A101052 -id:A101052 - cp's OEIS frontend

A186432 Triangle associated with the set S of squares {0,1,4,9,16,...}.

1, 1, 1, 1, 12, 1, 1, 30, 30, 1, 1, 56, 140, 56, 1, 1, 90, 420, 420, 90, 1, 1, 132, 990, 1848, 990, 132, 1, 1, 182, 2002, 6006, 6006, 2002, 182, 1, 1, 240, 3640, 16016, 25740, 16016, 3640, 240, 1, 1, 306, 6120, 37128, 87516, 87516, 37128, 6120, 306, 1, 1, 380, 9690, 77520, 251940, 369512, 251940, 77520, 9690, 380, 1

Offset: 0

Peter Bala, Feb 22 2011

Given a subset S of the integers Z, Bhargava [1] has shown how to associate with S a generalized factorial function, denoted n!_S, sharing many properties of the classical factorial function n! (which corresponds to the choice S = Z). In particular, he shows that the generalized binomial coefficients n!_S/(k!_S*(n-k)!_S) are always integral for any choice of S. Here we take S = {0,1,4,9,16,...}, the set of squares.
The associated generalized factorial function n!_S is given by the formula
n!S = Product{k=0..n} (n^2 - k^2), with the convention 0!S = 1. This should be compared with n! = Product{k=0..n} (n - k).
For n >= 1, n!_S = (2*n)!/2 = A002674(n).
Compare this triangle with A086645 and also A186430 - the generalized binomial coefficients for the set S of prime numbers {2,3,5,7,11,...}.

			Triangle begins
n/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1....12.....1
.3..|..1....30....30.....1
.4..|..1....56...140....56.....1
.5..|..1....90...420...420....90.....1
.6..|..1...132...990..1848...990...132.....1
.7..|..1...182..2002..6006..6006..2002...182.....1
...

M. Bhargava, The factorial function and generalizations, Amer. Math. Monthly, 107 (2000), 783-799.

Cf. A002114, A086645, A186430, A186433 (inverse).

Table[2 Binomial[2 n, 2 k] - Boole[Or[k == 0, k == n]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, May 23 2017 *)

TABLE ENTRIES
T(n,k) = n!_S/(k!_S*(n-k)!_S),
which simplifies to
T(n,k) = 2*binomial(2*n,2*k) for 1 <= k < n,
with boundary conditions T(n,0) = 1 and T(n,n) = 1 for n >= 0.
RELATIONS WITH OTHER SEQUENCES
Denote this triangle by T. The first column of the inverse T^-1 (see A186433) begins [1, -1, 11, -301, 15371, ...] and, apart from the initial 1, is a signed version of the Glaisher's H' numbers A002114.
The first column of (1/2)*T^2 begins [1/2, 1, 7, 31, 127, ...] and, apart from the initial term, equals A000225(2*n-1), counting the preferential arrangements on (2*n - 1) labeled elements having less than or equal to two ranks.
The first column of (1/3)*T^3 begins [1/3, 1, 13, 181, 1933, ...] and, apart from the initial term, is A101052(2*n-1), which gives the number of preferential arrangements on (2*n-1) labeled elements having less than or equal to three ranks.

A245023 Number of cases of tie (no winner) in the n-person rock-paper-scissors game.

Original entry on oeis.org

3, 3, 9, 39, 153, 543, 1809, 5799, 18153, 55983, 171009, 519159, 1569753, 4733823, 14250609, 42850119, 128746953, 386634063, 1160688609, 3483638679, 10454061753, 31368476703, 94118013009, 282379204839, 847187946153, 2541664501743, 7625194831809, 22875987148599, 68628766752153, 205887910869183, 617666953833009

Offset: 1

Jaeyool Park, Jul 10 2014

			R, P, S = each Rock, Paper, Scissors. For n = 2 RR, PP, SS. 3 cases. and for n = 3 RRR, PPP, SSS, RPS, PRS, RSP, PSR, SPR, SRP. 9 cases.
RRS (and RSR, SRR) is not a tie case because there are two winners. SPP (and PPS, PSP) is not a tie case because there is a winner even though the 2nd and 3rd places cannot be determined.  - _Wolfdieter Lang_, Jul 31 2014

Index entries for linear recurrences with constant coefficients, signature (6, -11, 6).

Cf. A101052.

[3^n - 3*(2^n-2) : n in [1..30]]; // Wesley Ivan Hurt, Jul 26 2014

A245023:=n->3^n - 3*(2^n-2): seq(A245023(n), n=1..30); # Wesley Ivan Hurt, Jul 26 2014

Table[3^n - 3 (2^n - 2), {n, 30}] (* Wesley Ivan Hurt, Jul 26 2014 *)

a(n) = 3^n - 3*(2^n-2); \\ Michel Marcus, Jul 16 2014

a(n) = 3^n - 3*(2^n-2) = 3 * A101052(n-1), n >= 1.
a(n) = 5*a(n-1)-5*a(n-2)-5*a(n-3)+6*a(n-4). - Colin Barker, Jul 26 2014
G.f.: -3*x*(8*x^2-5*x+1) / ((x-1)*(2*x-1)*(3*x-1)). - Colin Barker, Jul 26 2014
a(n) = 3 + 3!*S2(n, 3) with S2(n, k) = A008277(n, k) (Stirling numbers of the second kind). S2(n,3)= A000392(n). Proof: Use the formula given in A000392. Hint for an independent proof:  consider the partition array A036040 for the multinomial M_3 numbers. Only partitions of n with number of parts m = 1, 2 and 3 matter here. Each partition defines a pattern for a multilist, like 2^1,3^2 defines the n=8, m=3 pattern [..][...][...]. The corresponding M_3 number 280 = C(8,2)*C(6,3)/2 gives the number of possibilities to form from objects, here R,R,P,P,P,S,S,S, lists of length 8 (the order is relevant).  If m=1 then M_3 = 1 and for n one has 3 lists [n times R], [n times P] and [n times S] (no winner), If m=2 or 3 each of the M_3(n,m,j) (j=1..p(n,m), the number of partitions of n with m parts) comes 3! times from the permutation of the R, P and S symbols. The sum of the M_3 numbers over like m gives the Stirling2 numbers. If m=2 there are always winners for each n (only two symbols are present in each list). If m=1 or m=3 there is no winner.  - Wolfdieter Lang, Aug 01 2014

Typo in data fixed by Colin Barker, Jul 26 2014

A259569 Triangle T(n,k) read by rows, where T(n,k) is the number of k-dimensional faces of the polytope that is the convex hull of all permutations of the list (0,1,...,1,2), where there are n - 1 ones, for n > 0. T(0,0) is 1.

Original entry on oeis.org

1, 2, 1, 6, 6, 1, 12, 24, 14, 1, 20, 60, 70, 30, 1, 30, 120, 210, 180, 62, 1, 42, 210, 490, 630, 434, 126, 1, 56, 336, 980, 1680, 1736, 1008, 254, 1, 72, 504, 1764, 3780, 5208, 4536, 2286, 510, 1, 90, 720, 2940, 7560, 13020, 15120, 11430, 5100, 1022, 1, 110, 990, 4620, 13860, 28644, 41580, 41910, 28050, 11242, 2046, 1

Offset: 0

Vincent J. Matsko, Jun 30 2015

It appears that these integers, with sign changes, are also in A138106.

			Triangle begins:
   1;
   2,  1;
   6,  6,  1;
  12, 24, 14,  1;
  20, 60, 70, 30,  1;
  ...
Row 2 describes a regular hexagon.
Row 3 describes the cuboctahedron.

Row sums give A101052(n+1).

Cf. A138106.

T:= (n, k)-> `if`(n=k, 1, binomial(n+1, k+2)*(4*2^k-2)):
seq(seq(T(n,k), k=0..n), n=0..10);

Join @@ (CoefficientList[#,
     x] & /@ (Expand[
       D[((1 + 2 x) Exp[z (1 + 2 x)] - 2 (1 + x) Exp[z (1 + x)] + Exp[z] +
            x^2 Exp[z x])/x^2, {z, #}] /. z -> 0] & /@ Range[0, 10]))

T(n,n) = 1, n >= 0.
T(n,n-1) = 2^(n+1)-2, n > 0.
T(n,0) = n(n+1), n > 0.
T(n,k) = (n+1)*T(n-1,k)/(n-k-1), 0 <= k < n-1, n >= 2.
E.g.f.: ((2*x+1)*exp(z*(2*x+1)) - 2*(x+1)*exp(z*(x+1)) + x^2*exp(z*x)+exp(z))/x^2
Conjecture: Sum_{k=0..n-1} T(n,k)*x^(n-k-1) = x^(n+1) - 2(x+1)^(n+1) + (x+2)^(n+1). - Kevin J. Gomez, Jul 25 2017
T(n,n) = 1; T(n,k) = binomial(n+1,k+2)*(4*2^k - 2) for 0 <= k < n. - Aadesh Tikhe, May 25 2024

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A186432 Triangle associated with the set S of squares {0,1,4,9,16,...}.

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A245023 Number of cases of tie (no winner) in the n-person rock-paper-scissors game.

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A259569 Triangle T(n,k) read by rows, where T(n,k) is the number of k-dimensional faces of the polytope that is the convex hull of all permutations of the list (0,1,...,1,2), where there are n - 1 ones, for n > 0. T(0,0) is 1.

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