (20*C(5,k+l-4) + C(3,k+l-3) - C(3,k+l-5)) * C(n+2,k+l) + (10*C(3,k+l-6) + C(2,k+l-5) + C(3,k+l-5)) * C(n+3,k+l) + C(2,k+l-7) * C(n+4,k+l)) where C(n,k) = binomial(n,k). This formula can be obtained by a double sum rotation from Eq.(13) of Witek, Langner, Mos and Chou. Eq.(13) was first discovered heuristically as Eq.(37) of Chou, Li and Witek; a formal proof was given in Eqs.(66-71) on pp. 100-102 of Chou and Witek.

A338259 Triangle read by rows: T(n,k) is the coefficient of (1+x)^k in the ZZ polynomial of the hexagonal graphene flake O(3,4,n).

Original entry on oeis.org

1, 12, 18, 41, 24, 120, 200, 120, 24, 11, 36, 306, 996, 1446, 984, 303, 42, 21, 48, 576, 2800, 6525, 7848, 4957, 1644, 274, 22, 11, 60, 930, 6020, 19365, 33600, 32487, 17694, 5336, 858, 71, 21, 72, 1368, 11064, 45435, 103200, 134806, 102912, 45567, 11358, 1510, 86, 1
Offset: 1

Views

Author

Henryk A. Witek, Oct 19 2020

Keywords

Comments

The maximum k for which T(n,k) is nonzero, denoted by Cl(n), is usually referred to as the Clar number of O(3,4,n); one has: Cl(1)=3, Cl(2)=6, Cl(3)=8, Cl(4)=10, Cl(5)=11, and Cl(n)=12 for n>5.
T(n,k) denotes the number of perfect matchings (i.e., Kekulé structures) with k proper sextets for the hexagonal graphene flake O(3,4,n).
ZZ polynomials of hexagonal graphene flakes O(3,4,n) can be computed using ZZDecomposer (see link below), a graphical program to compute ZZ polynomials of benzenoids, or using ZZCalculator (see link below).

Examples

			Triangle begins:
   k=0 k=1 k=2   k=3    k=4    k=5    k=6    k=7   k=8   k=9 k=10 k=11 k=12
n=1: 1 12   18     4
n=2: 1 24  120   200    120     24      1
n=3: 1 36  306   996   1446    984    303     42     2
n=4: 1 48  576  2800   6525   7848   4957   1644   274    22    1
n=5: 1 60  930  6020  19365  33600  32487  17694  5336   858   71   2
n=6: 1 72 1368 11064  45435 103200 134806 102912 45567 11358 1510  86  1
   ...
Row n=4 corresponds to the polynomial 1 + 48*(1+x) + 576*(1+x)^2 + 2800*(1+x)^3 + 6525*(1+x)^4 + 7848*(1+x)^5 + 4957*(1+x)^6 + 1644*(1+x)^7 + 274*(1+x)^8 + 22*(1+x)^9 + (1+x)^10.
		

Crossrefs

Column k=0 is A000012.
Column k=1 is A008594.
Row n=3 is identical to row n=4 of A338217 owing to symmetry of hexagonal graphene flakes.
Row sums give A107915.
Row sums give column k=0 of A338244.

Programs

  • Maple
    (n,k) -> binomial(n,k)*binomial(12,k)+18*binomial(n+1,k)*binomial(10,k-2)+84*binomial(n+2,k)*binomial(8,k-4)+126*binomial(n+3,k)*binomial(6,k-6)+57*binomial(n+4,k)*binomial(4,k-8)+4*binomial(n+5,k)*binomial(2,k-10) +add(4*binomial(n+1+h,k)*binomial(9,k-3)+24*binomial(n+2+h,k)*binomial(7,k-5)+36*binomial(n+3+h,k)*binomial(5,k-7)+14*binomial(n+4+h,k)*binomial(3,k-9),h = 0 .. 1) +add(add(binomial(2,s)*binomial(2,h)*binomial(n+2+s+h,k)*binomial(6-2*s,k-6-2*s),s = 0 .. 2),h = 0 .. 2)

Formula

T(n,k) = binomial(n,k)*binomial(12,k) + 18*binomial(n+1,k)*binomial(10,k-2) + 84*binomial(n+2,k)*binomial(8,k-4) + 126*binomial(n+3,k)*binomial(6,k-6) + 57*binomial(n+4,k)*binomial(4,k-8) + 4*binomial(n+5,k)*binomial(2,k-10) + Sum_{h=0..1} (4*binomial(n+1+h,k)*binomial(9,k-3) + 24*binomial(n+2+h,k)*binomial(7,k-5) + 36*binomial(n+3+h,k)*binomial(5,k-7) + 14*binomial(n+4+h,k)*binomial(3,k-9)) + Sum_{s=0..2} Sum_{h=0..2} binomial(2,s)*binomial(2,h)*binomial(n+2+s+h,k)*binomial(6-2*s,k-6-2*s) (conjectured, explicitly confirmed for n=1..1000).
Showing 1-2 of 2 results.