A349839 -id:A349839 - 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).

A349840 The number of compositions of n using elements from the set {1,3,5,7,8}.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 8, 13, 22, 35, 56, 91, 147, 238, 385, 623, 1009, 1632, 2640, 4272, 6912, 11184, 18096, 29280, 47377, 76657, 124033, 200690, 324723, 525413, 850136, 1375549, 2225686, 3601235, 5826920, 9428155

Offset: 0

Michael A. Allen, Dec 05 2021

Number of ways to tile an n-board (an n X 1 array of 1 X 1 cells) using squares, trominoes, pentominoes, heptominoes, and octominoes.
Number of permutations (p(1),p(2),...,p(n)) of (1,2,...,n) such that p(i)-i is in {-1,0,2,4,6,7} for all i=1,...,n.
a(n) gives the sums of the antidiagonals of A349839.

D. H. Lehmer, Permutations with strongly restricted displacements. Combinatorial theory and its applications, II (Proc. Colloq., Balatonfured, 1969), North-Holland, Amsterdam, 1970, pp. 755-770.

Vladimir Baltic, On the number of certain types of strongly restricted permutations, Applicable Analysis and Discrete Mathematics, 4(1) (2010), 119-135.
K. Edwards and Michael A. Allen, Strongly restricted permutations and tiling with fences, Discrete Applied Mathematics, 187 (2015), 82-90.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,0,1,0,1,1).

Sums of antidiagonals of triangles in the same family as A349839: A000045, A006498, A079962, A349843.

CoefficientList[Series[1/(1-x-x^3-x^5-x^7-x^8),{x,0,35}],x]

a(n) = a(n-1) + a(n-3) + a(n-5) + a(n-7) + a(n-8) + delta(n,0), a(n<0)=0 (where delta(i,j) is the Kronecker delta).
a(n) = a(n-1) + a(n-2) + a(n-8) - a(n-9) - a(n-10) + delta(n,0) - delta(n,2), a(n<0)=0.
G.f.: 1/(1-x-x^3-x^5-x^7-x^8).

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).

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