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A008951
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Array read by columns: number of partitions of n into parts of 2 kinds.
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13
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1, 1, 1, 2, 2, 3, 4, 1, 5, 7, 2, 7, 12, 5, 11, 19, 9, 1, 15, 30, 17, 2, 22, 45, 28, 5, 30, 67, 47, 10, 42, 97, 73, 19, 1, 56, 139, 114, 33, 2, 77, 195, 170, 57, 5, 101, 272, 253, 92, 10, 135, 373, 365, 147, 20, 176, 508, 525, 227, 35, 1, 231, 684, 738, 345, 62, 2, 297
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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COMMENTS
| Fine-Riordan array S_n(m)=a(n,m) with extra row for n=0 added.
Row n of this triangle has length floor(1/2 + sqrt(2*(n+1))), n>=0. This is sequence A002024(n+1)=[1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,...].
Written as triangle this becomes A103923.
a(n,m) gives also the number of partitions of n-t(m), where t(m):=A000217(m) (triangular numbers), with two kinds of parts 1,2,..m. See the column o.g.f.'s in table A103923.
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REFERENCES
| H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 90.
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.
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LINKS
| W. Lang: First 20 rows and comments.
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FORMULA
| Riordan gives formula.
a(n, m)=sum over partitions of n of product(k[j], j=1..m), with k[j]=number of parts of size j (exponent of j in a given partition of n), if m>=1. If m=0 then a(n, 0)=p(n):=A000041(n) (number of partitions of n). O is counted as a part for n=0 and only for this n.
a(n, m)=sum over partitions of n of binomial(q(partition), m), with q the number of distinct parts of a given partition. m>=0.
a(n, m)=a(n-m, m-1) + a(n-m, m), n>=t(m):=m*(m+1)/2=A000217(m) (triangular numbers), else 0, with input a(n, 0)=p(n):=A000041(n).
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EXAMPLE
| Array begins:
m\n 0 1 2 3 4 .5 .6 .7 .8 ...
0 | 1 1 2 3 5 .7 11 15 22 ... (A000041)
1 | . 1 2 4 7 12 19 ... (A000070)
2 | . . . 1 2 .5 .9 ... (A000097)
3 | . . . . . .. .1 ... (A000098)
[1]; [1,1]; [2,2]; [3,4,1]; [5,7,2]; [7,12,5]; [11,19,9,1]...
a(3,1)=4 because the partitions (3), (1,2) and (1^3) have q values 1,2 and 1 which sum to 4.
a(3,1)=4 because the exponents of part 1 in the above given partitions of 3 are 0,1,3 and they sum to 4.
a(3,1)=4 because the partitions of 3-t(1)=2 with two kinds of part 1, say 1 and 1' and one kind of part 2 are (2),(1^2), (1'^2) and (11').
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CROSSREFS
| The first column (m=0) gives A000041(n). Columns m=1..10 are A000070 (partial sums of partition numbers), A000097, A000098, A000710, A103924-A103929.
Sequence in context: A159804 A104567 A087824 * A119473 A002122 A105689
Adjacent sequences: A008948 A008949 A008950 * A008952 A008953 A008954
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KEYWORD
| nonn,tabf,nice
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| More terms from Robert G Bearden (nem636(AT)myrealbox.com), Apr 27 2004
Correction, comments and Riordan formulae from W. Lang (wolfdieter.lang_AT_physik_DOT_uni-karlsruhe_DOT_de), Apr 28 2005
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