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%I #30 Jul 29 2020 19:32:36
%S 1,1,1,2,2,1,3,4,2,1,5,7,5,2,1,7,12,9,5,2,1,11,19,17,10,5,2,1,15,30,
%T 28,19,10,5,2,1,22,45,47,33,20,10,5,2,1,30,67,73,57,35,20,10,5,2,1,42,
%U 97,114,92,62,36,20,10,5,2,1,56,139,170,147,102,64,36,20,10,5,2,1,77,195
%N Triangle of partitions of n with parts of sizes 1,2,...,m, each of two different kinds, m>=1.
%C The corresponding Fine-Riordan triangle is A008951.
%C This is the array p_2(n,m) of Gupta et al. written as a triangle. p_2(n,m) is defined on p. x of this reference as the number of partitions of n into parts consisting of two varieties of each of the integers 1 to m and one variety of each larger integer. Therefore a(n,m) gives these numbers for the partitions of n-m.
%C a(n,m)= sum over partitions of n+t(m)-m of binomial(q(partition),m), with t(m):=A000217(m) and q the number of distinct parts of a given partition. m>=0.
%C a(n,m)= number of partitions of 2*n-m with exactly m odd parts.
%C a(n,m)= sum over partitions of n+t(m)-m of product(k[j],j=1..m), with t(m):=A000217(m) and 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). 0 is counted as a part for n=0 and only for this n.
%D H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958 (reprinted 1962), pp. 90-121.
%D J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.
%H Alois P. Heinz, <a href="/A103923/b103923.txt">Rows n = 0..140, flattened</a>
%H J. Huang, A. Senger, P. Wear, T. Wu, <a href="http://www.math.umn.edu/~reiner/REU/HuangSengerWearWu2013.pdf">Partition statistics equidistributed with the number of hook difference one cells</a>, 2013. See Remark 5.7. - _N. J. A. Sloane_, May 20 2014
%H W. Lang: <a href="/A103923/a103923.txt">First 16 rows.</a>
%F a(n, m) = a(n-1, m-1) + a(n-m, m), n>=m>=0, with a(n, 0)= A000041(n) (partition numbers), a(n, m)=0 if n<m.
%F a(n, m) = sum(a(n-1-j*m, m-1), j=0..floor((n-m)/m)), m>=1, input a(n, 0)= A000041(n).
%F G.f. column m: product(1/(1-x^j), j=1..m)*P(x), with P(x)= product(1/(1-x^j), j=1..infty), the o.g.f. for the partition numbers A000041.
%F G.f. column m>=1: (product(1/(1-x^k), k=1..m)^2)*product(1/(1-x^j), j=(m+1)..infty). For m=0 put the first product equal to 1.
%e Triangle starts:
%e [1];
%e [1,1];
%e [2,2,1];
%e [3,4,2,1];
%e [5,7,5,2,1];
%e ...
%e a(4,2)=5 from the partitions of 4-2=2 with two varieties of parts 1 and of 2, namely (2),(2'),(1^2),(1'^2) and (1,1').
%e a(4,2)=5 from the partitions of 4+t(2)-2=5 which have products of the exponents of parts 1 and 2: 0*0,1*0,0*1,2*1,1*2,5*0 and sum to 4.
%e a(4,2)=5 from the partitions of 4+t(2)-2=5 which have number of distinct parts (q values) 1,2,2,2,2,2,1. The corresponding binomial(q,2) values are 0,1,1,1,1,0 and sum to 4.
%e a(4,2)=5 from the partitions of 2*4-2=6 with exactly two odd parts, namely (1,5), (3^2), (1^2,4), (1,2,3) and (1^2,2^2), which are 5 in number.
%p with(numtheory):
%p b:= proc(n, k) option remember; `if`(n=0, 1, add(add(d*
%p `if`(d<=k, 2, 1), d=divisors(j)) *b(n-j, k), j=1..n)/n)
%p end:
%p A:= (n, k)-> b(n, k) -`if`(k=0, 0, b(n, k-1)):
%p seq(seq(A(n, k), k=0..n), n=0..14); # _Alois P. Heinz_, Sep 14 2014
%t a[n_, 0] := a[n, 0] = PartitionsP[n]; a[n_, m_] /; n<m = 0; a[n_, m_] /; n >= m >= 0 := a[n, m] = a[n-1, m-1] + a[n-m, m]; Table[a[n, m], {n, 0, 14}, {m, 0, n}] // Flatten (* _Jean-François Alcover_, Dec 09 2014 *)
%t Flatten@Table[Length@IntegerPartitions[n-m, All, Range@n~Join~Range@m], {n, 0, 12}, {m, 0, n}] (* _Robert Price_, Jul 29 2020 *)
%Y The column sequences (without leading 0's) are, for m=0..10: A000041, A000070, A000097, A000098, A000710, A103924-A103929.
%Y Cf. A000712, A124577.
%K nonn,easy,tabl
%O 0,4
%A _Wolfdieter Lang_, Mar 24 2005
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