A349842 -id:A349842 - cp's OEIS frontend

A349841 Triangle T(n,k) built by placing all ones on the left edge, [1,0,0,0,0] repeated on the right edge, and filling the body using the Pascal recurrence T(n,k) = T(n-1,k) + T(n-1,k-1).

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 1, 1, 5, 10, 10, 5, 2, 0, 1, 6, 15, 20, 15, 7, 2, 0, 1, 7, 21, 35, 35, 22, 9, 2, 0, 1, 8, 28, 56, 70, 57, 31, 11, 2, 0, 1, 9, 36, 84, 126, 127, 88, 42, 13, 2, 1

Offset: 0

Michael A. Allen, Dec 13 2021

This is the m=5 member in the sequence of triangles A007318, A059259, A118923, A349839, A349841 which have all ones on the left side, ones separated by m-1 zeros on the other side, and whose interiors obey Pascal's recurrence.
T(n,k) is the (n,n-k)-th entry of the (1/(1-x^5),x/(1-x)) Riordan array.
For n>0, T(n,n-1) = A002266(n+4).
For n>1, T(n,n-2) = A008732(n-2).
For n>2, T(n,n-3) = A122047(n-1).
Sums of rows give A349842.
Sums of antidiagonals give A349843.

			Triangle begins:
  1;
  1,   0;
  1,   1,   0;
  1,   2,   1,   0;
  1,   3,   3,   1,   0;
  1,   4,   6,   4,   1,   1;
  1,   5,  10,  10,   5,   2,   0;
  1,   6,  15,  20,  15,   7,   2,   0;
  1,   7,  21,  35,  35,  22,   9,   2,   0;
  1,   8,  28,  56,  70,  57,  31,  11,   2,   0;
  1,   9,  36,  84, 126, 127,  88,  42,  13,   2,   1;

Michael A. Allen and Kenneth Edwards, On Two Families of Generalizations of Pascal's Triangle, J. Int. Seq. 25 (2022) Article 22.7.1.

Other members of sequence of triangles: A007318, A059259, A118923, A349839.
Columns: A000012, A001477, A000217, A000292, A000332, A323228.
Diagonals: A079998, A002266, A008732, A122047.

Flatten[Table[CoefficientList[Series[(1 - x*y)/((1 - (x*y)^5)(1 - x - x*y)), {x, 0, 20}, {y, 0, 10}], {x, y}][[n+1,k+1]],{n,0,10},{k,0,n}]]

G.f.: (1-x*y)/((1-(x*y)^5)(1-x-x*y)) in the sense that T(n,k) is the coefficient of x^n*y^k in the series expansion of the g.f.
T(n,0) = 1.
T(n,n) = delta(n mod 5,0).
T(n,1) = n-1 for n>0.
T(n,2) = (n-1)*(n-2)/2 for n>1.
T(n,3) = (n-1)*(n-2)*(n-3)/6 for n>2.
T(n,4) = (n-1)*(n-2)*(n-3)*(n-4)/24 for n>3.
T(n,5) = C(n-1,5) + 1 for n>4.
T(n,6) = C(n-1,6) + n - 6 for n>5.
For 0 <= k < n, T(n,k) = (n-k)*Sum_{j=0..floor(k/5)} binomial(n-5*j,n-k)/(n-5*j).
The g.f. of the n-th subdiagonal is 1/((1-x^5)(1-x)^n).

A119610 Number of cases in which the first player is killed in a Russian roulette game where 5 players use a gun with n chambers and the number of bullets can be from 1 to n. Players do not rotate the cylinder after the game starts.

Original entry on oeis.org

1, 2, 4, 8, 16, 33, 66, 132, 264, 528, 1057, 2114, 4228, 8456, 16912, 33825, 67650, 135300, 270600, 541200, 1082401, 2164802, 4329604, 8659208, 17318416, 34636833, 69273666, 138547332, 277094664, 554189328, 1108378657, 2216757314

Offset: 1

Ryohei Miyadera, Jun 04 2006

Denote by U(p,n,m) the number of the cases in which the first player is killed in a Russian roulette game where p players use a gun with n chambers and m bullets. They never rotate the cylinder after the game starts. The chambers can be represented by the list {1,2,...,n}.
Here we let p = 5 to produce the above sequence, but p can be an arbitrary positive integer. By letting p = 2, 3, 4, 6, 7 we can produce sequences A000975, A033138, A083593, A195904 and A117302, respectively.
The number of cases for each of the situations identified below by (0), (1), ..., (t), where t = floor((n-m)/p), can be calculated separately:
(0) The first player is killed when one bullet is in the first chamber and the remaining m-1 bullets are in chambers {2,3,...,n}. There are binomial(n-1,m-1) cases for this situation.
(1) The first player is killed when one bullet is in the (p+1)-th chamber and the rest of the bullets are in chambers {p+2,...,n}. There are binomial(n-p-1,m-1) cases for this situation.
...
(t) The first player is killed when one bullet is in the (p*t+1)-th chamber and the remaining bullets are in chambers {p*t+2,...,n}. There are binomial(n-p*t-1,m-1) cases for this situation.
Therefore U(p,n,m) = Sum_{z=0..t} binomial(n-p*z-1,m-1), where t = floor((n-m)/p). Let A(p,n) be the number of the cases in which the first player is killed when p players use a gun with n chambers and the number of bullets can be from 1 to n. Then A(p,n) = Sum_{m=1..n} U(p,n,m).

			If the number of chambers is 3, then the number of the bullets can be 1, 2, or 3. The first player is killed when one bullet is in the first chamber, and the remaining bullets are in the second and third chambers. The only cases are {{1, 0, 0}, {1, 1, 0}, {1, 0, 1}, {1, 1, 1}}, where we denote by 1 a chamber that contains a bullet. Therefore a(3) = 4.

G. C. Greubel, Table of n, a(n) for n = 1..1000
R. Miyadera, General Theory of Russian Roulette, Mathematica source.
R. Miyadera, Interesting patterns of fractions, Archimedes' Laboratory.
Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,1,-2).

Cf. A000975, A033138, A083593, A195904, A117302.

Partial sums of A349842.

I:=[1,2,4,8,16,33]; [n le 6 select I[n] else 2*Self(n-1)+Self(n-5)-2*Self(n-6): n in [1..40]]; // Vincenzo Librandi, Mar 18 2015

seq(floor(2^(n+4)/31), n = 1..32); # Mircea Merca, Dec 22 2010

U[p_,n_,m_,v_]:=Block[{t},t=Floor[(1+p-m+n-v)/p]; Sum[Binomial[n-v-p*z,m-1], {z,0,t-1}]];
A[p_,n_,v_]:=Sum[U[p,n,k,v],{k,1,n}];
(* Here we let p = 5 to produce the above sequence, but this code can produce A000975, A033138, A083593, A195904, A117302 for p = 2, 3, 4, 6, 7. *)
Table[B[5,n,1],{n,1,20}] (* end of program *)
CoefficientList[ Series[ 1/(2x^6 - x^5 - 2x + 1), {x, 0, 32}], x] (* or *)
LinearRecurrence[{2, 0, 0, 0, 1, -2}, {1, 2, 4, 8, 16, 33}, 32] (* Robert G. Wilson v, Mar 12 2015 *)

for(n=1,50, print1(floor(2^(n+4)/31), ", ")) \\ G. C. Greubel, Oct 11 2017

a(n) = floor(2^(n+4)/31), which is obtained by letting p=5 in a_p(n) = (2^(n + p-1) - 2^((n-1) mod p))/(2^p - 1).
From Joerg Arndt, Jan 08 2011: (Start)
G.f.: x / ( (x-1)*(2*x-1)*(x^4+x^3+x^2+x+1) ).
a(n) = +2*a(n-1) +a(n-5) -2*a(n-6). (End)

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A349841 Triangle T(n,k) built by placing all ones on the left edge, [1,0,0,0,0] repeated on the right edge, and filling the body using the Pascal recurrence T(n,k) = T(n-1,k) + T(n-1,k-1).

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