A103919 Triangle of numbers of partitions of n with total number of odd parts equal to k from {0,...,n}.
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, 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, 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
Offset: 0
Examples
The triangle a(n,k) begins: n\k 0 1 2 3 4 5 6 7 8 9 10 0: 1 1: 0 1 2: 1 0 1 3: 0 2 0 1 4: 2 0 2 0 1 5: 0 4 0 2 0 1 6: 3 0 5 0 2 0 1 7: 0 7 0 5 0 2 0 1 8: 5 0 9 0 5 0 2 0 1 9: 0 12 0 10 0 5 0 2 0 1 10: 7 0 17 0 10 0 5 0 2 0 1 ... Reformatted - _Wolfdieter Lang_, Apr 28 2013 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. From _Gregory L. Simay_, Oct 31 2015: (Start) T(10,4) = T(2*3+4,4) = d(3,4) = A000712(3) = 10. 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. T(9,1) = T(8,0) + T(7,1) = 5 + 7 = 12. (End)
References
- J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.
Links
- Alois P. Heinz, Rows n = 0..140, flattened
- D. Kim, A. J. Yee, A note on partitions into distinct parts and odd parts, Ramanujan J. 3 (1999), 227-231. [_R. J. Mathar_, Nov 11 2008]
- Wolfdieter Lang, First 11 rows.
Crossrefs
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.
Cf. A066897.
The reverse version (without zeros) is the right half of A344612.
Removing all zeros gives A344651.
The strict reverse version (without zeros) is the right half of A344739.
Programs
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Maple
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
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Mathematica
T[n_, k_] := T[n, k] = Which[n
Jean-François Alcover, Mar 05 2014, after Paul D. Hanna *) Table[Length[Select[IntegerPartitions[n],Count[#,?OddQ]==k&]],{n,0,15},{k,0,n}] (* _Gus Wiseman, Jun 20 2021 *) -
PARI
{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)))))} for(n=0, 20, for(k=0, n, print1(T(n, k), ", ")); print("")) \\ Paul D. Hanna, Apr 27 2013
Formula
a(n, k) = number of partitions of n>=0, which have exactly k odd parts (and possibly even parts) for k from {0, ..., n}.
Sum_{k=0..n} k*T(n,k) = A066897(n). - Emeric Deutsch, Feb 17 2006
G.f.: G(t,x) = 1/Product_{j>=1} (1-t*x^(2*j-1))*(1-x^(2*j)). - Emeric Deutsch, Feb 17 2006
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
T(n,k) = T(n-1,k-1) + T(n-2k,k). - Gregory L. Simay, Nov 01 2015
Comments