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A239342
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Number of 1's in all compositions of n into odd parts.
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3
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0, 1, 2, 3, 6, 11, 20, 36, 64, 113, 198, 345, 598, 1032, 1774, 3039, 5190, 8839, 15016, 25452, 43052, 72685, 122502, 206133, 346346, 581136, 973850, 1630011, 2725254, 4551683, 7594748, 12660660, 21087448, 35094377, 58360134, 96979089, 161042110, 267248664
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OFFSET
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0,3
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COMMENTS
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a(n+1) is the number of ways to tile a strip of length n+1 using white tiles of only odd lengths, with total length n, and one red square of length one. - Gregory L. Simay, Aug 14 2016
A029907, the number of compositions of n with exactly one even part, is equal to a(n+1-2) + a(n+1-4) + a(n+1-6) + ... - Gregory L. Simay, Aug 14 2016
Apart from the initial 0 and 1, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = (1 - S)^2. See A291219. - Clark Kimberling, Sep 02 2017
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REFERENCES
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S. Heubach and T. Mansour, Combinatorics of Compositions and Words, Chapman and Hall, 2010, page 70.
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LINKS
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FORMULA
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For n >= 4, a(n) = a(n-1) + a(n-2) + A000045(n-2).
G.f.: x*(1 - x^2)^2/(1 - x - x^2)^2.
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EXAMPLE
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a(5) = 11 because in the compositions of 5 into odd parts there are a total of 11 1's: 5, 3+1+1, 1+3+1, 1+1+3, 1+1+1+1+1.
Let r represent the red square and 1,3,5 represent the possible odd lengths of the white squares for n=5. Then a(5+1) = a(6) = 20 because r combined with a tile of length 5 generates 2 compositions; r combined with 3,1,1 generates 12 compositions; and r combined with 1,1,1,1,1 generates 6 compositions. 2+12+6 = 20. - Gregory L. Simay, Aug 14 2016
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MATHEMATICA
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nn=30; CoefficientList[Series[x (1-x^2)^2/(1-x-x^2)^2, {x, 0, nn}], x]
(* or *)
Table[Count[Flatten[Level[Map[Permutations, IntegerPartitions[n, n, Table[2k+1, {k, 0, n/2}]]], {2}]], 1], {n, 0, 30}]
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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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