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%I #66 Jun 21 2021 00:05:38
%S 1,0,1,1,0,1,0,2,0,1,2,0,2,0,1,0,4,0,2,0,1,3,0,5,0,2,0,1,0,7,0,5,0,2,
%T 0,1,5,0,9,0,5,0,2,0,1,0,12,0,10,0,5,0,2,0,1,7,0,17,0,10,0,5,0,2,0,1,
%U 0,19,0,19,0,10,0,5,0,2,0,1,11,0,28,0,20,0,10,0,5,0,2,0,1,0,30,0,33,0,20,0,10,0,5,0,2,0,1
%N Triangle of numbers of partitions of n with total number of odd parts equal to k from {0,...,n}.
%C The partition (0) of n=0 is included. For n>0 no part 0 appears.
%C The first (k=0) column gives the number of partitions without odd parts, i.e., those with even parts only. See A035363.
%C Without the alternating zeros this becomes a triangle with columns given by the rows of the S_n(m) table shown in the Riordan reference.
%C From _Gregory L. Simay_, Oct 31 2015: (Start)
%C T(2n+k,k) = the number of partitions of n with parts 1..k of two kinds. If n<=k, then T(2n+k) = A000712(n), the number of partitions of n with parts of two kinds.
%C T(2n+k) = the convolution of A000041(n) and the number of partitions of n+k having exactly k parts.
%C T(2n+k) = d(n,k) where d(n,0) = p(n) and d(n,k) = d(n,k-1) + d(n-k,k-1) + d(n-2k,k-1) + ... (End)
%C From _Emeric Deutsch_, Oct 04 2016: (Start)
%C T(n,k) = number of partitions (p1 >= p2 >= p3 >= ...) of n having alternating sum p1 - p2 + p3 - ... = k. Example: T(5,3) = 2 because there are two partitions (3,1,1) and (4,1) of 5 with alternating sum 3.
%C The equidistribution of the partition statistics "alternating sum" and "total number of odd parts" follows by conjugation. (End)
%D J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.
%H Alois P. Heinz, <a href="/A103919/b103919.txt">Rows n = 0..140, flattened</a>
%H D. Kim, A. J. Yee, <a href="http://dx.doi.org/10.1023/A:1006905826378">A note on partitions into distinct parts and odd parts</a>, Ramanujan J. 3 (1999), 227-231. [_R. J. Mathar_, Nov 11 2008]
%H Wolfdieter Lang, <a href="/A103919/a103919.pdf"> First 11 rows.</a>
%F a(n, k) = number of partitions of n>=0, which have exactly k odd parts (and possibly even parts) for k from {0, ..., n}.
%F Sum_{k=0..n} k*T(n,k) = A066897(n). - _Emeric Deutsch_, Feb 17 2006
%F G.f.: G(t,x) = 1/Product_{j>=1} (1-t*x^(2*j-1))*(1-x^(2*j)). - _Emeric Deutsch_, Feb 17 2006
%F G.f. T(2n+k,k) = g.f. d(n,k) = (1/Product_{j=1..k} (1-x^j)) * g.f. p(n). - _Gregory L. Simay_, Oct 31 2015
%F T(n,k) = T(n-1,k-1) + T(n-2k,k). - _Gregory L. Simay_, Nov 01 2015
%e The triangle a(n,k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10
%e 0: 1
%e 1: 0 1
%e 2: 1 0 1
%e 3: 0 2 0 1
%e 4: 2 0 2 0 1
%e 5: 0 4 0 2 0 1
%e 6: 3 0 5 0 2 0 1
%e 7: 0 7 0 5 0 2 0 1
%e 8: 5 0 9 0 5 0 2 0 1
%e 9: 0 12 0 10 0 5 0 2 0 1
%e 10: 7 0 17 0 10 0 5 0 2 0 1
%e ... Reformatted - _Wolfdieter Lang_, Apr 28 2013
%e a(0,0) = 1 because n=0 has no odd part, only one even part, 0, by definition. a(5,3) = 2 because there are two partitions (1,1,3) and (1,1,1,2) of 5 with exactly 3 odd parts.
%e From _Gregory L. Simay_, Oct 31 2015: (Start)
%e T(10,4) = T(2*3+4,4) = d(3,4) = A000712(3) = 10.
%e T(10,2) = T(2*4+2,2) = d(4,2) = d(4,1)+d(2,1)+d(0,1) = d(4,0)+d(3,0)+d(2,0)+d(1,0)+d(0,0) + d(2,0)+d(1,0)+d(0,0) + d(0,0) = convolution sum p(4)+p(3)+2*p(2)+2*p(1)+3*p(0) = 5+3+2*2+2*1+3*1 = 17.
%e T(9,1) = T(8,0) + T(7,1) = 5 + 7 = 12.
%e (End)
%p g:=1/product((1-t*x^(2*j-1))*(1-x^(2*j)),j=1..20): gser:=simplify(series(g,x=0,22)): P[0]:=1: for n from 1 to 18 do P[n]:=coeff(gser,x^n) od: for n from 0 to 18 do seq(coeff(P[n],t,j),j=0..n) od; # yields sequence in triangular form # _Emeric Deutsch_, Feb 17 2006
%t T[n_, k_] := T[n, k] = Which[n<k, 0, n == k, 1, Mod[n-k+1, 2] == 0, 0, k == 0, Sum[T[Quotient[n, 2], m], {m, 0, n}], True, T[n-1, k-1]+T[n-2*k, k]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Mar 05 2014, after _Paul D. Hanna_ *)
%t Table[Length[Select[IntegerPartitions[n],Count[#,_?OddQ]==k&]],{n,0,15},{k,0,n}] (* _Gus Wiseman_, Jun 20 2021 *)
%o (PARI)
%o {T(n, k)=if(n>=k, if(n==k, 1, if((n-k+1)%2==0, 0, if(k==0, sum(m=0, n, T(n\2, m)), T(n-1, k-1)+T(n-2*k, k)))))}
%o for(n=0, 20, for(k=0, n, print1(T(n, k), ", ")); print(""))
%o \\ _Paul D. Hanna_, Apr 27 2013
%Y Row sums gives A000041 (partition numbers). Columns: k=0: A035363 (with zero entries) A000041 (without zero entries), k=1: A000070, k=2: A000097, k=3: A000098, k=4: A000710, 3k>=n: A000712.
%Y Cf. A066897.
%Y The strict version (without zeros) is A152146 interleaved with A152157.
%Y The rows (without zeros) are A239830 interleaved with A239829.
%Y The reverse version (without zeros) is the right half of A344612.
%Y Removing all zeros gives A344651.
%Y The strict reverse version (without zeros) is the right half of A344739.
%Y Cf. A006330, A027187, A116406, A344607, A344608, A344649, A344654.
%K nonn,easy,tabl
%O 0,8
%A _Wolfdieter Lang_, Mar 24 2005
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