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.

Previous Showing 11-13 of 13 results.

A322145 Number of permutations of 9 copies of 1..n with no element equal to another within a distance of 1.

Original entry on oeis.org

1, 0, 2, 17236524, 2511603532825176, 2829059722872229922701920, 17498057808683351584656839871450000, 459422439054082909311010463927575656038701920, 42176005899746902650961357272521722186133207293858938240
Offset: 0

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Author

Seiichi Manyama, Nov 28 2018

Keywords

Crossrefs

Row 9 of A322093.

Formula

a(n) = n! * A321669(n).
a(n) = Integral_{0..infinity} (Sum_{k=1..9} (-1)^(9-k) * binomial(8, 9-k) * x^k/k!)^n * exp(-x) dx.

A322146 Number of permutations of 10 copies of 1..n with no element equal to another within a distance of 1.

Original entry on oeis.org

1, 0, 2, 124948668, 174702663548149248, 2360719028641481267959955040, 211490077066069537208795610578715159120, 94446414948214202156311984061437135600678877848560
Offset: 0

Views

Author

Seiichi Manyama, Nov 28 2018

Keywords

Crossrefs

Row 10 of A322093.

Formula

a(n) = n! * A321670(n).
a(n) = Integral_{0..infinity} (Sum_{k=1..10} (-1)^(10-k) * binomial(9, 10-k) * x^k/k!)^n * exp(-x) dx.

A377586 Numbers of directed Hamiltonian paths in the complete 4-partite graph K_{n,n,n,n}.

Original entry on oeis.org

24, 13824, 53529984, 751480602624, 27917203599360000, 2267561150913576960000, 354252505303682314076160000, 97087054992658680467800719360000, 43551509948777170973522371396239360000, 30293653795894300342540281328749772800000000
Offset: 1

Views

Author

Zlatko Damijanic, Nov 02 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Table[n!^4 * SeriesCoefficient[1/(1 - Sum[x[i]/(1 + x[i]), {i, 1, 4}]), Sequence @@ Table[{x[i], 0, n}, {i, 1, 4}]], {n, 1, 10}]
  • Python
    from math import factorial as fact, comb
    from itertools import combinations_with_replacement
    def a(n):
        #  Using modified formula for counting sequences found in Eifler et al.
        result = 0
        fn = fact(n)
        for i, j, k in combinations_with_replacement(range(1, n+1), 3):
            patterns = [(3,0,0)] if i == j == k else \
              [(2,0,1)] if i == j != k else \
              [(1,2,0)] if i != j == k else [(1,1,1)]
            for a, b, c in patterns:
                s = a*i + b*j + c*k
                num = fact(3)
                den = fact(a) * fact(b) * fact(c)
                if a:
                    for _ in range(a): num, den = num * comb(n-1, i-1), den * fact(i)
                if b:
                    for _ in range(b): num, den = num * comb(n-1, j-1), den * fact(j)
                if c:
                    for _ in range(c): num, den = num * comb(n-1, k-1), den * fact(k)
                num *= comb(s + 1, n) * fact(s)
                result += (1 if (3*n - s) % 2 == 0 else -1) * (num // den)
        for _ in range(4): result *= fn
        return result
    print([a(n) for n in range(1,11)]) # Zlatko Damijanic, Nov 18 2024

Formula

a(n) = 24 * n!^4 * A190918(n).
a(n) = n!^4 * A322093(n,4).
Previous Showing 11-13 of 13 results.