>>395

a[n+2]-(n+4)a[n+1]=-{a[n+1]-(n+3)a[n]}
(公比-1の等比数列)
a[n+1]-(n+3)a[n]=(-1)^(n-1)

a[n+1]/(n+3)!-a[n]/(n+2)!=(-1)^(n-1) /(n+3)!
(階差数列形)
a[n]=(n+2)!Σ[k=1,n-1]{(-1)^(k-1) /(k+3)!}
(但しΣ[k=1,0]=0とする)