A004320 -id:A004320 - cp's OEIS frontend

A086754 Pascal's square pyramid read by slices, each slice being read by rows. Each entry in slice n is the sum of the 4 entries above it in slice n-1.

1, 1, 1, 1, 1, 1, 2, 1, 2, 4, 2, 1, 2, 1, 1, 3, 3, 1, 3, 9, 9, 3, 3, 9, 9, 3, 1, 3, 3, 1, 1, 4, 6, 4, 1, 4, 16, 24, 16, 4, 6, 24, 36, 24, 6, 4, 16, 24, 16, 4, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 5, 25, 50, 50, 25, 5, 10, 50, 100, 100, 50, 10, 10, 50, 100, 100, 50, 10, 5, 25, 50, 50, 25, 5, 1, 5, 10

Offset: 1

Jon Perry, Jul 31 2003

Element (i,j) of slice n is the coefficient of x^i * y^j in the expansion of ((1+x)*(1+y))^n. - Eitan Y. Levine, Sep 03 2023

			The first 4 slices are
  1..1 1..1 2 1..1 3 3 1
  ...1 1..2 4 2..3 9 9 3
  ........1 2 1..3 9 9 3
  ...............1 3 3 1

Rémy Sigrist, Table of n, a(n) for n = 1..10416
The Lost Math Lessons, Pascal's Pyramids, Friday, March 6, 2015.

Consider the sequence s[i, j](n) obtained by considering the (i, j)-th entry of the n-th slice. Then if [i, j]= [3, 2] we get A006002, if [3, 3] we get A000537, if [4, 2] we get A004320, if [4, 3] we get A004282.

Cf. A046816.

a086754 n = a086754_list !! (n-1)
a086754_list = concat $ concat $ iterate ([[1,1],[1,1]] *) [1]
instance Num a => Num [a] where
   fromInteger k = [fromInteger k]
   (p:ps) + (q:qs) = p + q : ps + qs
   ps + qs         = ps ++ qs
   (p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
    *                = []
-- Reinhard Zumkeller, Apr 02 2011

p:=n->seq(seq(binomial(n,i)*binomial(n,j),j=0..n),i=0..n): seq(p(n),n=0..5); # Emeric Deutsch, Nov 18 2004

A[m_]:=Module[{pt=Table[ConstantArray[1,{i,i}],{i,m}]},For[i=3,i<=m,i++,For[j=2,j<=i-1,j++,pt[[i,j,1]]=pt[[i-1,j-1,1]]+pt[[i-1,j,1]];pt[[i,1,j]]=pt[[i,j,1]];pt[[i,i,j]]=pt[[i,j,1]];pt[[i,j,i]]=pt[[i,j,1]];];For[j=2,j<=i-1,j++,For[k=2,k<=i-1,k++,pt[[i,j,k]]=pt[[i-1,j,k]]+pt[[i-1,j,k-1]]+pt[[i-1,j-1,k]]+pt[[i-1,j-1,k-1]];];];];pt//Flatten]; A[6] (* Robert P. P. McKone, Sep 14 2023, made from the PARI code *)

{ pt=vector(10,i,matrix(i,i,j,k,1)); for (i=3,10, for (j=2,i-1, pt[i][j,1]=pt[i-1][j-1,1]+pt[i-1][j,1]; pt[i][1,j]=pt[i][j,1]; pt[i][i,j]=pt[i][j,1]; pt[i][j,i]=pt[i][j,1]; ); for(j=2,i-1, for (k=2,i-1, pt[i][j,k]=pt[i-1][j,k]+pt[i-1][j,k-1]+pt[i-1][j-1,k]+pt[i-1][j-1,k-1]))); pt }

From Eitan Y. Levine, Sep 03 2023: (Start)
C(n,i)*C(n,j) gives the (i,j) element in slice n, where C(n,k) are the binomial coefficients A007318.
G.f.: 1/(1-z(1+x)(1+y)) = Sum_{n>=0,i=0..n,j=0..n} T(n,i,j) * z^n * x^i * y^j
G.f. for slice n: ((1+x)*(1+y))^n = Sum_{i=0..n,j=0..n} T(n,i,j) * x^i * y^j (End)

More terms from Emeric Deutsch, Nov 18 2004

A361099 a(n) = n + 2binomial(n,2) + 3binomial(n,3) + 4*binomial(n,4).

Original entry on oeis.org

0, 1, 4, 12, 32, 75, 156, 294, 512, 837, 1300, 1936, 2784, 3887, 5292, 7050, 9216, 11849, 15012, 18772, 23200, 28371, 34364, 41262, 49152, 58125, 68276, 79704, 92512, 106807, 122700, 140306, 159744, 181137, 204612, 230300, 258336, 288859, 322012, 357942, 396800, 438741

Offset: 0

Enrique Navarrete, Mar 01 2023

a(n) is the number of ordered set partitions of an n-set into 2 sets such that the first set has either 3, 2, 1 or no elements, the second set has no restrictions, and an element is selected from the second set.

			The 294 set partitions for n=7 are the following (where the element selected from the second set is in parentheses):
{ }, {(1),2,3,4,5,6,7}  (7 of these);
{1}, {(2),3,4,5,6,7}   (42 of these);
{1,2}, {(3),4,5,6,7}   (105 of these);
{1,2,3}, {(4),5,6,7}   (140 of these).

Paolo Xausa, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000290, A004320.

Table[n^2*(n*(n - 3) + 8)/6, {n, 0, 50}] (* Paolo Xausa, Jun 10 2024 *)

def A361099(n): return n**2*(n*(n - 3) + 8)//6 # Chai Wah Wu, Mar 24 2023

E.g.f.: (1 + x + x^2/2 + x^3/6)*x*exp(x).
From Stefano Spezia, Mar 04 2023: (Start)
O.g.f.: x*(1 - x + 2*x^2 + 2*x^3)/(1 - x)^5.
a(n) = A000290(n) + A004320(n-2). (End)

A338996 Numbers of squares and rectangles of all sizes in 3n(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.

Original entry on oeis.org

0, 5, 27, 85, 205, 420, 770, 1302, 2070, 3135, 4565, 6435, 8827, 11830, 15540, 20060, 25500, 31977, 39615, 48545, 58905, 70840, 84502, 100050, 117650, 137475, 159705, 184527, 212135, 242730, 276520

Offset: 0

Luce ETIENNE, Nov 18 2020

			a(1) = 2*3-1 = 5, a(2) = 2*16-5 = 27, a(3) = 2*50-15 = 85, a(4) = 2*120-35 = 205, a(5) = 2*245-70 = 420, a(6) = 2*448-126 = 770.

Luce ETIENNE, Illustration of a(1), a(2), a(3) and a(4)
Wikipedia, Aztec diamond.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A000332, A002417, A004320, A045943, A258440, A330805.

CoefficientList[Series[x (2 x + 5)/(1 - x)^5, {x, 0, 30}], x] (* Michael De Vlieger, Dec 12 2020 *)

G.f.: x*(2*x + 5)/(1 - x)^5.
E.g.f.: exp(x)*x*(120 + 204*x + 76*x^2 + 7*x^3)/24. - Stefano Spezia, Nov 18 2020
a(n) = 5*(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = n*(n + 1)*(n + 2)*(7*n + 13)/24.
a(n) = 2*A004320(n) - A000332(n+3).
a(n) = 2*A000332(n+2) + 5*A000332(n+3).

A347056 Triangle read by rows: T(n,k) = (n+1)(n+2)(k+3)*binomial(n,k)/6, 0 <= k <= n.

Original entry on oeis.org

1, 3, 4, 6, 16, 10, 10, 40, 50, 20, 15, 80, 150, 120, 35, 21, 140, 350, 420, 245, 56, 28, 224, 700, 1120, 980, 448, 84, 36, 336, 1260, 2520, 2940, 2016, 756, 120, 45, 480, 2100, 5040, 7350, 6720, 3780, 1200, 165, 55, 660, 3300, 9240, 16170, 18480, 13860, 6600, 1815, 220

Offset: 0

Luc Rousseau, Aug 14 2021

This triangle is T[3] in the sequence (T[p]) of triangles defined by: T[p](n,k) = (k+p)*(n+p-1)!/(k!*(n-k)!*p!) and T[0](0,0)=1.

Riordan triangle (1/(1-x)^3, x/(1-x)) with column k scaled with A000292(k+1) = binomial(k+3, 3), for k >= 0. - Wolfdieter Lang, Sep 30 2021

			T(6,2) = (6+1)*(6+2)*(2+3)*binomial(6,2)/6 = 7*8*5*15/6 = 700.
The triangle T begins:
n \ k  0   1    2     3     4     5     6     7     8    9  10 ...
0:     1
1:     3   4
2:     6  16   10
3:    10  40   50    20
4:    15  80  150   120    35
5:    21 140  350   420   245    56
6:    28 224  700  1120   980   448    84
7:    36 336 1260  2520  2940  2016   756   120
8:    45 480 2100  5040  7350  6720  3780  1200   165
9:    55 660 3300  9240 16170 18480 13860  6600  1815  220
10:   66 880 4950 15840 32340 44352 41580 26400 10890 2640 286
... - _Wolfdieter Lang_, Sep 30 2021

Luc Rousseau, Illustration: the A347056 triangle in a sequence of contiguous triangles.

Cf. A097805 (p=0), A103406 (p=1), A124932 (essentially p=2).
From Wolfdieter Lang, Sep 30 2021: (Start)
Columns (with leading zeros): A000217(n+1), 4*A000294, 10*A000332(n+2), 20*A000389(n+2), 35*A000579(n+2), 56*A000580(n+2), 84*A000581(n+2), 120*A000582(n+2), ...
Diagonals: A000292(k+1), A004320(k+1), 2*A006411(k+1), 10*A040977, ... (End)

T(p,n,k)=if(n==0&&p==0,1,((k+p)*(n+p-1)!)/(k!*(n-k)!*p!))
for(n=0,9,for(k=0,n,print1(T(3,n,k),", ")))

T(n,k) = (n+1)*(n+2)*(k+3)*binomial(n,k)/6.

G.f. column k: x^k*binomial(k+3, 3)/(1 - x)^(k+3), for k >= 0. - Wolfdieter Lang, Sep 30 2021

A182309 Triangle T(n,k) with 2 <= k <= floor(2(n+1)/3) gives the number of length-n binary sequences with exactly k zeros and with length two for the longest run of zeros.

Original entry on oeis.org

1, 2, 3, 2, 4, 6, 1, 5, 12, 6, 6, 20, 18, 3, 7, 30, 40, 16, 1, 8, 42, 75, 50, 10, 9, 56, 126, 120, 45, 4, 10, 72, 196, 245, 140, 30, 1, 11, 90, 288, 448, 350, 126, 15, 12, 110, 405, 756, 756, 392, 90, 5, 13, 132, 550, 1200, 1470, 1008, 357, 50, 1, 14, 156, 726

Offset: 2

Dennis P. Walsh, Apr 23 2012

Triangle T(n,k) captures several well known sequences. In particular, T(n,2)=(n-1), the natural numbers; T(n,3)=(n-2)(n-3)=A002378(n-3), the "oblong" numbers; T(n,4)=(n-3)(n-4)^2/2=A002411(n-4), "pentagonal pyramidal" numbers; and also T(n,5)=(n-4)C(n-4,3)=A004320(n-6). Furthermore, row sums=A000100(n+1).

			For n=6 and k=3, T(6,3)=12 since there are 12 binary sequences of length 6 that contain 3 zeros and that have a maximum run of zeros of length 2, namely, 011100, 101100, 110100, 011001, 101001, 110010, 010011, 100110, 100101, 001110, 001101, and 001011.
Triangle T(n,k) begins
   1,
   2,
   3,   2,
   4,   6,   1,
   5,  12,   6,
   6,  20,  18,    3,
   7,  30,  40,   16,    1,
   8,  42,  75,   50,   10,
   9,  56, 126,  120,   45,    4,
  10,  72, 196,  245,  140,   30,    1,
  11,  90, 288,  448,  350,  126,   15,
  12, 110, 405,  756,  756,  392,   90,    5,
  13, 132, 550, 1200, 1470, 1008,  357,   50,   1,
  14, 156, 726, 1815, 2640, 2268, 1106,  266,  21,
  15, 182, 936, 2640, 4455, 4620, 2898, 1016, 161, 6,

Dennis Walsh, Notes on binary sequences with a maximum run length of two

Row sums of triangle T(n,k)=A000100(n+1);
T(n,3)=A002378(n-3); T(n,4)=A002411(n-4);
T(n,5)=A004320(n-6).

seq(seq(sum(binomial(n-k+1,j)*binomial(n-k+1-j,k-2*j),j=1..floor(k/2)),k=2..floor(2*(n+1)/3)),n=2..20);

t[n_, k_] := Sum[ Binomial[n-k+1, j]*Binomial[n-k-j+1, k-2*j], {j, 1, k/2}]; Table[t[n, k], {n, 2, 15}, {k, 2, 2*(n+1)/3}] // Flatten (* Jean-François Alcover, Jun 06 2013 *)

T(n,k) = Sum_{j=1..k/2} binomial(n-k+1,j)*binomial(n-k-j+1,k-2j) for 2 <= k <= 2(n+1)/3.

cp's OEIS Frontend

A086754 Pascal's square pyramid read by slices, each slice being read by rows. Each entry in slice n is the sum of the 4 entries above it in slice n-1.

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A361099 a(n) = n + 2*binomial(n,2) + 3*binomial(n,3) + 4*binomial(n,4).

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A338996 Numbers of squares and rectangles of all sizes in 3*n*(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.

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A347056 Triangle read by rows: T(n,k) = (n+1)*(n+2)*(k+3)*binomial(n,k)/6, 0 <= k <= n.

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A182309 Triangle T(n,k) with 2 <= k <= floor(2(n+1)/3) gives the number of length-n binary sequences with exactly k zeros and with length two for the longest run of zeros.

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Comments

Examples

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Programs

Maple

Mathematica

Formula

A361099 a(n) = n + 2binomial(n,2) + 3binomial(n,3) + 4*binomial(n,4).

A338996 Numbers of squares and rectangles of all sizes in 3n(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.

A347056 Triangle read by rows: T(n,k) = (n+1)(n+2)(k+3)*binomial(n,k)/6, 0 <= k <= n.