Stirling Number of the First Kind

Author

Eric W. Weisstein
August 29, 2009

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

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

©2009 Wolfram Research, Inc. except for portions noted otherwise

Traditional Form

StirlingS1[n,m]//TraditionalForm

Table

Expand[Table[Product[1+k x,{k,n}],{n,5}]]//ColumnForm

Table[Sum[(-1)^(n+1-k) StirlingS1[n+1,k]x^(n+1-k),{k,n+1}],{n,5}]

Identities

Table[FunctionExpand[StirlingS1[n,n-k],n∈Integers&&n>k],{k,5}]//Column//TraditionalForm

(StirlingS1[n, n-1])==(Binomial[n, 2])

FunctionExpand[StirlingS1[n,n-1],n∈Integers&&n>0]

FullSimplify[%]

Subtract@@@Table[{StirlingS1[n,n-1],-Binomial[n,2]},{n,10}]//Union

(StirlingS1[n, n-2])==(1/4) (3 n+5) (Binomial[n+2, 3])

FunctionExpand[StirlingS1[n,n-2],n∈Integers&&n>2]

Factor[%]

Table[{n,StirlingS1[n,n-2]},{n,2,10}]

FindSequenceFunction[%,n]//Factor

A000914

Table[Binomial[n+2,3]*(3*n+5)/4,{n,10}]

(StirlingS1[n, n-3])

FunctionExpand[StirlingS1[n,n-3],n∈Integers&&n>2]

Factor[%]

Table[{n,StirlingS1[n,n-3]},{n,3,10}]

FindSequenceFunction[%,n]//Factor

(StirlingS1[n, n-4])

FunctionExpand[StirlingS1[n,n-4],n∈Integers&&n>3]

Factor[%]

Table[{n,StirlingS1[n,n-4]},{n,4,15}]

FindSequenceFunction[%,n]//Factor

(StirlingS1[n, n-5])==-(1/8) (3 (n^2)+23 n+38) (Binomial[n+5, 6]) (Binomial[n+5, 2])

FunctionExpand[StirlingS1[n,n-5],n∈Integers&&n>4]

Factor[%]

Table[{n,StirlingS1[n,n-5]},{n,5,15}]

Table[-Binomial[n+5,6]*Binomial[n+5,2]*(3*n^2+23*n+38)/8,{n,10}]

Relationship to StirlingS2

(G. Helms, pers. comm., Apr. 28, 2006).

Table[StirlingS2[n,m],{n,6},{m,6}]//TraditionalForm

Inverse[Table[StirlingS1[n,m],{n,6},{m,6}]]//TraditionalForm

Sums

(∑_(k=0)^n)(x^k) (StirlingS1[n, k])==(FactorialPower[x, n])

Equal@@@Expand[Table[{
Sum[StirlingS1[n,k]x^k,{k,0,n}],
(((-1))^n) Pochhammer[-x,n],
FallingFactorial[x,n],
RisingFactorial[1+x-n,n],
Pochhammer[1+x-n,n],
n! Binomial[x,n],
(-1)^n n! Binomial[n-x-1,n]
},{n,5}]]

V7

Sum[StirlingS1[n,k]x^k,{k,0,n}]

FullSimplify[(((-1))^n) Pochhammer[-x,n]-FactorialPower[x,n],n∈Integers&&n>0]

V8

Sum[StirlingS1[n,k]x^k,{k,0,n}]

(∑_(k=m)^∞)(((x^k) (StirlingS1[k, m]))/(k!))==(((log^m)(x+1))/(m!))

Sum[((StirlingS1[k,m])/(k!))(x^k),{k,m,∞}]

(∑_k^(n+1))(((-1))^(-k+n+1)) (StirlingS1[n+1, k]) (x^(-k+n+1))==(x^n) (Pochhammer[1+(1/x), n])

Sum[(((-1))^(n+1-k))StirlingS1[n+1,k](x^(n+1-k)),{k,n+1}]

Product[1+k x,{k,n}]//FullSimplify

Subtract@@@Flatten[Table[{Sum[(((-1))^(n+1-k))StirlingS1[n+1,k](x^(n+1-k)),{k,n+1}],Product[1+k x,{k,n}]},{n,10},{k,n}],1]//Expand//Union

(∑_(k=m)^n)(n^(k-m)) (StirlingS1[n+1, k+1])==(StirlingS1[n, m])

Sum[(n^(k-m))StirlingS1[n+1,k+1],{k,m,n}]

Table[Sum[(n^(k-m))StirlingS1[n+1,k+1],{k,m,n}]-StirlingS1[n,m],{n,5},{m,n}]

(∑_(k=m-r)^(n-r))(Binomial[n, k]) (StirlingS1[k, m-r]) (StirlingS1[n-k, r])==(Binomial[m, r]) (StirlingS1[n, m])

Sum[Binomial[n,k]StirlingS1[n-k,r]StirlingS1[k,m-r],{k,m-r,n-r}]

Table[Subtract@@{Sum[Binomial[n,k]StirlingS1[n-k,r]StirlingS1[k,m-r],{k,m-r,n-r}],Binomial[m,r]StirlingS1[n,m]},{n,5},{m,n},{r,m}]

(∑_(j=i)^n)(∑_(j=0)^k)(StirlingS2[j, i]) (StirlingS1[k, j]) (StirlingS1[n, k])==(StirlingS1[n, i])

Sum[StirlingS1[n,k]StirlingS1[k,j]StirlingS2[j,i],{k,i,n},{j,0,k}]

Table[Subtract@@{Sum[StirlingS1[n,k]StirlingS1[k,j]StirlingS2[j,i],{k,i,n},{j,0,k}],StirlingS1[n,i]},{n,5},{k,n},{i,k}]

(∑_(k=i)^n)(∑_(j=0)^k)(StirlingS1[j, i]) (StirlingS2[k, j]) (StirlingS2[n, k])==(StirlingS2[n, i])

Sum[StirlingS2[n,k]StirlingS2[k,j]StirlingS1[j,i],{k,i,n},{j,0,k}]

Table[Subtract@@{Sum[StirlingS2[n,k]StirlingS2[k,j]StirlingS1[j,i],{k,i,n},{j,0,k}],
StirlingS2[n,i]},{n,5},{k,n},{i,k}]

(∑_(k=0)^(m-n))(((-1))^k) (Binomial[k+n-1, k-m+n]) (Binomial[2 n-m, -k-m+n]) (StirlingS1[k-m+n, k])==(StirlingS2[n, m])

Sum[(((-1))^k)Binomial[n+k-1,k+n-m]Binomial[2n-m,n-k-m]StirlingS1[k-m+n,k],{k,0,m-n}]

Table[Subtract@@{Sum[(((-1))^k)Binomial[n+k-1,k+n-m]Binomial[2n-m,n-k-m]StirlingS1[k-m+n,k],{k,0,n-m}],
StirlingS2[n,m]},{n,5},{m,n}]

(∑_(k=0)^(m-n))(((-1))^k) (Binomial[k+n-1, k-m+n]) (Binomial[2 n-m, -k-m+n]) (StirlingS2[k-m+n, k])==(StirlingS1[n, m])

Sum[(((-1))^k)Binomial[n+k-1,k+n-m]Binomial[2n-m,n-k-m]StirlingS2[k-m+n,k],{k,0,m-n}]

Table[Subtract@@{Sum[(((-1))^k)Binomial[n+k-1,k+n-m]Binomial[2n-m,n-k-m]StirlingS2[k-m+n,k],{k,0,n-m}],
StirlingS1[n,m]},{n,5},{m,n}]

(StirlingS1[n, n-k])==(Binomial[n-1, k]) (∑_(j=0)^k)(((-1))^j) (Binomial[k+1, j+1])((Binomial[k+j n, k])^(-1)) (StirlingS2[k+j n, j n])

Sum[(((-1))^j)Binomial[k+1,j+1]((Binomial[k+j n,k])^(-1))StirlingS2[k+j n,j n],{j,0,k}]

Refine[(((-1))^k)StirlingCarlitzS1[n-1,k],k∈Integers]

StirlingCarlitzS2[j n,k]

Binomial[n-1,k]Sum[(((-1))^j)Binomial[k+1,j+1]((Binomial[k+j n,k])^(-1))StirlingS2[k+j n,j n],{j,0,k}]

Subtract@@@Flatten[Table[{StirlingS1[n,n-k],Binomial[n-1,k]Sum[(((-1))^j)Binomial[k+1,j+1]((Binomial[k+j n,k])^(-1))StirlingS2[k+j n,j n],{j,0,k}]},{k,10},{n,k,10}],1]//Union

(∑_(l=0)^(max(j,k)+1))(StirlingS1[l, j]) (StirlingS2[k, l])==(KroneckerDeltaSeq[j,k])

Sum[StirlingS1[l,j]StirlingS2[k,l],{l,0,Max[k,j]+1}]

Table[Subtract@@{Sum[StirlingS1[l,j]StirlingS2[k,l],{l,0,Max[k,j]+1}],
KroneckerDelta[j,k]},{j,5},{k,5}]

(∑_(l=0)^(max(k,j)+1))(StirlingS2[l, j]) (StirlingS1[k, l])==(KroneckerDeltaSeq[j,k])

Sum[StirlingS1[k,l]StirlingS2[l,j],{l,0,Max[k,j]+1}]

Table[Subtract@@{Sum[StirlingS1[k,l]StirlingS2[l,j],{l,0,Max[k,j]+1}],
KroneckerDelta[j,k]},{j,5},{k,5}]

Recurrence

(StirlingS1[n, m-1])-n (StirlingS1[n, m])==(StirlingS1[n+1, m])

FullSimplify[StirlingS1[n,m-1]-n StirlingS1[n,m]]

FullSimplify[Subtract@@{StirlingS1[n+1,m],StirlingS1[n,m-1]-n StirlingS1[n,m]}]

Table[-StirlingS1[n,-1+m]+n StirlingS1[n,m]+StirlingS1[1+n,m],{n,5},{m,n}]

RSolve[s[n+1,m]==s[n,m-1]-n s[n,m],s[n,m],{m,n}]

Unsigned Stirling Numbers

(Abs[(StirlingS1[n+1, k])])==(Abs[(StirlingS1[n, k-1])])+n (Abs[(StirlingS1[n, k])])

FullSimplify[Subtract@@{
Abs[StirlingS1[n+1,k]],
n Abs[StirlingS1[n,k]]+Abs[StirlingS1[n,k-1]]
},(k|n)∈Integers&&n>0&&k≤n]

Table[Subtract@@{Abs[StirlingS1[n+1,k]],n Abs[StirlingS1[n,k]]+Abs[StirlingS1[n,k-1]]},{n,10},{k,n}]

(∑_(k=0)^h)((Abs[(StirlingS1[h, h-k])])/(j^k))==((Gamma[h+j])/((j^h) (Gamma[j])))

Sum[((Abs[StirlingS1[h,h-k]])/(j^k)),{k,0,h}]

SeriesCoefficient[((Gamma[h+j])/((j^h)Gamma[j])),{j,∞,n}]

Factor//@Simplify[Series[((Gamma[j+h])/((j^h)Gamma[j])),{j,∞,4}],j>0]