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 1-3 of 3 results.

A279662 a(n) = (2/3)^n*Gamma(n+3/4)*Gamma(n+1)*Gamma(n+2)/Gamma(3/4).

Original entry on oeis.org

1, 1, 7, 154, 7700, 731500, 117771500, 29678418000, 11040371496000, 5796195035400000, 4144279450311000000, 3920488359994206000000, 4790836775912919732000000, 7411424492337286825404000000, 14266992147749277138902700000000, 33670101468688294047810372000000000
Offset: 0

Views

Author

Ilya Gutkovskiy, Dec 16 2016

Keywords

Comments

Hexagonal pyramidal factorial numbers.
More generally, the m-gonal pyramidal factorial numbers is 6^(-n)*(m-2)^n*Gamma(n+1)*Gamma(n+2)*Gamma(n+3/(m-2))/Gamma(3/(m-2)), m>2.

Links

Crossrefs

Cf. A002412.
Cf. A000680 (hexagonal factorial numbers).
Cf. A087047 (tetrahedral factorial numbers), A135438 (square pyramidal factorial numbers), A167484 (pentagonal pyramidal factorial numbers), A279663 (heptagonal pyramidal factorial numbers).

Programs

  • Magma
    [Round((2/3)^n*Gamma(n+3/4)*Gamma(n+1)*Gamma(n+2) / Gamma(3/4)): n in [0..20]]; // Vincenzo Librandi, Dec 17 2016
  • Mathematica
    FullSimplify[Table[(2/3)^n Gamma[n + 3/4] Gamma[n + 1] Gamma[n + 2]/Gamma[3/4], {n, 0, 15}]]

Formula

a(n) = Product_{k=1..n} k*(k + 1)*(4*k - 1)/6, a(0)=1.
a(n) = Product_{k=1..n} A002412(k), a(0)=1.
a(n) ~ (2*Pi)^(3/2)*(2/3)^n*n^(3*n+9/4)/(Gamma(3/4)*exp(3*n)).

A279663 a(n) = (5/6)^n*Gamma(n+3/5)*Gamma(n+1)*Gamma(n+2)/Gamma(3/5).

Original entry on oeis.org

1, 1, 8, 208, 12480, 1435200, 281299200, 86640153600, 39507910041600, 25482601976832000, 22424689739612160000, 26147188236387778560000, 39429959860472770068480000, 75350653293363463600865280000, 179334554838205043370059366400000, 523656900127558726640573349888000000
Offset: 0

Views

Author

Ilya Gutkovskiy, Dec 16 2016

Keywords

Comments

Heptagonal pyramidal factorial numbers.

Links

Crossrefs

Cf. A002413.
Cf. A084940 (heptagonal factorial numbers).
Cf. A087047 (tetrahedral factorial numbers), A135438 (square pyramidal factorial numbers), A167484 (pentagonal pyramidal factorial numbers), A279662 (hexagonal pyramidal factorial numbers).

Programs

  • Magma
    [Round((5/6)^n*Gamma(n+3/5)*Gamma(n+1)*Gamma(n+2)/Gamma(3/5)): n in [0..20]]; // Vincenzo Librandi Dec 17 2016
  • Mathematica
    FullSimplify[Table[(5/6)^n Gamma[n + 3/5] Gamma[n + 1] Gamma[n + 2]/Gamma[3/5], {n, 0, 15}]]

Formula

a(n) = Product_{k=1..n} k*(k + 1)*(5*k - 2)/6, a(0)=1.
a(n) = Product_{k=1..n} A002413(k), a(0)=1.
a(n) ~ (2*Pi)^(3/2)*(5/6)^n*n^(3*n+21/10)/(Gamma(3/5)*exp(3*n)).

A356480 a(n) is the minimal number of river crossings necessary to solve the missionaries and cannibals problem for n missionaries and n cannibals where the boat capacity is the minimum necessary to allow a solution.

Original entry on oeis.org

1, 5, 11, 9, 11, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 133, 135
Offset: 1

Views

Author

Sela Fried, Aug 09 2022

Keywords

Comments

The problem is: n missionaries and n cannibals must cross the river using a boat. The missionaries must not be outnumbered by the cannibals at either river bank, or on the boat, and the boat cannot cross the river by itself.
It turns out that the necessary boat capacity is two people for n=1,2,3; three people for n=4,5; and four people for n > 5.
This problem is a generalization of the classical missionaries and cannibals problem, in which n = 3 and the boat capacity is two people. In this case the minimal number of crossings is a(3) = 11 (see example).

Examples

			Suppose n = 3 and that all the people must cross from the left river side to the right. Let m and c denote the number of missionaries and the number of the cannibals on the left bank of the river at any time. Let b=L if the boat is on the left bank, b=R if the boat is on the right bank. Then (m, c, b) fully captures the condition of the system. A solution of minimal length is then given by (3, 3, L)-->(2, 2, R)-->(3, 2, L)-->(3, 0, R)-->(3, 1, L)-->(1, 1, R)-->(2, 2, L)-->(0, 2, R)-->(0, 3, L)-->(0, 1, R)-->(1, 1, L)-->(0, 0, R).

References

  • P. Norvig and S. J. Russell, Artificial Intelligence: A Modern Approach, Third Edition, 2010. Exercise 3.9.

Links

Crossrefs

Cf. A060747, A167484.

Formula

G.f.: x*(4*x^6 - 4*x^5 + 4*x^4 - 8*x^3 + 2*x^2 + 3*x + 1)/(x-1)^2.
Showing 1-3 of 3 results.

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