A208741 -id:A208741 - cp's OEIS frontend

A102866 Number of finite languages over a binary alphabet (set of nonempty binary words of total length n).

1, 2, 5, 16, 42, 116, 310, 816, 2121, 5466, 13937, 35248, 88494, 220644, 546778, 1347344, 3302780, 8057344, 19568892, 47329264, 114025786, 273709732, 654765342, 1561257968, 3711373005, 8797021714, 20794198581, 49024480880, 115292809910, 270495295636

Offset: 0

Philippe Flajolet, Mar 01 2005

nonn

Analogous to A034899 (which also enumerates multisets of words)

			a(2) = 5 because the sets are {a,b}, {aa}, {ab}, {ba}, {bb}.
a(3) = 16 because the sets are {a,aa}, {a,ab}, {a,ba}, {a,bb}, {b,aa}, {b,ab}, {b,ba}, {b,bb}, {aaa}, {aab}, {aba}, {abb}, {baa}, {bab}, {bba}, {bbb}.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 64
Stefan Gerhold, Counting finite languages by total word length, INTEGERS 11 (2011), #A44.
Vaclav Kotesovec, A method of finding the asymptotics of q-series based on the convolution of generating functions, arXiv:1509.08708 [math.CO], Sep 30 2015, p. 27.

Cf. A000079, A034899, A256142.
Column k=2 of A292804.
Row sums of A208741 and of A360634.

series(exp(add((-1)^(j-1)/j*(2*z^j)/(1-2*z^j),j=1..40)),z,40);

nn = 20; p = Product[(1 + x^i)^(2^i), {i, 1, nn}]; CoefficientList[Series[p, {x, 0, nn}], x] (* Geoffrey Critzer, Mar 07 2012 *)
CoefficientList[Series[E^Sum[(-1)^(k-1)/k*(2*x^k)/(1-2*x^k), {k,1,30}], {x, 0, 30}], x] (* Vaclav Kotesovec, Sep 13 2014 *)

G.f.: exp(Sum((-1)^(j-1)/j*(2*z^j)/(1-2*z^j), j=1..infinity)).
Asymptotics (Gerhold, 2011): a(n) ~ c * 2^(n-1)*exp(2*sqrt(n)-1/2) / (sqrt(Pi) * n^(3/4)), where c = exp( Sum_{k>=2} (-1)^(k-1)/(k*(2^(k-1)-1)) ) = 0.6602994483152065685... . - Vaclav Kotesovec, Sep 13 2014
Weigh transform of A000079. - Alois P. Heinz, Jun 25 2018

A293815 Number T(n,k) of sets of exactly k nonempty words with a total of n letters over n-ary alphabet such that within each prefix of a word every letter of the alphabet is at least as frequent as the subsequent alphabet letter; triangle T(n,k), n>=0, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 2, 0, 4, 2, 0, 10, 5, 0, 26, 18, 1, 0, 76, 52, 8, 0, 232, 168, 30, 0, 764, 533, 114, 4, 0, 2620, 1792, 411, 22, 0, 9496, 6161, 1462, 116, 0, 35696, 22088, 5237, 482, 6, 0, 140152, 81690, 18998, 1966, 48, 0, 568504, 313224, 70220, 7682, 274

Offset: 0

Alois P. Heinz, Oct 16 2017

The smallest nonzero term in column k is A291057(k).

			T(0,0) = 1: {}.
T(3,1) = 4: {aaa}, {aab}, {aba}, {abc}.
T(3,2) = 2: {a,aa}, {a,ab}.
T(4,2) = 5: {a,aaa}, {a,aab}, {a,aba}, {a,abc}, {aa,ab}.
T(5,3) = 1: {a,aa,ab}.
Triangle T(n,k) begins:
  1;
  0,      1;
  0,      2;
  0,      4,     2;
  0,     10,     5;
  0,     26,    18,     1;
  0,     76,    52,     8;
  0,    232,   168,    30;
  0,    764,   533,   114,    4;
  0,   2620,  1792,   411,   22;
  0,   9496,  6161,  1462,  116;
  0,  35696, 22088,  5237,  482,  6;
  0, 140152, 81690, 18998, 1966, 48;
  ...

Alois P. Heinz, Rows n = 0..300, flattened

Columns k=0-10 give: A000007, A000085 (for n>0), A293964, A293965, A293966, A293967, A293968, A293969, A293970, A293971, A293972.
Row sums give A293114.
Cf. A208741, A293808, A291057, A294129.

g:= proc(n) option remember; `if`(n<2, 1, g(n-1)+(n-1)*g(n-2)) end:
b:= proc(n, i) option remember; expand(`if`(n=0, 1, `if`(i<1, 0,
      add(b(n-i*j, i-1)*binomial(g(i), j)*x^j, j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2)):
seq(T(n), n=0..15);

g[n_] := g[n] = If[n < 2, 1, g[n - 1] + (n - 1)*g[n - 2]];
b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i<1, 0, Sum[b[n - i*j, i-1]* Binomial[g[i], j]*x^j, {j, 0, n/i}]]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][ b[n, n]];
Table[T[n], {n, 0, 15}] // Flatten (* Jean-François Alcover, Jun 04 2018, from Maple *)

G.f.: Product_{j>=1} (1+y*x^j)^A000085(j).

A209406 Triangular array read by rows: T(n,k) is the number of multisets of exactly k nonempty binary words with a total of n letters.

Original entry on oeis.org

2, 4, 3, 8, 8, 4, 16, 26, 12, 5, 32, 64, 44, 16, 6, 64, 164, 132, 62, 20, 7, 128, 384, 376, 200, 80, 24, 8, 256, 904, 1008, 623, 268, 98, 28, 9, 512, 2048, 2632, 1792, 870, 336, 116, 32, 10, 1024, 4624, 6624, 5040, 2632, 1117, 404, 134, 36, 11

Offset: 1

Geoffrey Critzer, Mar 08 2012

Equivalently, T(n,k) is the number of partitions of the integer n with two types of 1's, four types of 2's, ..., 2^i types of i's...; having exactly k summands (of any type).

Row sums = A034899.

			Triangle T(n,k) begins:
    2;
    4,    3;
    8,    8,    4;
   16,   26,   12,    5;
   32,   64,   44,   16,   6;
   64,  164,  132,   62,  20,   7;
  128,  384,  376,  200,  80,  24,   8;
  256,  904, 1008,  623, 268,  98,  28,  9;
  512, 2048, 2632, 1792, 870, 336, 116, 32, 10;
  ...

Alois P. Heinz, Rows n = 1..141, flattened
Index entries for triangles generated by the Multiset Transformation

Cf. A034899, A055375, A208741, A290222, A292506.

T(2n,n) gives A359962.

b:= proc(n, i, p) option remember; `if`(p>n, 0, `if`(n=0, 1,
      `if`(min(i, p)<1, 0, add(b(n-i*j, i-1, p-j)*
       binomial(2^i+j-1, j), j=0..min(n/i, p)))))
    end:
T:= (n, k)-> b(n$2, k):
seq(seq(T(n, k), k=1..n), n=1..14);  # Alois P. Heinz, Apr 13 2017

nn = 10; p[x_, y_] := Product[1/(1 - y x^i)^(2^i), {i, 1, nn}]; f[list_] := Select[lst, # > 0 &]; Map[f, Drop[CoefficientList[Series[p[x, y], {x, 0, nn}], {x, y}], 1]] // Flatten

O.g.f.: Product_{i>=1} 1/(1-y*x^i)^(2^i).

A360634 Number T(n,k) of sets of nonempty words over binary alphabet with a total of n letters of which k are the first letter; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 2, 6, 6, 2, 2, 11, 16, 11, 2, 3, 18, 37, 37, 18, 3, 4, 28, 73, 100, 73, 28, 4, 5, 42, 133, 228, 228, 133, 42, 5, 6, 61, 227, 470, 593, 470, 227, 61, 6, 8, 86, 370, 899, 1370, 1370, 899, 370, 86, 8, 10, 119, 580, 1617, 2894, 3497, 2894, 1617, 580, 119, 10

Offset: 0

Alois P. Heinz, Feb 14 2023

			T(4,0) = 2: {bbbb}, {b,bbb}.
T(4,1) = 11: {abbb}, {babb}, {bbab}, {bbba}, {a,bbb}, {ab,bb}, {abb,b}, {b,bab}, {b,bba}, {ba,bb}, {a,b,bb}.
T(4,2) = 16: {aabb}, {abab}, {abba}, {baab}, {baba}, {bbaa}, {a,abb}, {a,bab}, {a,bba}, {aa,bb}, {aab,b}, {ab,ba}, {aba,b}, {b,baa}, {a,ab,b}, {a,b,ba}.
Triangle T(n,k) begins:
   1;
   1,   1;
   1,   3,   1;
   2,   6,   6,    2;
   2,  11,  16,   11,    2;
   3,  18,  37,   37,   18,    3;
   4,  28,  73,  100,   73,   28,    4;
   5,  42, 133,  228,  228,  133,   42,    5;
   6,  61, 227,  470,  593,  470,  227,   61,   6;
   8,  86, 370,  899, 1370, 1370,  899,  370,  86,   8;
  10, 119, 580, 1617, 2894, 3497, 2894, 1617, 580, 119, 10;
  ...

Alois P. Heinz, Rows n = 0..200, flattened

Columns k=0-2 give: A000009, A095944, A360650.
Row sums give A102866.
T(2n,n) gives A360638.
Cf. A055375 (the same for multisets), A200751, A208741.

g:= proc(n, i, j) option remember; expand(`if`(j=0, 1, `if`(i<0, 0, add(
      g(n, i-1, j-k)*x^(i*k)*binomial(binomial(n, i), k), k=0..j))))
    end:
b:= proc(n, i) option remember; expand(`if`(n=0, 1,
     `if`(i<1, 0, add(b(n-i*j, i-1)*g(i$2, j), j=0..n/i))))
    end:
T:= (n, k)-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2)):
seq(T(n), n=0..15);

g[n_, i_, j_] := g[n, i, j] = Expand[If[j == 0, 1, If[i < 0, 0, Sum[g[n, i - 1, j - k]*x^(i*k)*Binomial[Binomial[n, i], k], {k, 0, j}]]]];
b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i < 1, 0, Sum[b[n - i*j, i - 1]*g[i, i, j], {j, 0, n/i}]]]];
T[n_] := CoefficientList[b[n, n], x];
Table[T[n], {n, 0, 15}] // Flatten (* Jean-François Alcover, Dec 05 2023, after Alois P. Heinz *)

T(n,k) = T(n,n-k).

Sum_{k=0..2n} (-1)^k*T(2n,k) = A200751(n). - Alois P. Heinz, Sep 09 2023

A208742 Number of subsets of the set {1,2,...,n} which do not contain two elements whose difference is 5.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 48, 72, 108, 162, 243, 405, 675, 1125, 1875, 3125, 5000, 8000, 12800, 20480, 32768, 53248, 86528, 140608, 228488, 371293, 599781, 968877, 1565109, 2528253, 4084101, 6612354, 10705716, 17333064, 28063056, 45435424, 73498480, 118894600

Offset: 0

David Nacin, Mar 01 2012

			If n=6 then we must count all subsets not containing both 1 and 6.  There are 2^4 subsets containing 1 and 6, giving us 2^6 - 2^4 = 48.  Thus a(6) = 48.

M. El-Mikkawy, T. Sogabe, A new family of k-Fibonacci numbers, Appl. Math. Comput. 215 (2010) 4456-4461 doi:10.1016/j.amc.2009.12.069, Table 1 k=5.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Tetiva, Subsets that make no difference d, Mathematics Magazine 84 (2011), no. 4, 300-301.
Index entries for linear recurrences with constant coefficients, signature (1, 1, 0, 0, -3, 3, 3, 0, 0, 6, -6, -6, 0, 0, 3, -3, -3, 0, 0, -1, 1, 1).

Cf. A006498, A006500, A208741, A208743.

Table[Fibonacci[Floor[n/5] + 3]^Mod[n, 5] * Fibonacci[Floor[n/5] + 2]^(5 - Mod[n, 5]), {n, 1, 40}]
LinearRecurrence[{1, 1, 0, 0, -3, 3, 3, 0, 0, 6, -6, -6, 0, 0, 3, -3, -3, 0, 0, -1, 1, 1}, {2, 4, 8, 16, 32, 48, 72, 108, 162, 243, 405, 675, 1125, 1875, 3125, 5000, 8000, 12800, 20480, 32768, 53248, 86528, 140608, 228488, 371293, 599781, 968877}, 80]

a(n)=fibonacci(n\5+3)^(n%5)*fibonacci(n\5+2)^(5-n%5) \\ Charles R Greathouse IV, Mar 05 2012

a(n) = F(floor(n/5) + 3)^(n mod 5)*F(floor(n/5) + 2)^(5 - (n mod 5)) where F(n) is the n-th Fibonacci number.
a(n) = a(n-1) + a(n-2) - 3*a(n-5) + 3*a(n-6) + 3*a(n-7) + 6*a(n-10) - 6*a(n-11) - 6*a(n-12) + 3*a(n-15) - 3*a(n-16) - 3*a(n-17) - a(n-20) + a(n-21) + a(n-22).
G.f.: 1-x*(x^21 +2*x^20 +x^19 +x^18 +x^17 -2*x^16 -6*x^15 -4*x^14 -3*x^13 -3*x^12 -9*x^11 -12*x^10 -3*x^9 -6*x^8 -6*x^7 -2*x^6 +6*x^5 +8*x^4 +4*x^3 +2*x^2 +2*x +2) / ((x^2 +x -1) * (x^10 -4*x^5 -1) * (x^10 +x^5 -1)). - Colin Barker, Jun 02 2013

a(0)=1 prepended by Alois P. Heinz, Sep 17 2024

A208743 Number of subsets of the set {1,2,...,n} which do not contain two elements whose difference is 6.

Original entry on oeis.org

2, 4, 8, 16, 32, 64, 96, 144, 216, 324, 486, 729, 1215, 2025, 3375, 5625, 9375, 15625, 25000, 40000, 64000, 102400, 163840, 262144, 425984, 692224, 1124864, 1827904, 2970344, 4826809, 7797153, 12595401, 20346417, 32867289, 53093313, 85766121, 138859434

Offset: 1

David Nacin, Mar 01 2012

			If n=7 then we must count all subsets not containing both 1 and 7.  There are 2^5 subsets containing 1 and 7, giving us 2^7 - 2^5 = 48.  Thus a(7) = 96.

M. El-Mikkawy, T. Sogabe, A new family of k-Fibonacci numbers, Appl. Math. Comput. 215 (2010) 4456-4461 doi:10.1016/j.amc.2009.12.069, Table 1 k=6.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Katharine A. Ahrens, Combinatorial Applications of the k-Fibonacci Numbers: A Cryptographically Motivated Analysis, Ph. D. thesis, North Carolina State University (2020).
M. Tetiva, Subsets that make no difference d, Mathematics Magazine 84 (2011), no. 4, 300-301.
Index entries for linear recurrences with constant coefficients, signature (1, 1, 0, 0, 0, -5, 5, 5, 0, 0, 0, 15, -15, -15, 0, 0, 0, 15, -15, -15, 0, 0, 0, -5, 5, 5, 0, 0, 0, -1, 1, 1).

Cf. A006498, A006500, A208741, A208742.

Table[Fibonacci[Floor[n/6] + 3]^Mod[n, 6] * Fibonacci[Floor[n/6] + 2]^(6 - Mod[n, 6]), {n, 1, 80}]
LinearRecurrence[{1, 1, 0, 0, 0, -5, 5, 5, 0, 0, 0, 15, -15, -15, 0, 0, 0, 15, -15, -15, 0, 0, 0, -5, 5, 5, 0, 0, 0, -1, 1, 1}, {2, 4, 8, 16, 32, 64, 96, 144, 216, 324, 486, 729, 1215, 2025, 3375, 5625, 9375, 15625, 25000, 40000, 64000, 102400, 163840, 262144, 425984, 692224, 1124864, 1827904, 2970344, 4826809, 7797153, 12595401}, 80]

Vec(x*(2 + 2*x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 10*x^6 - 6*x^7 - 14*x^8 - 16*x^9 - 14*x^10 - x^11 - 30*x^12 - 29*x^13 - 15*x^14 - 15*x^15 - 15*x^16 - 20*x^17 - 30*x^18 - 10*x^19 + 5*x^20 + 5*x^21 + 5*x^22 + 4*x^23 + 10*x^24 + 6*x^25 + x^26 + x^27 + x^28 + x^29 + 2*x^30 + x^31) / ((1 + x^2)*(1 - x - x^2)*(1 - x^2 + x^4)*(1 + x^3 - x^6)*(1 - x^3 - x^6)*(1 + 7*x^6 + x^12)) + O(x^30)) \\ Colin Barker, Feb 23 2017

a(n) = F(floor(n/6) + 3)^(n mod 6)*F(floor(n/6) + 2)^(6 - (n mod 6)) where F(n) is the n-th Fibonacci number.
a(n) = a(n-1) + a(n-2) - 5*a(n-6) + 5*a(n-7) + 5*a(n-8) + 15*a(n-12) - 15*a(n-13) - 15*a(n-14) + 15*a(n-18) - 15*a(n-19) - 15*a(n-20) - 5*a(n-24) + 5*a(n-25) + 5*a(n-26) - a(n-30) + a(n-31) + a(n-32).
G.f.: x*(2 + 2*x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 10*x^6 - 6*x^7 - 14*x^8 - 16*x^9 - 14*x^10 - x^11 - 30*x^12 - 29*x^13 - 15*x^14 - 15*x^15 - 15*x^16 - 20*x^17 - 30*x^18 - 10*x^19 + 5*x^20 + 5*x^21 + 5*x^22 + 4*x^23 + 10*x^24 + 6*x^25 + x^26 + x^27 + x^28 + x^29 + 2*x^30 + x^31) / ((1 + x^2)*(1 - x - x^2)*(1 - x^2 + x^4)*(1 + x^3 - x^6)*(1 - x^3 - x^6)*(1 + 7*x^6 + x^12)). - Colin Barker, Feb 23 2017

A209408 Number of subsets of {1,...,n} containing {a,a+4} for some a.

Original entry on oeis.org

0, 0, 0, 0, 0, 8, 28, 74, 175, 377, 799, 1673, 3471, 7192, 14784, 30208, 61440, 124416, 251328, 506712, 1020015, 2051015, 4119775, 8268215, 16582735, 33239558, 66599068, 133392344, 267099120, 534709192, 1070244924, 2141826898, 4285816671, 8575127217

Offset: 0

David Nacin, Mar 08 2012

For n=5, subsets containing {a,a+4} occur only when a=1. There are 2^3 subsets including {1,5}, thus a(5) = 8.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-1,-2,-2,6,-2,-4,2,-6,2,4,1,-3, 1,2).

Cf. A031923, A209409, A209410.

[2^n - Fibonacci(Floor(n/4) + 2)*Fibonacci(Floor((n + 1)/4) + 2)*Fibonacci(Floor((n + 2)/4) + 2)*Fibonacci(Floor((n + 3)/4) + 2): n in [0..30]]; // G. C. Greubel, Jan 03 2018

Table[2^n - Product[Fibonacci[Floor[(n + i)/4] + 2], {i, 0, 3}], {n, 0, 30}]
LinearRecurrence[{3, -1, -2, -2, 6, -2, -4, 2, -6, 2, 4, 1, -3, 1, 2}, {0, 0, 0, 0, 0, 8, 28, 74, 175, 377, 799, 1673, 3471, 7192, 14784}, 30]

for(n=0,20, print1(2^n - fibonacci(floor(n/4) + 2)*fibonacci( floor((n + 1)/4) + 2)*fibonacci(floor((n + 2)/4) + 2)*fibonacci( floor((n + 3)/4) + 2), ", ")) \\ G. C. Greubel, Jan 03 2018

#Returns the actual list of valid subsets
def contains10001(n):
 patterns=list()
 for start in range (1,n-3):
  s=set()
  for i in range(5):
   if (1,0,0,0,1)[i]:
    s.add(start+i)
  patterns.append(s)
 s=list()
 for i in range(2,n+1):
  for temptuple in comb(range(1,n+1),i):
   tempset=set(temptuple)
   for sub in patterns:
    if sub <= tempset:
     s.append(tempset)
     break
 return s
#Counts all such sets
def countcontains10001(n):
 return len(contains10001(n))
#From recurrence
def a(n, adict={0:0, 1:0, 2:0, 3:0, 4:0, 5:8, 6:28, 7:74, 8:175, 9:377, 10:799, 11:1673, 12:3471, 13:7192, 14:14784}):
 if n in adict:
  return adict[n]
 adict[n]=3*a(n-1)-a(n-2)-2*a(n-3)-2*a(n-4)+6*a(n-5)-2*a(n-6)-4*a(n-7)+2*a(n-8)-6*a(n-9)+2*a(n-10)+4*a(n-11)+a(n-12)-3*a(n-13)+a(n-14)+2*a(n-15)
 return adict[n]

a(n) = 2^n - A208741(n-1).
a(n) = 2^n - Product_{i=0..3} Fibonacci(floor((n + i)/4) + 2).
a(n) = 3*a(n-1) - a(n-2) -2*a(n-3) -2*a(n-4) + 6*a(n-5) - 2*a(n-6) - 4*a(n-7) + 2*a(n-8) - 6*a(n-9) + 2*a(n-10) + 4*a(n-11) + a(n-12) - 3*a(n-13) + a(n-14) + 2*a(n-15).
G.f.: x^5*(8 + 4 x - 2 x^2 - 3 x^3 - 2 x^4 - x^5 - x^6 - x^7 - 2 x^8 - x^9) / ((1 - x) (1 + x) (1 - 2 x) (1 + x^2) (1 - x - x^2) (1 + 3 x^4 + x^8)).

A216158 The total number of nonempty words in all length n finite languages on an alphabet of two letters.

Original entry on oeis.org

0, 2, 6, 24, 72, 220, 652, 1848, 5160, 14130, 38102, 101296, 266328, 692740, 1785524, 4563888, 11577888, 29170128, 73032808, 181793136, 450100760, 1108868820, 2719167020, 6639085968, 16144137800, 39107596850, 94393612782, 227062741160, 544439640328, 1301446217244

Offset: 0

Geoffrey Critzer, Sep 03 2012

nonn

A finite language is a set of distinct words with size being the total number of letters in all words.

			a(3) = 24 because the sets (languages) are {a,aa}; {a,ab}; {a,ba}; {a,bb}; {b,aa}; {b,ab}; {b,ba}; {b,bb}; {aaa}; {aab}; {aba}; {abb}; {baa}; {bab}; {bba}; {bbb} where the distinct words are separated by commas.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009, page 64.

Cf. A102866.

h:= proc(n, i) option remember; `if`(n=0, [1, 0], `if`(i<1, 0, add(
      (p-> p+[0, p[1]*j])(binomial(2^i, j)*h(n-i*j, i-1)), j=0..n/i)))
    end:
a:= n-> h(n$2)[2]:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 24 2017

nn=30;p=Product[(1+y x^i)^(2^i),{i,1,nn}];CoefficientList[Series[D[p,y]/.y->1,{x,0,nn}],x]

a(n) = Sum_{k>0} k * A208741(n,k).

cp's OEIS Frontend

A102866 Number of finite languages over a binary alphabet (set of nonempty binary words of total length n).

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A293815 Number T(n,k) of sets of exactly k nonempty words with a total of n letters over n-ary alphabet such that within each prefix of a word every letter of the alphabet is at least as frequent as the subsequent alphabet letter; triangle T(n,k), n>=0, read by rows.

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A209406 Triangular array read by rows: T(n,k) is the number of multisets of exactly k nonempty binary words with a total of n letters.

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A360634 Number T(n,k) of sets of nonempty words over binary alphabet with a total of n letters of which k are the first letter; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A208742 Number of subsets of the set {1,2,...,n} which do not contain two elements whose difference is 5.

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A208743 Number of subsets of the set {1,2,...,n} which do not contain two elements whose difference is 6.

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A209408 Number of subsets of {1,...,n} containing {a,a+4} for some a.

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