cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A343577 Number of generalized polyforms on the truncated square tiling with n cells.

Original entry on oeis.org

1, 2, 2, 7, 22, 93, 413, 2073, 10741, 57540, 312805, 1722483, 9564565, 53489304, 300840332, 1700347858, 9650975401
Offset: 0

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Author

Peter Kagey, Apr 20 2021

Keywords

Comments

This sequence counts "free" polyforms where holes are allowed. This means that two polyforms are considered the same if one is a rigid transformation (translation, rotation, reflection or glide reflection) of the other.
a(n) >= A343417(n), the number of (n-k)-polyominoes with k distinguished vertices.

Crossrefs

Cf. A121197 (one-sided).
Analogous for other tilings: A000105 (square), A000228 (hexagonal), A000577 (triangular), A197156 (prismatic pentagonal), A197159 (floret pentagonal), A197459 (rhombille), A197462 (kisrhombille), A197465 (tetrakis square), A309159 (snub square), A343398 (trihexagonal), A343406 (truncated hexagonal).

Extensions

a(11) from Drake Thomas, May 02 2021
a(12)-a(16) from John Mason, Mar 20 2022

A373418 Triangle read by rows: T(n,k) is the number of permutations in symmetric group S_n with (n-k) fixed points and an odd number of non-fixed point cycles. Equivalent to the number of cycles of n items with cycle type defined by non-unity partitions of k <= n that contain an odd number of parts.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 3, 2, 0, 0, 6, 8, 6, 0, 0, 10, 20, 30, 24, 0, 0, 15, 40, 90, 144, 135, 0, 0, 21, 70, 210, 504, 945, 930, 0, 0, 28, 112, 420, 1344, 3780, 7440, 7420, 0, 0, 36, 168, 756, 3024, 11340, 33480, 66780, 66752, 0, 0, 45, 240, 1260, 6048, 28350, 111600, 333900, 667520, 667485
Offset: 0

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Author

Keywords

Comments

a(n) + A343417(n) = A098825(n) = partial derangement "rencontres" triangle.
a(n) - A343417(n) = (k-1) * binomial(n,k) = A127717(n-1,k-1).
Difference of 2nd and 1st leading diagonals (n > 0):
T(n,n-1) - T(n,n) = 0,-1,1,2,6,9,15,20,28,35,45,54,...
= (0-1) + (2+1) + (4+3) + (6+5) + (8+7) + (10+9) + ...
Cf. A084265(n) with 2 terms 0,-1 prepended (moving its offset from 0 to -2).

Examples

			Triangle begins:
   n: {k<=n}
   0: {0}
   1: {0, 0}
   2: {0, 0,  1}
   3: {0, 0,  3,   2}
   4: {0, 0,  6,   8,    6}
   5: {0, 0, 10,  20,   30,   24}
   6: {0, 0, 15,  40,   90,  144,   135}
   7: {0, 0, 21,  70,  210,  504,   945,    930}
   8: {0, 0, 28, 112,  420, 1344,  3780,   7440,   7420}
   9: {0, 0, 36, 168,  756, 3024, 11340,  33480,  66780,  66752}
  10: {0, 0, 45, 240, 1260, 6048, 28350, 111600, 333900, 667520, 667485}
T(n,0) = 0 because the sole permutation in S_n with n fixed points, namely the identity permutation, has 0 non-fixed point cycles, not an odd number.
T(n,1) = 0 because there are no permutations in S_n with (n-1) fixed points.
Example:
T(3,3) = 2 since S_3 contains 3 permutations with 0 fixed points and an odd number of non-fixed point cycles, namely the derangements (123) and (132).
Worked Example:
T(7,6) = 945 permutations in S_7 with 1 fixed point and an odd number of non-fixed point cycles;
T(7,6) = 945 possible 6- and (2,2,2)-cycles of 7 items.
N(n,y) = possible y-cycles of n items;
N(n,y) = (n!/(n-k)!) / (M(y) * s(y)).
y = partition of k<=n with q parts = (p_1, p_2, ..., p_i, ..., p_q) such that k = Sum_{i=1..q} p_i.
Or:
y = partition of k<=n with d distinct parts, each with multiplicity m_j = (y_1^m_1, y_2^m_2, ..., y_j^m_j, ..., y_d^m_d) such that k = Sum_{j=1..d} m_j*y_j.
M(y) = Product_{i=1..q} p_i = Product_{j=1..d} y_j^m_j.
s(y) = Product_{j=1..d} m_j! ("symmetry of repeated parts").
Note: (n!/(n-k)!) / s(y) = multinomial(n, {m_j}).
Therefore:
T(7,6) = N(7,y=(6)) + N(7,y=(2^3))
       = (7!/6) + (7!/(2^3)/3!)
       = 7! * (1/6 + 1/48)
       = 5040 * (3/16);
T(7,6) = 945.
		

Crossrefs

Cf. A373417 (even case), A373340 (row sums), A216779 (main diagonal).

Programs

  • Maple
    b:= proc(n, t) option remember; `if`(n=0, t, add(expand(`if`(j>1, x^j, 1)*
          b(n-j, irem(t+signum(j-1), 2)))*binomial(n-1, j-1)*(j-1)!, j=1..n))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n, 0)):
    seq(T(n), n=0..10);
  • Mathematica
    Table[Table[n!/(n-k)!/2 * (Sum[(-1)^j/j!, {j, 0, k}] - ((k - 1)/k!)),{k,1,n}], {n,1,10}]

Formula

T(n,k) = (n!/(n-k)!/2) * ((Sum_{j=0..k} (-1)^j/j!) + (k-1)/k!). Cf. Sum_{j=0..k} (-1)^j/j! = A053557(k) / A053556(k).
Showing 1-2 of 2 results.