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A074761 Number of partitions of n of order n. 37
1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 9, 1, 4, 5, 1, 1, 12, 1, 27, 7, 6, 1, 81, 1, 7, 1, 54, 1, 407, 1, 1, 11, 9, 13, 494, 1, 10, 13, 423, 1, 981, 1, 137, 115, 12, 1, 1309, 1, 59, 17, 193, 1, 240, 21, 1207, 19, 15, 1, 47274, 1, 16, 239, 1, 25, 3284, 1, 333, 23, 3731, 1, 42109, 1, 19 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,6
COMMENTS
Order of partition is lcm of its parts.
a(n) is the number of conjugacy classes of the symmetric group S_n such that a representative of the class has order n. Here order means the order of an element of a group. Note that a(n) = 1 if and only if n is a prime power. - W. Edwin Clark, Aug 05 2014
LINKS
Joerg Arndt and Alois P. Heinz, Table of n, a(n) for n = 1..4000 (first 1025 terms from Joerg Arndt)
FORMULA
Coefficient of x^n in expansion of Sum_{i divides n} A008683(n/i)*1/Product_{j divides i} (1-x^j).
EXAMPLE
The a(15) = 5 partitions are (15), (5,3,3,3,1), (5,5,3,1,1), (5,3,3,1,1,1,1), (5,3,1,1,1,1,1,1,1). - Gus Wiseman, Aug 01 2018
MAPLE
A:= proc(n)
uses numtheory;
local S;
S:= add(mobius(n/i)*1/mul(1-x^j, j=divisors(i)), i=divisors(n));
coeff(series(S, x, n+1), x, n);
end proc:
seq(A(n), n=1..100); # Robert Israel, Aug 06 2014
MATHEMATICA
a[n_] := With[{s = Sum[MoebiusMu[n/i]*1/Product[1-x^j, {j, Divisors[i]}], {i, Divisors[n]}]}, SeriesCoefficient[s, {x, 0, n}]]; Array[a, 80}] (* Jean-François Alcover, Feb 29 2016 *)
Table[Length[Select[IntegerPartitions[n], LCM@@#==n&]], {n, 50}] (* Gus Wiseman, Aug 01 2018 *)
PROG
(PARI)
pr(k, x)={my(t=1); fordiv(k, d, t *= (1-x^d) ); return(t); }
a(n) =
{
my( x = 'x+O('x^(n+1)) );
polcoeff( Pol( sumdiv(n, i, moebius(n/i) / pr(i, x) ) ), n );
}
vector(66, n, a(n) )
\\ Joerg Arndt, Aug 06 2014
CROSSREFS
Main diagonal of A256067, A256554.
Sequence in context: A254655 A344971 A077254 * A037861 A145037 A267115
KEYWORD
easy,nonn
AUTHOR
Vladeta Jovovic, Sep 28 2002
STATUS
approved

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Last modified May 18 15:24 EDT 2024. Contains 372664 sequences. (Running on oeis4.)