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A180249
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a(n) is the total number of k-reverses of n.
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4
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1, 2, 4, 8, 16, 26, 50, 80, 130, 212, 342, 518, 820, 1276, 1864, 2960, 4336, 6704, 9710, 15068, 21368, 33420, 47082, 72950, 102316, 158888, 220882, 342616, 475108, 734816, 1015778, 1569680, 2161944, 3337952, 4587200, 7069748, 9699292, 14932444, 20445520
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OFFSET
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1,2
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COMMENTS
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See sequence A180171 for the definition of a k-reverse of n.
Briefly, a k-reverse of n is a k-composition of n whose reverse is cyclically equivalent to itself.
This sequence is the total number of k-reverses of n for k=1,2,...,n.
It is the row sums of the 'R(n,k)' triangle from sequence A180171.
For example a(6)=26 because there are 26 k-reverses of n=6 for k=1,2,3,4,5, or 6.
They are, in cyclically equivalent, classes: {6}, {15,51}, {24,42},{33},{114,411,141},{222} {1113,3111,1311,1131}, {1122,2112,2211,1221}, {1212,2121}, {11112,21111,12111,11211,11121}, {111111}.
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REFERENCES
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John P. McSorley: Counting k-compositions with palindromic and related structures. Preprint, 2010.
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LINKS
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FORMULA
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a(n) = (n/2)*Sum_{d|n} (phi^(-1)(d)/d)*b(n/d), where phi^(-1)(n) = A023900(n) is the Dirichlet inverse of the Euler totient function and b(n) = A029744(n+1) (= 3*2^((n/2)-1), if n is even, and = 2^((n+1)/2), if n is odd).
G.f.: Sum_{n>=1} phi^(-1)(n)*g(x^n), where phi^(-1)(n) = A023900(n) and g(x) = x*(x+1)*(2*x+1)/(1-2*x^2)^2.
(End)
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MATHEMATICA
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f[n_Integer] := Block[{c = 0, k = 1, ip = IntegerPartitions@ n, lmt = 1 + PartitionsP@ n, ipk}, While[k < lmt, c += g[ ip[[k]]]; k++ ]; c]; g[lst_List] := Block[{c = 0, len = Length@ lst, per = Permutations@ lst}, While[ Length@ per > 0, rl = Union[ RotateLeft[ per[[1]], # ] & /@ Range@ len]; If[ MemberQ[rl, Reverse@ per[[1]]], c += Length@ rl]; per = Complement[ per, rl]]; c]; Array[f, 24] (* Robert G. Wilson v, Aug 25 2010 *)
b[n_] := Sum[MoebiusMu[n/d] * If[OddQ[d], 2, 3] * 2^Quotient[d-1, 2], {d, Divisors[n]}]; a[n_] := Sum[d*b[d], {d, Divisors[n]}] / 2; Array[a, 39] (* Jean-François Alcover, Nov 04 2017, after Andrew Howroyd *)
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PROG
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b(n) = sumdiv(n, d, moebius(n/d) * if(d%2, 2, 3) * 2^((d-1)\2));
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CROSSREFS
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If we ask for the number of cyclically equivalent classes we get sequence A052955.
For example the 6th term of A052955 is 11, corresponding to the 11 classes in the example above.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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