1,4

Also number of partitions of n in which the largest part occurs exactly k times. Example: T(6,2)=2 because we have [3,3] and [2,2,1,1]. G.f. of column k is x^k/product(1-x^j, j=k..infinity) (k=1,2,...). Row sums yield the partition numbers (A000041). T(n,1)=A000041(n-1) (the partition numbers). T(n,2)=A002865(n-2) (n>=2). T(n,3)=A026796(n). T(n,4)=A026797(n). T(n,5)=A026798(n). T(n,6)=A026799(n). T(n,7)=A026800(n). T(n,8)=A026801(n). T(n,9)=A026802(n). T(n,10)=A026803(n). Sum(k*T(n,k),k=1..n)=A046746(n). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 19 2006

T(n, k)=Sum{T(n-k, i), k<=i<=n-k} for k=1, 2, ..., m, T(n, k)=0 for k=m+1, ..., n-1, where m=[ (n+1)/2 ]; T(n, n)=1 for n >= 1.

G.f.=G(t,x)=sum(t^i*x^i/product(1-x^j, j=i..infinity), i=1..infinity). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 19 2006

G.f.=sum(tx^k/(1-tx^k)/product(1-x^j,j=1..k-1), k=1..infinity). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Mar 13 2006

T(6,2)=2 because we have [4,2] and [2,2,2].

Triangle starts:

1;

1,1;

2,0,1;

3,1,0,1;

5,1,0,0,1;

7,2,1,0,0,1;

g:=sum(t^i*x^i/product(1-x^j, j=i..30), i=1..30): gser:=simplify(series(g, x=0, 19)): for n from 1 to 15 do P[n]:=coeff(gser, x^n) od: for n from 1 to 15 do seq(coeff(P[n], t^j), j=1..n) od; - Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 19 2006

(PARI) T(n, k)=if(k<1|k>n, 0, if(n==k, 1, sum(i=k, n-k, T(n-k, i))))

Row sums give A026806.

Cf. A000041, A002865, A026796, A026797, A026798, A026799, A026800, A026801, A026802, A026803, A046746, A008284.

Sequence in context: A099557 A079217 A079221 this_sequence A137712 A093555 A065432

Adjacent sequences: A026791 A026792 A026793 this_sequence A026795 A026796 A026797

nonn,tabl

Clark Kimberling (ck6(AT)evansville.edu)

More terms from Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 19 2006