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A111133
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Number of partitions of n into at least two distinct parts.
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35
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0, 0, 0, 1, 1, 2, 3, 4, 5, 7, 9, 11, 14, 17, 21, 26, 31, 37, 45, 53, 63, 75, 88, 103, 121, 141, 164, 191, 221, 255, 295, 339, 389, 447, 511, 584, 667, 759, 863, 981, 1112, 1259, 1425, 1609, 1815, 2047, 2303, 2589, 2909, 3263, 3657, 4096, 4581, 5119, 5717, 6377
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OFFSET
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0,6
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COMMENTS
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Old name: Number of sets of natural numbers less than n which sum to n.
(1) Number of partitions of n into at least two distinct parts.
(2) Also, number of partitions of 2n into distinct parts having maximal part n; see Example section. (End)
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LINKS
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FORMULA
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G.f.: Sum_{k>=0} (x^((k^2+k)/2) / Product_{j=1..k} (1-x^j)) - 1/(1-x). - Joerg Arndt, Sep 17 2012
G.f.: -1/(1 - x) + Product_{k>=1} (1 + x^k). - Ilya Gutkovskiy, Aug 10 2018
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EXAMPLE
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a(6) = 3 because 1+5, 2+4 and 1+2+3 each sum to 6. That is, the three sets are {1,5},{2,4},{1,2,3}.
For n=6, the partitions of 2n into distinct parts having maximum 6 are 6+5+1, 6+4+2, 6+3+2+1, so that a(6)=3, as an example for Comment (2). - Clark Kimberling, Mar 13 2012
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MAPLE
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seq(coeff(series(mul((1+x^k), k=1..n)-1/(1-x), x, n+1), x, n), n=0..60); # Muniru A Asiru, Aug 10 2018
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MATHEMATICA
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Needs["DiscreteMath`Combinatorica`"]
f[n_] := Block[{lmt = Floor[(Sqrt[8n + 1] - 1)/2] + 1, t}, Sum[ Length[ Select[Plus @@@ KSubsets[ Range[n - k(k - 1)/2 + 1], k], # == n &]], {k, 2, lmt}]]; Array[f, 55] (* Robert G. Wilson v, Oct 17 2005 *)
(* Next program shows the partitions (sets) *)
d[n_] := Select[IntegerPartitions[n], Max[Length /@ Split@ #] == 1 &]; Table[d[n], {n, 1, 12}]
TableForm[%]
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PROG
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(PARI) N=66; x='x+O('x^N);
gf=sum(k=0, N, x^((k^2+k)/2) / prod(j=1, k, 1-x^j)) - 1/(1-x);
concat( [0, 0, 0], Vec(gf) ) /* Joerg Arndt, Sep 17 2012 */
(Haskell)
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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David Sharp (davidsharp(AT)rcn.com), Oct 17 2005
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EXTENSIONS
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STATUS
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approved
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