A105693 -id:A105693 - cp's OEIS frontend

A318945 Triangle read by rows: T(n,k) (n>=2, 0 <= k <= n-2) = number of Dyck paths with k valleys of altitude k.

1, 4, 1, 13, 5, 1, 39, 19, 6, 1, 112, 64, 26, 7, 1, 313, 201, 97, 34, 8, 1, 859, 603, 331, 139, 43, 9, 1, 2328, 1752, 1064, 512, 191, 53, 10, 1

Offset: 2

N. J. A. Sloane, Sep 18 2018

			Triangle begins:
1,
4,1,
13,5,1,
39,19,6,1,
112,64,26,7,1,
313,201,97,34,8,1,
859,603,331,139,43,9,1,
2328,1752,1064,512,191,53,10,1,
...

Czabarka, É., Flórez, R., Junes, L., & Ramírez, J. L., Enumerations of peaks and valleys on non-decreasing Dyck paths, Discrete Mathematics (2018), 341(10), 2789-2807.

Columns 0, 1, 2. 3 are A105693, A318946, A318947, A319405.

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

A125172 Triangle T(n,k) with partial column sums of the even-indexed Fibonacci numbers.

Original entry on oeis.org

1, 3, 1, 8, 4, 1, 21, 12, 5, 1, 55, 33, 17, 6, 1, 144, 88, 50, 23, 7, 1, 377, 232, 138, 73, 30, 8, 1, 987, 609, 370, 211, 103, 38, 9, 1, 2584, 1596, 979, 581, 314, 141, 47, 10, 1

Offset: 1

Gary W. Adamson, Nov 22 2006

"Partial column sums" means the 1st column consists of the even-indexed Fibonacci numbers, the 2nd column shows the partial sums of the first column, the 3rd column the partial sums of the 2nd, etc. - R. J. Mathar, Sep 06 2011

Mirror of the fission triangle A193667, as in the Mathematica program below. - Clark Kimberling, Aug 11 2011

			First few rows of the triangle:
    1;
    3,  1;
    8,  4,  1;
   21, 12,  5,  1;
   55, 33, 17,  6,  1;
  144, 88, 50, 23,  7,  1;
  ...

Cf. A105693 (row sums), A125171, A193667.

z = 11;
p[n_, x_] := (x + 1)^n;
q[n_, x_] := Sum[Fibonacci[k + 1]*x^(n - k), {k, 0, n}];
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A193667 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]]  (* this sequence *)
(* Clark Kimberling, Aug 11 2011 *)

T(n,1) = A001906(n).
T(n,k) = T(n-1,k-1) + T(n-1,k), k > 1.
From R. J. Mathar, Sep 06 2011: (Start)
T(n,k) = A125171(n,k), i.e., A125171 without column k=0.
Conjecture: T(n,k) = T(n,k-1) - A121460(n+1,k). (End)

A079285 First differences of A079284.

Original entry on oeis.org

0, 2, 1, 5, 4, 13, 13, 34, 39, 89, 112, 233, 313, 610, 859, 1597, 2328, 4181, 6253, 10946, 16687, 28657, 44320, 75025, 117297, 196418, 309619, 514229, 815656, 1346269, 2145541, 3524578, 5637351, 9227465, 14799280, 24157817

Offset: 0

Paul Barry, Feb 08 2003

			a(2n) = Fib(2n+2) - 2^(n/2). a(2n+1) = Fib(2n+3)

Index entries for linear recurrences with constant coefficients, signature (1,3,-2,-2)

Cf. A000045, A008949, A001519 (bisection), A105693 (bisection).

a(n)=Fib(n+2)-2^floor(n/2)(1-(1)^(n+1))/2

G.f.: -x*(-2+x+2*x^2) / ( (x^2+x-1)*(2*x^2-1) ). - R. J. Mathar, Feb 13 2015

A320904 T(n, k) = binomial(2n + 1 - k, k)hypergeom([1, 1, -k], [1, 2*(n - k + 1)], -1), triangle read by rows, T(n, k) for n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 5, 7, 1, 7, 16, 15, 1, 9, 29, 42, 31, 1, 11, 46, 93, 99, 63, 1, 13, 67, 176, 256, 219, 127, 1, 15, 92, 299, 562, 638, 466, 255, 1, 17, 121, 470, 1093, 1586, 1486, 968, 511, 1, 19, 154, 697, 1941, 3473, 4096, 3302, 1981, 1023

Offset: 0

Peter Luschny, Oct 28 2018

			Triangle starts:
[0] 1
[1] 1,  3
[2] 1,  5,   7
[3] 1,  7,  16,  15
[4] 1,  9,  29,  42,   31
[5] 1, 11,  46,  93,   99,   63
[6] 1, 13,  67, 176,  256,  219,  127
[7] 1, 15,  92, 299,  562,  638,  466, 255
[8] 1, 17, 121, 470, 1093, 1586, 1486, 968, 511

Row sums are A105693(n-1).

Cf. A097750.

T := (n, k) -> binomial(2*n + 1 - k, k)*hypergeom([1, 1, -k], [1, 2*(n-k+1)], -1):
for n from 0 to 11 do seq(simplify(T(n, k)), k = 0..n) od;

s={};For[n=0,n<19,n++,For[k=0,kDetlef Meya, Oct 03 2023 *)

cp's OEIS Frontend

A318945 Triangle read by rows: T(n,k) (n>=2, 0 <= k <= n-2) = number of Dyck paths with k valleys of altitude k.

Views

Author

Keywords

Examples

Links

Crossrefs

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A125172 Triangle T(n,k) with partial column sums of the even-indexed Fibonacci numbers.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A079285 First differences of A079284.

Views

Author

Keywords

Examples

Links

Crossrefs

Formula

A320904 T(n, k) = binomial(2*n + 1 - k, k)*hypergeom([1, 1, -k], [1, 2*(n - k + 1)], -1), triangle read by rows, T(n, k) for n >= 0 and 0 <= k <= n.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Mathematica

A320904 T(n, k) = binomial(2n + 1 - k, k)hypergeom([1, 1, -k], [1, 2*(n - k + 1)], -1), triangle read by rows, T(n, k) for n >= 0 and 0 <= k <= n.