Schmidt's Problem

Author

Eric W. Weisstein
October 4, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Combinatorics/SchmidtsProblem.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/SchmidtsProblem.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Code

LHS[r_,n_]:=Sum[Binomial[n,k]^r Binomial[n+k,k]^r,{k,0,n}]
RHS[r_,n_]:=Sum[Binomial[n,k]Binomial[n+k,k]c[r,k],{k,0,n}]

c[2,n_]:=Sum[Binomial[n,j]^3,{j,0,n}]
c[2,n_]:=Sum[Binomial[n,j]^2Binomial[2j,n],{j,0,n}]
c[3,n_]:=Sum[Binomial[2j,j]^2Binomial[2j,n-j]Binomial[n,j]^2,{j,0,n}]
c[4,n_]:=Sum[Binomial[2j,j]^3Binomial[n,j]Binomial[k+j,k-j]Binomial[j,n-k]Binomial[k,j]Binomial[2j,k-j],{k,0,n},{j,0,n}]
c[5,n_]:=Sum[Binomial[2j,j]^4Binomial[n,j]^2Binomial[k+j,k-j]^2Binomial[2j,n-k]Binomial[2j,k-j],{j,0,n},{k,0,n}]
c[r_Integer?EvenQ,n_]:=Module[{s=r/2,i,k},
Sum[Binomial[2j,j]^(2s-1)Binomial[n,j]*
Binomial[j,n-k[1]]Binomial[k[1],j]Binomial[k[1]+j,k[1]-j]*
Times@@Table[Binomial[2j,k[i-1]-k[i]]Binomial[k[i]+j,k[i]-j],{i,2,s-2}]*
Binomial[2j,k[s-2]-k[s-1]]Binomial[k[s-1]+j,k[s-1]-j]^2Binomial[2j,k[s-1]-j],{j,0,n},Evaluate[Sequence@@({#,0,n}&/@Array[k,s-1])]]
]
c[r_Integer?OddQ,n_]:=Module[{s=(r-1)/2,i,k},
Sum[Binomial[2j,j]^(2s)Binomial[n,j]^2*
Binomial[2j,n-k[1]]Binomial[k[1]+j,k[1]-j]^2*
Times@@Table[Binomial[2j,k[i-1]-k[i]]Binomial[k[i]+j,k[i]-j]^2,{i,2,s-2}]*
Binomial[2j,k[s-2]-k[s-1]]Binomial[k[s-1]+j,k[s-1]-j]^2Binomial[2j,k[s-1]-j],{j,0,n},Evaluate[Sequence@@({#,0,n}&/@Array[k,s-1])]]
]

LHS

Table[LHS[r,n],{r,11},{n,0,7}]//ColumnForm

RHS

Table[RHS[r,n],{r,2,11},{n,0,7}]//ColumnForm

(Subscript[c, n]^((r)))

(t=Table[c[r,n],{r,2,10},{n,0,10}]);//Timing

ColumnForm[t]

Individual cases

<<zb_alg.m

r==1

First[Zb[Binomial[n,k] Binomial[n+k,k],k,n,2]/.{F[k_,n_]:>a[n],Delta[__]:>0}/.n->n-2]/.a[n_]:>Subscript[a,n]//FullSimplify//TraditionalForm

RSolve[(n-1)a[n-2]+(3-6n)a[n-1]+n a[n]==0,a[n],n]

r==2

First[Zb[Binomial[n,k]^2 Binomial[n+k,k]^2,k,n,2]/.{F[k_,n_]:>a[n],Delta[__]:>0}/.n->n-2]/.a[n_]:>Subscript[a,n]//Simplify//TraditionalForm

Zb[Binomial[n,k]^2 Binomial[n+k,k]^2,k,n,2][[1]]/.{F[k_,n_]:>a[n],Delta[__]:>0}/.n->n-2

RSolve[{(((-1+n))^3) a[-2+n]-(3+2 (-2+n)) (39+51 (-2+n)+17 (((-2+n))^2)) a[-1+n]+(n^3) a[n]==0,a[0]==1,a[1]==2},a[n],n]

Sum[((Binomial[n,j])^3),{j,0,n}]

Sum[((Binomial[n,j])^2)Binomial[2j,n],{j,0,n}]

Table[HypergeometricPFQ[{-n,-n,-n},{1,1},-1],{n,0,10}]

Table[c[2,n],{n,0,10}]

Franel numbers http://www.research.att.com/projects/OEIS?Anum=A000172

r==3

Zb[Binomial[n,k]^3 Binomial[n+k,k]^3,k,n,2]

Zb[Binomial[n,k]^3 Binomial[n+k,k]^3,k,n,3]

Zb[Binomial[n,k]^3 Binomial[n+k,k]^3,k,n,4]

Sum[((Binomial[2j,j])^2)Binomial[2j,n-j]((Binomial[n,j])^2),{j,0,n}]

...wrong...

FullSimplify[%,n>0&&n∈Integers]

Sum[((Binomial[2j,j])^2)Binomial[2j,n-j]((Binomial[n,j])^2),{j,n}]

FullSimplify[%,n∈Integers]

Table[Sum[((Binomial[2j,j])^2)Binomial[2j,n-j]((Binomial[n,j])^2),{j,n}],{n,5}]

Table[c[3,n],{n,0,10}]

Strehl's Sequence http://www.research.att.com/projects/OEIS?Anum=A000658

r==4

Zb[Binomial[n,k]^4Binomial[n+k,k]^4,k,n,2]

Zb[Binomial[n,k]^4 Binomial[n+k,k]^4,k,n,3]

Zb[Binomial[n,k]^4 Binomial[n+k,k]^4,k,n,4]

Sum[((Binomial[2j,j])^3)Binomial[n,j]Sum[Binomial[k+j,k-j]Binomial[j,n-k]Binomial[k,j]Binomial[2j,k-j],{k,0,n}],{j,0,n}]

Table[c[4,n],{n,0,10}]

r==5

Sum[((Binomial[2j,j])^4)((Binomial[n,j])^2)Sum[((Binomial[k+j,k-j])^2)Binomial[2j,n-k]Binomial[2j,k-j],{k,0,n}],{j,0,n}]

Table[Sum[((Binomial[2j,j])^4)((Binomial[n,j])^2)Sum[((Binomial[k+j,k-j])^2)Binomial[2j,n-k]Binomial[2j,k-j],{k,0,n}],{j,0,n}],{n,0,10}]

(∑_(j=0)^n)(∑_(k=0)^n)((Binomial[2 j, j])^4) ((Binomial[j+k, k-j])^2) (Binomial[2 j, k-j]) ((Binomial[n, j])^2) (Binomial[2 j, n-k])

Table[c[5,n],{n,0,10}]