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A239871
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Number of strict partitions of n having 1 more even part than odd, so that there is at least one ordering of the parts in which the even and odd parts alternate, and the first and last terms are even.
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7
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0, 0, 1, 0, 1, 0, 1, 1, 1, 2, 1, 4, 1, 6, 1, 9, 2, 12, 3, 16, 6, 20, 10, 25, 17, 30, 26, 36, 40, 43, 57, 51, 81, 61, 110, 74, 148, 91, 193, 113, 250, 144, 316, 184, 397, 239, 491, 311, 603, 407, 732, 530, 885, 692, 1061, 895, 1268, 1155, 1508, 1478, 1790
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OFFSET
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0,10
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COMMENTS
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Let c(n) be the number of strict partitions (that is, every part has multiplicity 1) of n having 1 more odd part than even, so that there is an ordering of parts for which the odd and even parts alternate and the first and last terms are odd. Then c(n) = a(n+1) for n >= 0.
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LINKS
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FORMULA
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a(n) = [x^n y^(-1)] Product_{i>=1} 1+x^i*y^(2*(i mod 2)-1). - Alois P. Heinz, Apr 03 2014
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EXAMPLE
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a(11) counts these 4 partitions: 812, 614, 632, 452.
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MAPLE
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b:= proc(n, i, t) option remember; `if`(n>i*(i+1)/2 or
abs(t)>n, 0, `if`(n=0, 1, b(n, i-1, t)+
`if`(i>n, 0, b(n-i, i-1, t+(2*irem(i, 2)-1)))))
end:
a:= n-> b(n$2, 1):
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MATHEMATICA
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d[n_] := Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &];
p[n_] := p[n] = Select[d[n], Count[#, _?OddQ] == -1 + Count[#, _?EvenQ] &]; t = Table[p[n], {n, 0, 20}]
TableForm[t] (* shows the partitions *)
u = Table[Length[p[n]], {n, 0, 70}] (* A239871 *)
b[n_, i_, t_] := b[n, i, t] = If[n > i*(i + 1)/2 || Abs[t] > n, 0, If[n == 0, 1, b[n, i - 1, t] + If[i > n, 0, b[n - i, i - 1, t + (2*Mod[i, 2] - 1)]]]]; a[n_] := b[n, n, 1]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Nov 16 2015, after Alois P. Heinz *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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