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 A008951 Array read by columns: number of partitions of n into parts of 2 kinds. 14
 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; text; internal format)
 OFFSET 0,4 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. In general, column m is asymptotic to exp(Pi*sqrt(2*n/3)) * 6^(m/2) * n^((m-2)/2) / (4*sqrt(3) * m! * Pi^m), equivalently to 6^(m/2) * n^(m/2) / (m! * Pi^m) * p(n), where p(n) is the partition function A000041. - Vaclav Kotesovec, Aug 28 2015 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. LINKS Alois P. Heinz, Columns n = 0..500, flattened William K. Keith, Restricted k-color partitions, arXiv preprint arXiv:1408.4089, 2014 Wolfdieter Lang, First 20 rows and comments. 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). 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]; [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'). MAPLE a:= proc(n, m) option remember; `if`(n<0, 0,       `if`(m=0, combinat[numbpart](n), a(n-m, m-1) +a(n-m, m)))     end: seq(seq(a(n, m), m=0..round(sqrt(2*n+2))-1), n=0..20);  # Alois P. Heinz, Nov 16 2012 MATHEMATICA a[n_, 0] := PartitionsP[n]; a[n_, m_] /; (n >= m*(m+1)/2) := a[n, m] = a[n-m, m-1] + a[n-m, m]; a[n_, m_] = 0; Flatten[ Table[ a[n, m], {n, 0, 18}, {m, 0, Floor[1/2 + Sqrt[2*(n+1)]] - 1}]](* Jean-François Alcover, May 02 2012, after recurrence formula *) 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 KEYWORD nonn,tabf,nice AUTHOR EXTENSIONS More terms from Robert G Bearden (nem636(AT)myrealbox.com), Apr 27 2004 Correction, comments and Riordan formulas from Wolfdieter Lang, Apr 28 2005 STATUS approved

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Last modified October 19 21:28 EDT 2019. Contains 328244 sequences. (Running on oeis4.)