A028243 -id:A028243 - cp's OEIS frontend

A293617 Array of triangles read by ascending antidiagonals, T(m, n, k) = Pochhammer(m, k) * Stirling2(n + m, k + m) with m >= 0, n >= 0 and 0 <= k <= n.

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 2, 1, 0, 1, 10, 3, 7, 3, 0, 1, 15, 4, 25, 12, 2, 0, 1, 21, 5, 65, 30, 6, 1, 0, 1, 28, 6, 140, 60, 12, 15, 7, 0, 1, 36, 7, 266, 105, 20, 90, 50, 12, 0, 1, 45, 8, 462, 168, 30, 350, 195, 60, 6, 0, 1, 55, 9, 750, 252, 42, 1050, 560, 180, 24, 1, 0

Offset: 0

Peter Luschny, Oct 20 2017

			Array starts:
m\j| 0   1  2     3       4       5       6       7       8       9      10
---|-----------------------------------------------------------------------
m=0| 1,  0, 0,    0,      0,      0,      0,      0,      0,      0,      0
m=1| 1,  1, 1,    1,      3,      2,      1,      7,     12,      6,      1
m=2| 1,  3, 2,    7,     12,      6,     15,     50,     60,     24,     31
m=3| 1,  6, 3,   25,     30,     12,     90,    195,    180,     60,    301
m=4| 1, 10, 4,   65,     60,     20,    350,    560,    420,    120,   1701
m=5| 1, 15, 5,  140,    105,     30,   1050,   1330,    840,    210,   6951
m=6| 1, 21, 6,  266,    168,     42,   2646,   2772,   1512,    336,  22827
m=7| 1, 28, 7,  462,    252,     56,   5880,   5250,   2520,    504,  63987
m=8| 1, 36, 8,  750,    360,     72,  11880,   9240,   3960,    720, 159027
m=9| 1, 45, 9, 1155,    495,     90,  22275,  15345,   5940,    990, 359502
   A000217, A001296,A027480,A002378,A001297,A293475,A033486,A007531,A001298
.
m\j| ...      11      12      13      14
---|-----------------------------------------
m=0| ...,      0,      0,      0,      0, ... [A000007]
m=1| ...,     15,     50,     60,     24, ... [A028246]
m=2| ...,    180,    390,    360,    120, ... [A053440]
m=3| ...,   1050,   1680,   1260,    360, ... [A294032]
m=4| ...,   4200,   5320,   3360,    840, ...
m=5| ...,  13230,  13860,   7560,   1680, ...
m=6| ...,  35280,  31500,  15120,   3024, ...
m=7| ...,  83160,  64680,  27720,   5040, ...
m=8| ..., 178200, 122760,  47520,   7920, ...
m=9| ..., 353925, 218790,  77220,  11880, ...
         A293476,A293608,A293615,A052762, ...
.
The parameter m runs over the triangles and j indexes the triangles by reading them by rows. Let T(m, n) denote the row [T(m, n, k) for 0 <= k <= n] and T(m) denote the triangle [T(m, n) for n >= 0]. Then for instance T(2) is the triangle A053440, T(3, 2) is row 2 of A294032 (which is [25, 30, 12]) and T(3, 2, 1) = 30.
.
Remark: To adapt the sequences A028246 and A053440 to our enumeration use the exponential generating functions exp(x)/(1 - y*(exp(x) - 1)) and exp(x)*(2*exp(x) - y*exp(2*x) + 2*y*exp(x) - 1 - y)/(1 - y*(exp(x) - 1))^2 instead of those indicated in their respective entries.

Eric Weisstein's World of Mathematics, Nørlund Polynomial.

A000217(n) = T(n, 1, 0), A001296(n) = T(n, 2, 0), A027480(n) = T(n, 2, 1),
A002378(n) = T(n, 2, 2), A001297(n) = T(n, 3, 0), A293475(n) = T(n, 3, 1),
A033486(n) = T(n, 3, 2), A007531(n) = T(n, 3, 3), A001298(n) = T(n, 4, 0),
A293476(n) = T(n, 4, 1), A293608(n) = T(n, 4, 2), A293615(n) = T(n, 4, 3),
A052762(n) = T(n, 4, 4), A052787(n) = T(n, 5, 5), A000225(n) = T(1, n, 1),
A028243(n) = T(1, n, 2), A028244(n) = T(1, n, 3), A028245(n) = T(1, n, 4),
A032180(n) = T(1, n, 5), A228909(n) = T(1, n, 6), A228910(n) = T(1, n, 7),
A000225(n) = T(2, n, 0), A007820(n) = T(n, n, 0).
A028246(n,k) = T(1, n, k), A053440(n,k) = T(2, n, k), A294032(n,k) = T(3, n, k),
A293926(n,k) = T(n, n, k), A124320(n,k) = T(n, k, k), A156991(n,k) = T(k, n, n).
Cf. A293616.

A293617 := proc(m, n, k) option remember:
if m = 0 then 0^n elif k < 0 or k > n then 0 elif n = 0 then 1 else
(k+m)*A293617(m,n-1,k) + k*A293617(m,n-1,k-1) + A293617(m-1,n,k) fi end:
for m in [$0..4] do for n in [$0..6] do print(seq(A293617(m, n, k), k=0..n)) od od;
# Sample uses:
A027480 := n -> A293617(n, 2, 1): A293608 := n -> A293617(n, 4, 2):
# Flatten:
a := proc(n) local w; w := proc(k) local t, s; t := 1; s := 1;
while t <= k do s := s + 1; t := t + s od; [s - 1, s - t + k] end:
seq(A293617(n - k, w(k)[1], w(k)[2]), k=0..n) end: seq(a(n), n = 0..11);

T[m_, n_, k_] := Pochhammer[m, k] StirlingS2[n + m, k + m];
For[m = 0, m < 7, m++, Print[Table[T[m, n, k], {n,0,6}, {k,0,n}]]]
A293617Row[m_, n_] := Table[T[m, n, k], {k,0,n}];
(* Sample use: *)
A293926Row[n_] := A293617Row[n, n];

T(m,n,k) = (k + m)*T(m, n-1, k) + k*T(m, n-1, k-1) + T(m-1, n, k) with boundary conditions T(0, n, k) = 0^n; T(m, n, k) = 0 if k<0 or k>n; and T(m, 0, k) = 0^k.

T(m,n,k) = Pochhammer(m, k)*binomial(n + m, k + m)*NorlundPolynomial(n - k, -k - m).

A298668 Number T(n,k) of set partitions of [n] into k blocks such that the absolute difference between least elements of consecutive blocks is always > 1; triangle T(n,k), n>=0, 0<=k<=ceiling(n/2), read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 3, 0, 1, 7, 2, 0, 1, 15, 12, 0, 1, 31, 50, 6, 0, 1, 63, 180, 60, 0, 1, 127, 602, 390, 24, 0, 1, 255, 1932, 2100, 360, 0, 1, 511, 6050, 10206, 3360, 120, 0, 1, 1023, 18660, 46620, 25200, 2520, 0, 1, 2047, 57002, 204630, 166824, 31920, 720

Offset: 0

Alois P. Heinz, Jan 24 2018

			T(5,1) = 1: 12345.
T(5,2) = 7: 1234|5, 1235|4, 123|45, 1245|3, 124|35, 125|34, 12|345.
T(5,3) = 2: 124|3|5, 12|34|5.
T(7,4) = 6: 1246|3|5|7, 124|36|5|7, 124|3|56|7, 126|34|5|7, 12|346|5|7, 12|34|56|7.
T(9,5) = 24: 12468|3|5|7|9, 1246|38|5|7|9, 1246|3|58|7|9, 1246|3|5|78|9, 1248|36|5|7|9, 124|368|5|7|9, 124|36|58|7|9, 124|36|5|78|9, 1248|3|56|7|9, 124|38|56|7|9, 124|3|568|7|9, 124|3|56|78|9, 1268|34|5|7|9, 126|348|5|7|9, 126|34|58|7|9, 126|34|5|78|9, 128|346|5|7|9, 12|3468|5|7|9, 12|346|58|7|9, 12|346|5|78|9, 128|34|56|7|9, 12|348|56|7|9, 12|34|568|7|9, 12|34|56|78|9.
Triangle T(n,k) begins:
  1;
  0, 1;
  0, 1;
  0, 1,    1;
  0, 1,    3;
  0, 1,    7,     2;
  0, 1,   15,    12;
  0, 1,   31,    50,     6;
  0, 1,   63,   180,    60;
  0, 1,  127,   602,   390,    24;
  0, 1,  255,  1932,  2100,   360;
  0, 1,  511,  6050, 10206,  3360,  120;
  0, 1, 1023, 18660, 46620, 25200, 2520;
  ...

Alois P. Heinz, Rows n = 0..200, flattened
Wenqin Cao, Emma Yu Jin, and Zhicong Lin, Enumeration of inversion sequences avoiding triples of relations, Discrete Applied Mathematics (2019); see also author's copy

Columns k=0-11 give (offsets may differ): A000007, A057427, A168604, A028243, A028244, A028245, A032180, A228909, A228910, A228911, A228912, A228913.
Row sums give A229046(n-1) for n>0.
T(2n+1,n+1) gives A000142.
T(2n,n) gives A001710(n+1).
Cf. A000629, A000670, A008277, A028246, A048993, A076726, A110654, A136011, A173018.

b:= proc(n, m, t) option remember; `if`(n=0, x^m, add(
      b(n-1, max(m, j), `if`(j>m, 1, 0)), j=1..m+1-t))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)):
seq(T(n), n=0..14);
# second Maple program:
T:= (n, k)-> `if`(k=0, `if`(n=0, 1, 0), (k-1)!*Stirling2(n-k+1, k)):
seq(seq(T(n, k), k=0..ceil(n/2)), n=0..14);
# third Maple program:
T:= proc(n, k) option remember; `if`(k<2, `if`(n=0 xor k=0, 0, 1),
      `if`(k>ceil(n/2), 0, add((k-j)*T(n-1-j, k-j), j=0..1)))
    end:
seq(seq(T(n, k), k=0..ceil(n/2)), n=0..14);

T[n_, k_] := T[n, k] = If[k < 2, If[Xor[n == 0, k == 0], 0, 1],
     If[k > Ceiling[n/2], 0, Sum[(k-j) T[n-1-j, k-j], {j, 0, 1}]]];
Table[Table[T[n, k], {k, 0, Ceiling[n/2]}], {n, 0, 14}] // Flatten (* Jean-François Alcover, Mar 08 2021, after third Maple program *)

T(n,k) = (k-1)! * Stirling2(n-k+1,k) for k>0, T(n,0) = A000007(n).
T(n,k) = Sum_{j=0..k-1} (-1)^j*C(k-1,j)*(k-j)^(n-k) for k>0, T(n,0) = A000007(n).
T(n,k) = (k-1)! * A136011(n,k) for n, k >= 1.
Sum_{j>=0} T(n+j,j) = A076726(n) = 2*A000670(n) = A000629(n) + A000007(n).

A341617 Repair factors for Stirling numbers of the second kind.

Original entry on oeis.org

1, 1, 2, 6, 12, 60, 30, 210, 840, 2520, 1260, 13860, 13860, 180180, 90090, 30030, 240240, 4084080, 6126120, 116396280, 58198140, 58198140

Offset: 1

Thomas Ward, Feb 16 2021

a(k) is the smallest number with the property that the sequence (a(k)*S(n+k-1, k))_{n>=1}, viewed as a sequence in n, counts the periodic points of some map. Here S(n, k) denotes the Stirling numbers of the second kind, so S(n, k) is the number of ways to partition a set of n elements into k nonempty subsets.
Because these numbers relate to properties of the Stirling number sequence S(n, k) viewed as a sequence in n, it is natural to think of these values as indexed by k.
The values generated by a finite calculation are inherently empirical, calculated as the least common multiple of the first 3000 terms and then checked for the repair property for period up to 50,000. For smaller values of k (those listed here) various ad hoc arguments can be used to check the empirical candidate values.
The empirical values can be calculated using a simple PARI code, and this only gives a possible candidate value for a(k). If that value happens to be (k-1)! then it is correct. If it is a strict divisor of (k-1)! then further checks are needed for each prime in the quotient (candidate value)/(k-1)!. The potentially unbounded nature of the calculation required would make a proof of a closed formula particularly interesting.
(EMPIRICAL PARI CODE TO GENERATE CANDIDATE VALUES) for(k=1, 40, a=vector(3000, n, stirling(n+k-1, k, 2)); b=vector(length(a), n, (1/n)*sumdiv(n, d, moebius(n/d)*a[d])); print1(denominator(b)", "))

			The statement that a(3) = 2 means that the sequence of Stirling numbers S_3 = (1, 6, 25, 90, ...) (that is, the sequence A000392 with an offset of 3) does not have the property of counting periodic points for some map, but does have this property after multiplication by 2 (which gives A028243 with an offset of 2), and 2 is the smallest integer with this property. This specific value is immediately known to be exact, because a(3) divides (3-1)! = 2.

P. Miska and T. Ward, Stirling numbers and periodic points, arXiv:2102.07561 [math.NT], 2021.

Cf. A000392, A028243, A008277, A341991.

It is known that a(k) divides (k-1)!.

There is an explicit formula for a(k), but it involves an in principle infinite calculation as follows: compute the set of rational numbers {(1/n) Sum_{d|n} mu(n/d)*S(d+k-1, k): n>=1}, and then define a(k) to be the least common multiple of the denominators of that (possibly infinite) set of rational numbers. Here mu denotes the classical Möbius function, and the sum is taken over the divisors d of n. Strictly speaking, any calculation only gives candidate values which are initially only known to be a factor of the real value, which in turn is a factor of (k-1)!. For small values of k listed above ad hoc arguments are needed to check the candidate values evaluated as the least common multiple of the denominators of the first 3000 terms.

A091913 Triangle read by rows: a(n,k) = C(n,k)*(2^(n-k) - 1) for k= n, where k=0..max(n-1,0).

Original entry on oeis.org

0, 1, 3, 2, 7, 9, 3, 15, 28, 18, 4, 31, 75, 70, 30, 5, 63, 186, 225, 140, 45, 6, 127, 441, 651, 525, 245, 63, 7, 255, 1016, 1764, 1736, 1050, 392, 84, 8, 511, 2295, 4572, 5292, 3906, 1890, 588, 108, 9, 1023, 5110, 11475, 15240, 13230, 7812, 3150, 840, 135, 10, 2047

Offset: 0

Ross La Haye, Mar 10 2004

Row lengths are 1,1,2,3,4,... = A028310. - M. F. Hasler, Jul 21 2012
Rows: Sum of the n-th row = A001047(n); Sum of the n-th row excluding column 0 = A028243(n+1). Columns: a(n,0) = A000225(n); a(n,1) = A058877(n). Diagonals: a(n,n-2) = A045943(n-1). Also note that the sums of the antidiagonals = A006684.
As an infinite lower triangular matrix * the Bernoulli numbers as a vector (Cf. A027641) = the natural numbers: [1, 2, 3, ...]. The same matrix * the Bernoulli number version starting [1, 1/2, 1/6, ...] = A001787: (1, 4, 12, 32, ...). - Gary W. Adamson, Mar 13 2012

			Triangle begins
   0;
   1;
   3,   2;
   7,   9,   3;
  15,  28,  18,   4;
  31,  75,  70,  30,   5;
  63, 186, 225, 140,  45,   6;
  ...
a(5,3) = 30 because C(5,3) = 10, 2^(5 - 3) - 1 = 3 and 10 * 3 = 30.

Cf. A007318, A038207.

Cf. A001787, A027641, A027642.

For k>=n, a(n, k) = 0; for k < n, a(n, k) = C(n, k) * (2^(n-k) - 1) = Sum [C(n,k) * C(n-k, m), {m=1 to n-k}]. [Formula corrected Aug 22 2006]
The triangle (1; 3,2; 7,9,3; ...) = A007318^2 - A007318, then delete the right border of zeros. - Gary W. Adamson, Nov 16 2007
O.g.f.: 1/( (1 - (1 + x)*t)*(1 - (2 + x)*t) ) = 1 + (3 + 2*x)*t + (7 + 9*x + 3*x^2)*t^2 + .... - Peter Bala, Jul 16 2013

More terms from Pab Ter (pabrlos(AT)yahoo.com), May 25 2004

A134168 Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which either 0) x and y are disjoint and for which either x is a subset of y or y is a subset of x, or 1) x and y are intersecting but for which x is not a subset of y and y is not a subset of x, or 2) x and y are intersecting and for which either x is a proper subset of y or y is a proper subset of x, or 3) x = y.

Original entry on oeis.org

1, 3, 9, 30, 111, 438, 1779, 7290, 29871, 121998, 496299, 2011650, 8129031, 32769558, 131850819, 529745610, 2126058591, 8525561118, 34166421339, 136858609170, 548013994551, 2193796224678, 8780408783859, 35137313082330, 140596298752911, 562526359448238, 2250528981434379, 9003386657325090

Offset: 0

Ross La Haye, Jan 12 2008

			a(2) = 9 because for P(A) = {{},{1},{2},{1,2}} we have for case 0 {{},{1}}, {{},{2}}, {{},{1,2}} and we have for case 2 {{1},{1,2}}, {{2},{1,2}} and we have for case 3 {{},{}}, {{1},{1}}, {{2},{2}}, {{1,2},{1,2}}. There are 0 {x,y} of P(A) in this example that fall under case 1.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
Index entries for linear recurrences with constant coefficients, signature (10,-35,50,-24).

Cf. A000225, A032263, A028243, A000079.

LinearRecurrence[{10, -35, 50, -24}, {1, 3, 9, 30}, 50] (* or *) Table[(1/2)*(4^n - 3^n + 3*2^n - 1), {n,0,50}] (* G. C. Greubel, May 30 2016 *)

a(n) = (1/2)*(4^n - 3^n + 3*2^n - 1).
a(n) = 3*StirlingS2(n+1,4) +2*StirlingS2(n+1,3) +2*StirlingS2(n+1,2) +1.
G.f.: -(5*x^3 - 14*x^2 + 7*x - 1)/((x-1)*(2*x-1)*(3*x-1)*(4*x-1)). - Colin Barker, Jul 30 2012

A285867 Triangle T(n, k) read by rows: T(n, k) = S2(n, k)k! + S2(n, k-1)(k-1)! with the Stirling2 triangle S2 = A048993.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 1, 7, 12, 0, 1, 15, 50, 60, 0, 1, 31, 180, 390, 360, 0, 1, 63, 602, 2100, 3360, 2520, 0, 1, 127, 1932, 10206, 25200, 31920, 20160, 0, 1, 255, 6050, 46620, 166824, 317520, 332640, 181440, 0, 1, 511, 18660, 204630, 1020600, 2739240, 4233600, 3780000, 1814400, 0, 1, 1023, 57002, 874500, 5921520, 21538440, 46070640, 59875200, 46569600, 19958400

Offset: 0

Wolfdieter Lang, May 03 2017

This triangle T(n, k) appears in the e.g.f. of the sum of powers SP(n, m) = Sum_{j=0..m} j^n, n >= 0, m >= 0 with 0^0:=1 as ESP(n, t) = exp(t)*(Sum_{k=0..n} T(n, k)*t^k/k! + t^(n+1)/(n+1)), n >= 0.
The sub-triangle T(n, k) for 1 <= k <=n, see A028246(n+1,k) (diagonal not needed).
For S2(n, m)*m! see A131689.
The columns (starting sometimes with n=k) are A000007, A000012, A000225, A028243(n-1), A028244(n-1), A028245(n-1), A032180(n-1), A228909, A228910, A228911, A228912, A228913. See below for the e.g.f.s and o.g.f.s.
The row sums are 1 for n=1 and A000629(n) - n! for n >= 1, See A285868.

			The triangle T(n, k) begins:
n\k 0  1    2     3      4       5        6        7        8        9  ...
0:  1
1:  0  1
2:  0  1    3
3:  0  1    7    12
4:  0  1   15    50     60
5:  0  1   31   180    390     360
6:  0  1   63   602   2100    3360     2520
7:  0  1  127  1932  10206   25200    31920    20160
8:  0  1  255  6050  46620  166824   317520   332640   181440
9:  0  1  511 18660 204630 1020600  2739240  4233600  3780000  1814400
...

Cf. A000629, A028246, A048993, A131689, A285868.

Table[If[k == 0, Boole[n == 0], StirlingS2[n, k] k! + StirlingS2[n, k - 1] (k - 1)!], {n, 0, 10}, {k, 0, n}] (* Michael De Vlieger, May 08 2017 *)

T(n, k) =  A131689(n, k) + A131689(n, k-1), 0 <= k <= n, with A131689(n, -1) = 0.
T(0, 0) = 1 and T(n, k) = Stirling2(n+1, k)*(k-1)! for n >= k >= 1. For Stirling2 see A048993. Stirling2(n, k)*(k-1)! = A028246(n, k) for n >= k >= 1.
Recurrence: T(0, 0) = 1, T(n, n) = (n+1)!/2, T(n, -1) = 0, T(n, k) = 0 if n < k, and T(n, k) = (k-1)*T(n-1, k-1) + k*T(n-1, k), for n > k >= 0.
E.g.f. for column k=0 is 1, and for  k >= 1: Sum_{j=1..k}((-1)^(k-j) * binomial(k-1, j-1) * exp(j*x)) - x^(k-1).
O.g.f. for column k = 0 is 1, and for  k >= 1: ((k-1)!*x^(k-1) / Product_{j=1..k} (1-j*x)) - (k-1)!*x^(k-1).

A350280 Number of bracelets describing topological configurations of points and lines formed by the perpendicular bisectors of the sides of a convex cyclic n-gon.

Original entry on oeis.org

0, 1, 1, 5, 9, 30, 69, 203, 519, 1466, 3933, 11025, 30345, 85190, 238063, 671651, 1895265, 5376856, 15279117, 43568435, 124478129, 356537150, 1023113061, 2941713513, 8472215013, 24439992746, 70604898953, 204253079165, 591631927785, 1715743930880, 4981202429973

Offset: 1

Sanjay Ramassamy, Dec 23 2021

nonn

For n>=3, a(n) is the number of topological configurations (up to cyclic shifts and reversal) of n points and n lines, where the points lie at the vertices of a convex cyclic n-gon and the lines are the perpendicular bisectors of its sides. Counting such configurations without quotienting out by cyclic shifts and reversal gives the sequence A028243.

a(n) is also the number of equivalence classes (up to cyclic shifts and reversal) of 2n-tuples composed of n 0's and n 1's which have an interlacing signature. The signature of a 2n-tuple (v_1,...,v_{2n}) is the n-tuple (s_1,...,s_n) defined by s_i=v_i+v_{i+n}. The signature is called interlacing if after deleting the 1's, there are letters remaining and the remaining 0's and 2's are alternating.

			For n=3, drawing the three perpendicular bisectors of a triangle divides the plane into 6 regions. Three of these regions contain one vertex of the triangle and the other three contain none. Up to cyclic shifts and reversal, the only possible configuration is (nonempty, nonempty, empty, empty, nonempty, empty), thus a(3)=1.
For n=3, the only 6-tuple (up to cyclic shift and reversal) which has interlacing signature is (1,1,0,0,1,0). Its signature is (1,2,0).
For n=4, the a(4)=5 equivalence classes of 8-tuples with interlacing signature are (0,1,0,1,0,1,0,1), (0,0,0,1,0,1,1,1), (0,1,0,1,0,0,1,1), (0,1,1,1,0,0,1,0) and (0,0,1,1,0,1,1,0).

Andrew Howroyd, Table of n, a(n) for n = 1..1000
P. Melotti, S. Ramassamy and P. Thévenin, Points and lines configurations for perpendicular bisectors of convex cyclic polygons, arXiv:2003.11006 [math.CO], 2020.

Cf. A028243.

\\ here c(n) is up to rotations only.
c(n)={(n%2==0) + sumdiv(n, d, if(n/d%2==1, eulerphi(n/d)*((3^d - (-1)^d)/2 - 2^d)))/n}
a(n)={(c(n) + if(n%2==0, 3^(n/2-1)))/2} \\ Andrew Howroyd, Dec 25 2021

seq(n)=Vec((x^2/(1-x^2) + x^2/(1-3*x^2))/2 + sum(k=0, (n-1)\2, my(d=2*k+1); eulerphi(d)*log((1+x^d)*(1-2*x^d)^2/(1-3*x^d) + O(x*x^n))/d)/4, -n) \\ Andrew Howroyd, Dec 25 2021

From Andrew Howroyd, Dec 25 2021: (Start)
a(n) = (b(n) + (Sum_{d|n, n/d==1 (mod 2)} phi(n/d)*((3^d - (-1)^d)/2 - 2^d))/n)/2 where b(n) = 1 + 3^(n/2-1) for even n and 0 otherwise.
G.f.: (1/2)*(x^2/(1-x^2) + x^2/(1-3*x^2)) + (1/4)*Sum_{k>=0} phi(2*k+1)*log(B(x^(2*k+1)))/(2*k+1) where B(x) = (1+x)*(1-2*x)^2/(1-3*x).
(End)

Terms a(11) and beyond from Andrew Howroyd, Dec 25 2021

A118979 O.g.f: -12x^3/(-1+x)/(-1+2x)/(-1+3x) = -2-2/(-1+3x)-6/(-1+x)+6/(-1+2*x) .

Original entry on oeis.org

12, 72, 300, 1080, 3612, 11592, 36300, 111960, 342012, 1038312, 3139500, 9467640, 28501212, 85700232, 257493900, 773268120, 2321377212, 6967277352, 20908123500, 62736953400, 188236026012, 564758409672, 1694375892300

Offset: 3

Roger L. Bagula, May 25 2006

Negative of the determinant of a series of 3 X 3 matrices, related to Stirling's numbers of the second kind by a factor of 12 (cf. A000392, A028243).

Index entries for linear recurrences with constant coefficients, signature (6,-11,6).

Cf. A000392, A028243.

M = {{1, 1, 1}, {2^n, 4, 2}, {3^n, 9, 3}} a = Table[ -Det[M], {n, 3, 30}]

Let M = {{1, 1, 1}, {2^n, 4, 2}, {3^n, 9, 3}}. Then a(n) = -Det[M]

a(n) = 6*(1-2^n)+2*3^n = 12*A000392(n).

Edited by N. J. A. Sloane, Dec 13 2007

A134063 a(n) = (1/2)*(3^n - 2^(n+1) + 3).

Original entry on oeis.org

1, 1, 2, 7, 26, 91, 302, 967, 3026, 9331, 28502, 86527, 261626, 788971, 2375102, 7141687, 21457826, 64439011, 193448102, 580606447, 1742343626, 5228079451, 15686335502, 47063200807, 141197991026, 423610750291, 1270865805302, 3812664524767, 11438127792026

Offset: 0

Ross La Haye, Jan 11 2008

Let P(A) be the power set of an n-element set A. Then a(n-1) = the number of pairs of elements {x,y} of P(A) for which either 0) x and y are disjoint and for which x is not a subset of y and y is not a subset of x, or 1) x and y are intersecting and for which either x is a proper subset of y or y is a proper subset of x, or 2) x = y.

The inverse binomial transform yields A033484 with another leading 1. - R. J. Mathar, Jul 06 2009

			a(3) = 7 because for P(A) = {{},{1},{2},{1,2}} we have: case 0 {{1},{2}}, case 1 {{1},{1,2}}, {{2},{1,2}}, case 2 {{},{}}, {{1},{1}}, {{2},{2}}, {{1,2},{1,2}}.

Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6. [_Ross La Haye_, Feb 22 2009]
Index entries for linear recurrences with constant coefficients, signature (6,-11,6)

Cf. A000392, A028243, A000079.

Maple
```
f := n -> (1/2)*(3^n - 2^(n+1) + 3);
```

Table[(3^n-2^(n+1)+3)/2,{n,0,30}] (* or *) LinearRecurrence[{6,-11,6},{1,1,2},30] (* Harvey P. Dale, May 05 2020 *)

a(n) = 3*StirlingS2(n,3) + StirlingS2(n,2) + 1.
a(n) = StirlingS2(n+1,3) + 1. - Ross La Haye, Jan 21 2008
a(n) = 6 a(n-1)-11 a(n-2) +6 a(n-3) (n >= 3). Also a(n) = 4 a(n-1)-3 a(n-2)+ 2^{n-2} (n >= 3). - Tian-Xiao He (the(AT)iwu.edu), Jul 02 2009
G.f.: -(1-4*x+6*x^2)/((x-1)*(3*x-1)*(2*x-1)). a(n+1)-a(n)=A001047(n+1). [R. J. Mathar, Jul 06 2009]

Edited by N. J. A. Sloane, Jul 06 2009

A134064 Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which either 0) x and y are intersecting but for which x is not a subset of y and y is not a subset of x, or 1) x and y are intersecting and for which either x is a proper subset of y or y is a proper subset of x, or 2) x = y.

Original entry on oeis.org

1, 2, 6, 23, 96, 407, 1716, 7163, 29616, 121487, 495276, 2009603, 8124936, 32761367, 131834436, 529712843, 2125993056, 8525430047, 34166159196, 136858084883, 548012945976, 2193794127527, 8780404589556, 35137304693723, 140596281975696, 562526325893807, 2250528914325516

Offset: 0

Ross La Haye, Jan 11 2008

			a(2) = 6 because for P(A) = {{},{1},{2},{1,2}} we have for case 1 {{1},{1,2}}, {{2},{1,2}} and we have for case 2 {{},{}}, {{1},{1}}, {{2},{2}}, {{1,2},{1,2}}. There are 0 {x,y} of P(A) in this example that fall under case 0.

Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
Index entries for linear recurrences with constant coefficients, signature (10,-35,50,-24).

Cf. A032263, A028243, A000079.

LinearRecurrence[{10,-35,50,-24},{1,2,6,23},30] (* Harvey P. Dale, Jul 04 2023 *)

Vec((1-8*x+21*x^2-17*x^3)/((1-x)*(1-2*x)*(1-3*x)*(1-4*x)) + O(x^30)) \\ Michel Marcus, Oct 30 2015

a(n) = (1/2)(4^n - 3^n + 2^n + 1) = 3*StirlingS2(n+1,4) + 2*StirlingS2(n+1,3) + StirlingS2(n+1,2) + 1.
a(n) = C(2^n + 1,2) - (1/2)(3^n - 1) = StirlingS2(2^n + 1,2^n) - StirlingS2(n+1,3) - StirlingS2(n+1,2). - Ross La Haye, Jan 21 2008
G.f.: (1-8*x+21*x^2-17*x^3)/((1-x)*(1-2*x)*(1-3*x)*(1-4*x)). - Colin Barker, Jul 30 2012

cp's OEIS Frontend

A293617 Array of triangles read by ascending antidiagonals, T(m, n, k) = Pochhammer(m, k) * Stirling2(n + m, k + m) with m >= 0, n >= 0 and 0 <= k <= n.

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A298668 Number T(n,k) of set partitions of [n] into k blocks such that the absolute difference between least elements of consecutive blocks is always > 1; triangle T(n,k), n>=0, 0<=k<=ceiling(n/2), read by rows.

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A341617 Repair factors for Stirling numbers of the second kind.

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A091913 Triangle read by rows: a(n,k) = C(n,k)*(2^(n-k) - 1) for k= n, where k=0..max(n-1,0).

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A285867 Triangle T(n, k) read by rows: T(n, k) = S2(n, k)*k! + S2(n, k-1)*(k-1)! with the Stirling2 triangle S2 = A048993.

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A350280 Number of bracelets describing topological configurations of points and lines formed by the perpendicular bisectors of the sides of a convex cyclic n-gon.

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A118979 O.g.f: -12*x^3/(-1+x)/(-1+2*x)/(-1+3*x) = -2-2/(-1+3*x)-6/(-1+x)+6/(-1+2*x) .

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A285867 Triangle T(n, k) read by rows: T(n, k) = S2(n, k)k! + S2(n, k-1)(k-1)! with the Stirling2 triangle S2 = A048993.

A118979 O.g.f: -12x^3/(-1+x)/(-1+2x)/(-1+3x) = -2-2/(-1+3x)-6/(-1+x)+6/(-1+2*x) .