1,2

Multiplying by 1; 1,2; 1,3,6; 1,4,6,12,24; ... (A036038) yields 1; 2,2; 3,18,6; 4,48,36,144,24; ... in which the groups sum to 1; 4; 27; 256; .... (A000312).

a(n,k) enumerates distributions of n identical objects (balls) into m of alltogether n distinguishable boxes. The k-th partition of n, taken in the Abramowitz-Stegun (A-St) order, specifies the occupation of the m =m(n,k)= A036043(n,k) boxes. m = m(n,k) is the number of parts of the k-th partition of n. For the A-St ordering see pp.831-2 of the reference given in A117506. W. Lang, Nov 13 2007.

The sequence of row lengths is p(n)= A000041(n) (partition numbers) [1, 2, 3, 5, 7, 11, 15, 22, 30, 42,...].

For the A-St order of partitions see the Abramowitz-Stegun reference given in A117506.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

W. Lang, First 10 rows and more.

a(n,k) = A048996(n,k)*binomial(n,m(n,k)),n>=1, k=1,...,p(n) and m(n,k):=A036043(n,k) gives the number of parts of the k-th partition of n.

1; 2,1; 3,6,1; 4,12,6,12,1; 5,20,20,30,30,20,1; ...

a(5,5) relates to the partition (1,2^2) of n=5. Here m=3 and 5 indistinguishable (identical)

balls are put into boxes b1,...,b5 with m=3 boxes occupied; one with one ball and two with two balls.

Therefore a(5,5) = binomial(5,3)*3!/(1!*2!) = 10*3 = 30. W. Lang, Nov 13 2007.

Cf. A036038, A048996, A049009.

Cf. A001700 (row sums).

Cf. A103371(n-1, m-1) (triangle obtained after summing in every row the numbers with like part numbers m).

Sequence in context: A078760 A103280 A046899 this_sequence A115196 A093346 A115597

Adjacent sequences: A035203 A035204 A035205 this_sequence A035207 A035208 A035209

nonn,tabf,easy

Alford Arnold (Alford1940(AT)aol.com)

More terms from Joshua Zucker (joshua.zucker(AT)stanfordalumni.org), Jul 27 2006