■カタラン数と漸化式(その11)

(n+1)^k−(n+1)=(k,1)Sk-1+(k,2)Sk-2+・・・+(k,k−1)S1

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[1]k=2を代入

(n+1)^2−(n+1)=2S1

S1=n(n+1)/2

[2]k=3を代入

(n+1)^3−(n+1)=3S2+3S1

n(n^2+3n+2)=3S2+3S1

S2={2n(n+1)(n+2)−3n(n+1)}/6

S2=n(n+1)(2n+1)/6

[3]k=4を代入

(n+1)^4−(n+1)=4S3+6S2+4S1

n(n^3+4n^2+6n+3)=4S3+n(n+1)(2n+1)+2n(n+1)

n(n+1)(n^2+3n+3)=4S3+n(n+1)(2n+1)+2n(n+1)

n(n+1)(n^2+3n+3−2n−1−2)=4S3

n^2(n+1)^2/4=S3

S1=Σk=n(n+1)/2

S2=Σk^2=n(n+1)(2n+1)/6

S3=Σk^3=n^2(n+1)^2/4

は多くの読者にとってお馴染みの公式であろう.

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さらに,

S4=Σk^4=n(n+1)(2n+1)(3n^2+3n−1)/30

S5=Σk^5=n^2(n+1)^2(2n^2+2n−1)/12

S6=Σk^6=n(n+1)(2n+1)(3n^4+6n^3−3n+1)/42

S7=Σk^7=n^2(n+1)^2(3n^4+6n^3−n^2−4n+2)/24

S8=Σk^8=n(n+1)(2n+1)(5n^6+15n^5+5n^4−15n^3−n^2+9n−3)/90

と続く.

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