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A239342
Number of 1's in all compositions of n into odd parts.
3
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
OFFSET
0,3
COMMENTS
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
REFERENCES
S. Heubach and T. Mansour, Combinatorics of Compositions and Words, Chapman and Hall, 2010, page 70.
LINKS
Mengmeng Liu, Andrew Yezhou Wang, The Number of Designated Parts in Compositions with Restricted Parts, J. Int. Seq., Vol. 23 (2020), Article 20.1.8.
FORMULA
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.
EXAMPLE
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
MATHEMATICA
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}]
CROSSREFS
Sequence in context: A090167 A352500 A002985 * A093608 A320328 A079976
KEYWORD
nonn
AUTHOR
Geoffrey Critzer, Mar 16 2014
STATUS
approved