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A103919
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Triangle of numbers of partitions of n with total number of odd parts equal to k from {0,...,n}.
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136
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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
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OFFSET
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0,8
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COMMENTS
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The partition (0) of n=0 is included. For n>0 no part 0 appears.
The first (k=0) column gives the number of partitions without odd parts, i.e., those with even parts only. See A035363.
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.
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.
T(2n+k) = the convolution of A000041(n) and the number of partitions of n+k having exactly k parts.
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)
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.
The equidistribution of the partition statistics "alternating sum" and "total number of odd parts" follows by conjugation. (End)
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REFERENCES
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J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.
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LINKS
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FORMULA
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a(n, k) = number of partitions of n>=0, which have exactly k odd parts (and possibly even parts) for k from {0, ..., n}.
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
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EXAMPLE
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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
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.
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)
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MAPLE
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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
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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 *)
Table[Length[Select[IntegerPartitions[n], Count[#, _?OddQ]==k&]], {n, 0, 15}, {k, 0, n}] (* Gus Wiseman, Jun 20 2021 *)
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PROG
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(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(""))
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CROSSREFS
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The strict version (without zeros) is A152146 interleaved with A152157.
The reverse version (without zeros) is the right half of A344612.
The strict reverse version (without zeros) is the right half of A344739.
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KEYWORD
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AUTHOR
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STATUS
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approved
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