OFFSET
1,4
COMMENTS
Number of ways to partition n elements into pie slices each with at least 2 elements.
Hackl and Prodinger (2018) indirectly refer to this sequence because their Proposition 2.1 contains the g.f. of this sequence. In the paragraph before this proposition, however, they refer to sequence A000358(n) = a(n) + 1. - Petros Hadjicostas, Jun 04 2019
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..1000
Ricardo Gómez Aíza, Symbolic dynamical scales: modes, orbitals, and transversals, arXiv:2009.02669 [math.DS], 2020.
C. G. Bower, Transforms (2)
Daryl DeFord, Enumerating distinct chessboard tilings, Fibonacci Quart. 52 (2014), 102-116; see formula (5.3) in Theorem 5.2, p. 111.
Benjamin Hackl and Helmut Prodinger, The Necklace Process: A Generating Function Approach, arXiv:1801.09934 [math.PR], 2018.
Benjamin Hackl and Helmut Prodinger, The Necklace Process: A Generating Function Approach, Statistics and Probability Letters 142 (2018), 57-61.
P. Hadjicostas, Cyclic Compositions of a Positive Integer with Parts Avoiding an Arithmetic Sequence, Journal of Integer Sequences, 19 (2016), Article 16.8.2.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 764
FORMULA
"CIK" (necklace, indistinct, unlabeled) transform of 0, 1, 1, 1...
From Petros Hadjicostas, Sep 10 2017: (Start)
For all the formulas below, assume n >= 1. Here, phi(n) = A000010(n) is Euler's totient function.
a(n) = (1/n) * Sum_{d|n} b(d)*phi(n/d), where b(n) = A001610(n-1).
a(n) = (1/n) * Sum_{d|n} phi(n/d)*(Fibonacci(d-1) + Fibonacci(d+1) - 1) (because of the equation a(n) = A000358(n) - 1 stated in the CROSSREFS section below).
G.f.: -x/(1-x) + Sum_(k>=1} phi(k)/k * log(1/(1-B(x^k))) where B(x) = x*(1+x). (This is a modification of a formula due to Joerg Arndt.)
G.f.: Sum_{k>=1} phi(k)/k * log((1-x^k)/(1-B(x^k))), which agrees with the one in the Encyclopedia of Combinatorial Structures, #764, above. (We have Sum_{n>=1} (phi(n)/n)*log(1-x^n) = -x/(1-x), which follows from the Lambert series Sum_{n>=1} phi(n)*x^n/(1-x^n) = x/(1-x)^2.)
Sum_{d|n} a(d)*d = n*Sum_{d|n} b(d)/d, where b(n) = A001610(n-1).
(End)
a(n) = Sum_{1 <= i <= ceiling((n-1)/2)} [ (1/(n - i)) * Sum_{d|gcd(i, n-i)} phi(d) * binomial((n - i)/d, i/d) ]. (This is a slight variation of DeFord's formula for the number of distinct Lucas tilings of a 1 X n bracelet up to symmetry, where we exclude the case with i = 0 dominoes.) - Petros Hadjicostas, Jun 07 2019
MAPLE
# formula (5.3) of Daryl Deford for "Number of distinct Lucas tilings of a 1 X n
# bracelet up to symmetry" in "Enumerating distinct chessboard tilings"
A032190 := proc(n)
local a, i, d ;
a := 0 ;
for i from 0 to ceil((n-1)/2) do
for d in numtheory[divisors](i) do
if modp(igcd(i, n-i), d) = 0 then
a := a+(numtheory[phi](d)*binomial((n-i)/d, i/d))/(n-i) ;
end if;
end do:
end do:
a ;
end proc:
seq(A032190(n), n=1..60) ; # R. J. Mathar, Nov 27 2014
MATHEMATICA
nn=40; Apply[Plus, Table[CoefficientList[Series[CycleIndex[CyclicGroup[n], s]/.Table[s[i]->x^(2i)/(1-x^i), {i, 1, n}], {x, 0, nn}], x], {n, 1, nn/2}]] (* Geoffrey Critzer, Aug 10 2013 *)
A032190[n_] := Module[{a=0, i, d, j, dd}, For[i=1, i <= Ceiling[(n-1)/2], i++, For[dd = Divisors[i]; j=1, j <= Length[dd], j++, d=dd[[j]]; If[Mod[GCD[i, n-i], d] == 0, a = a + (EulerPhi[d]*Binomial[(n-i)/d, i/d])/(n-i)]]]; a]; Table[A032190[n], {n, 1, 60}] (* Jean-François Alcover, Nov 27 2014, after R. J. Mathar *)
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Better name from Geoffrey Critzer, Aug 10 2013
STATUS
approved