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A116596
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Number of partitions of n having exactly 1 part that appears exactly once.
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1
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1, 1, 1, 2, 4, 4, 8, 8, 12, 16, 23, 24, 40, 45, 59, 72, 99, 108, 153, 171, 224, 263, 341, 377, 504, 567, 711, 821, 1035, 1153, 1467, 1648, 2028, 2317, 2841, 3171, 3923, 4403, 5308, 6014, 7250, 8095, 9778, 10949, 13018, 14672, 17400, 19405, 23061, 25769, 30243
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OFFSET
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1,4
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COMMENTS
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LINKS
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FORMULA
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G.f.=sum(x^j*(1-x^j)/(1-x^j+x^(2j)), j=1..infinity)product((1-x^j+x^(2j))/(1-x^j), j=1..infinity).
G.f. for number of partitions of n having exactly 1 part that appears exactly m times is sum(x^(m*j)*(1-x^j)/(1-x^(m*j)+x^((m+1)*j)), j=1..infinity)*product((1-x^(m*j)+x^((m+1)*j))/(1-x^j), j=1..infinity). - Vladeta Jovovic, May 01 2006
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EXAMPLE
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a(5)=4 because we have [5],[3,1,1],[2,2,1] and [2,1,1,1] ([4,1],[3,2] and [1,1,1,1,1] do not qualify).
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MAPLE
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f:=sum(x^j*(1-x^j)/(1-x^j+x^(2*j)), j=1..75)*product((1-x^j+x^(2*j))/(1-x^j), j=1..75): fser:=series(f, x=0, 73): seq(coeff(fser, x^n), n=1..55);
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MATHEMATICA
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z = 30; u[p_] := Length[DeleteDuplicates[Select[p, Count[p, #] == 1 &]]]; m1[p_] := Min[Map[Length, Split[p]]]; Table[Count[IntegerPartitions[n], p_ /; u[p] == m1[p]], {n, 0, z}] (* Clark Kimberling, Apr 23 2014 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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