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.

A304553 Number of total dominating sets in the n-triangular honeycomb bishop graph.

Original entry on oeis.org

0, 3, 29, 578, 22361, 1640001, 230504570, 62711805029, 33349745638405, 34939148029874116, 72519613634620549271, 299330052017563038619309, 2462757900094871658430161896, 40447266466705082294217088869999, 1327138203952448478777157651757710177, 87038459949342074077614931718690581743298
Offset: 1

Views

Author

Eric W. Weisstein, May 14 2018

Keywords

Crossrefs

Programs

Extensions

a(8)-a(10) from Andrew Howroyd, May 19 2018
a(11) onwards from Andrew Howroyd, May 16 2025

A304564 Number of minimum total dominating sets in the n-triangular honeycomb bishop graph.

Original entry on oeis.org

0, 2, 2, 6, 75, 21, 208, 3950, 540, 11220, 314880, 25740, 917280, 36029700, 1965600, 107100000, 5627890800, 219769200, 16995484800, 1153034190000, 33844456800, 3525796058400, 300234909744000, 6868433880000, 927359072640000, 96883959332160000, 1776393899280000, 301733192320560000
Offset: 1

Views

Author

Eric W. Weisstein, May 14 2018

Keywords

Crossrefs

Programs

  • PARI
    T(n, k)=binomial(2*n-k, k)*binomial(n+k, n-k)*(2*(n-k))!*(2*k)!/(2^n)
    b1(n) = sum(k=0, n, T(n,k))
    b2(n) = sum(k=0, n, T(n,k)*(2*binomial(n+k+3,3)*(2*n-k+1) + 4*binomial(n+k+2,2)*binomial(2*n-k+2,2)))
    b3(n) = sum(k=0, n, T(n,k)*(n+k)*(n+k+1)*(7*n-2*k+5)/3)
    b4(n) = sum(k=0, n, T(n,k)*(2*binomial(n+k+4,4)*(2*n-k+1) + 24*binomial(n+k+2,2)*binomial(2*n-k+3,3)))
    b5(n) = sum(k=0, n, T(n,k)*(40*binomial(n+k+6,6)*binomial(2*n-k+2,2) + 240*binomial(n+k+5,5)*binomial(2*n-k+3,3) + 304*binomial(n+k+4,4)*binomial(2*n-k+4,4)))
    a(n) = my(t=n\3); if(n%3==0, b1(t), if(n%3==1, b2(t-1), b1(t+1) + b3(t) + b4(t-1) + b5(t-2))) \\ Andrew Howroyd, Apr 09 2025

Formula

From Andrew Howroyd, Apr 04 2025: (Start)
a(3*n) = A382777(n).
a(3*n+4) = Sum_{k=0..n} A382776(n,k)*(4*binomial(n+k+2,2) * binomial(2*n-k+2,2) + 2*binomial(n+k+3,3) * (2*n-k+1)).
See the PARI program for a(3*n+2). (End)

Extensions

a(8)-a(10) from Andrew Howroyd, May 19 2018
a(11) onwards from Andrew Howroyd, May 16 2025
Showing 1-2 of 2 results.