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A171398 `X(n,k)' triangle read by rows. X(n,k) is the number of k-subsets of Z_n up to (u,z)-equivalence. 1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 3, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 3, 4, 6, 4, 3, 1, 1, 1, 1, 2, 3, 4, 4, 3, 2, 1, 1, 1, 1, 3, 4, 9, 9, 9, 4, 3, 1, 1, 1, 1, 1, 2, 4, 6, 6, 4, 2, 1, 1, 1, 1, 1, 5, 9, 21, 25, 34, 25, 21, 9, 5, 1, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,13
COMMENTS
Let Z_n={0,1,...,n-1} denote the integers mod n,
let U(n) denote the units mod n, the elements in Z_n relatively prime to n.
Let S and S' be two k-subsets of Z_n.
Define an equivalence relation on the set of k-subsets as follows:
S is (u,z)-equivalent to S' iff there is a u in U(n) and a z in Z_n such that S=uS'+z.
Then define X(n,k) to be the number of such (u,z)-equivalence classes.
This sequence is the `X(n,k)' triangle read by rows.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50).
EXAMPLE
The triangle begins
1;
1,1;
1,1,1;
1,1,1,1;
1,1,2,1,1;
1,1,1,1,1,1;
1,1,3,3,3,1,1;
1,1,1,2,2,1,1,1;
1,1,3,4,6,4,3,1,1;
...
For example row 8 is 1,1,3,4,6,4,3,1,1.
We have X(8,3)=4 because there are 4 (u,z)-equivalence classes of 3-subsets in Z_8, their representatives are: {0,1,2}, {0,1,3}, {0,1,4}, and {0,2,4}.
PROG
(PARI)
Follow(s, f)={my(t=f(s), k=1); while(t>s, k++; t=f(t)); if(s==t, k, 0)}
C(n, k, t, x)=prod(u=0, n-1, my(t=Follow(u, v->(v*k+t)%n)); 1 + if(t, x^t));
row(n)=Vecrev(if(n==0, 1, sum(t=0, n-1, sum(k=1, n, if (gcd(k, n)==1, C(n, k, t, 'x), 0)))/(n * eulerphi(n))));
{ for(n=0, 10, print(row(n))) } \\ Andrew Howroyd, Apr 04 2023
CROSSREFS
Row sums are A002729.
Sequence in context: A161606 A300362 A248145 * A113607 A351352 A082586
KEYWORD
nonn,tabl
AUTHOR
John P. McSorley, Dec 07 2009
EXTENSIONS
Offset corrected and terms a(45) and beyond from Andrew Howroyd, Apr 04 2023
STATUS
approved

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Last modified April 18 16:22 EDT 2024. Contains 371780 sequences. (Running on oeis4.)