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

Author

Jon Perry, Jul 31 2003

Keywords

• nonn
• easy
• look

Comments

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

Examples

			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
		

Links

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

Crossrefs

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.

Programs

Haskell

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

Maple

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

Mathematica

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 *)

PARI

{ 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 }

Formula

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)

Extensions

More terms from Emeric Deutsch, Nov 18 2004