What is Sm(n) if S0(n)=1 and Sm+1(n)=∑k=1nSm(k) and m and n are integers?

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Elijah Hunt · Accepted Answer

Consider for each m≥0 the generating function fn(u,v)=um∑n≥0Sm(n)vn. Then your first condition on S0 implies that f0(u,v)=1−v)−1. The condition on Sm+1(n) implies thatu1−vfm(u,v)=fm+1(u,v)Indeed, multiplying by (1−v)−1 sums the v coefficients, an u just accounts for the shift from m to m+1.It follows that fm(u,v)=um(1−v)m+1. If you now sum over m, you will get thatf(u,v)=∑m≥0fm(u,v)=11−v11−u1−v=11−u−vExpanding now as geometric series in u+v, you will recognize thatf(u,v)=∑N≥0(u+v)N=∑N≥0(Nj)ujvN−j=∑m,n≥0(m+nn)umvn

asserena3wx · Accepted Answer

Proposition:Sm(n)=(n+m−1)!m!(n−1)!=(n+m−1)(n+m−2)…nm!for m≥0, n≥1Proof: for m=0Sm(n)=S0(n)=(n−1)!(n−1)!=1For m&amp;gt;0Sm+1(n+1)−Sm+1(n)=(n+m+1)!(m+1)!n!−(n+m)!(m+1)!(n−1)!=(n+m+1)!−(n−m)!⋅n(m+1)!n!=(n+m)!m!n!=Sm(n+1)Then we haveSm+1(n+1)=Sm(n+1)+Sm+1(n)=Sm(n+1)+Sm(n)+Sm+1(n−1)=Sm(n+1)+Sm(n)+Sm(n−1)+Sm+1(n−2)=…=∑k=1n+1Sm(k)ThereforeSm(n)=(n+m−1)!m!(n−1)!is the solution. To prove it&#039;s only solution, you can assume there is another solution Sm′(n), and prove by induction that Sm′(n)−Sm(n)=0 for m≥0 and n≥1.You are actually not too far from the answer. You already haveS2(n)=n(n+1)2Then you easily get

What is S_m(n) if S_0(n)=1 and S_{m+1}(n)=\sum_{k=1}^n S_m(k) and m

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