A057711 -id:A057711 - cp's OEIS frontend

A229338 Product of products of elements of subsets of divisors of n.

1, 4, 9, 4096, 25, 2821109907456, 49, 281474976710656, 531441, 10000000000000000, 121, 39939223824273992215667642551956428337968885602521915290518994217942463316460321327052965050967304175616, 169, 2177953337809371136, 6568408355712890625, 1461501637330902918203684832716283019655932542976

Offset: 1

Jaroslav Krizek, Sep 20 2013

nonn

			For n = 4; divisors of 4: {1, 2, 4}; subsets of divisors of n: {}, {1}, {2}, {4}, {1, 2}, {1, 4}, {2, 4}, {1, 2, 4}; product of products of elements of subsets = 1*1*2*4*2*4*8*8 = 4096.

Cf. A229335 (sum of sums of the elements), A229336 (product of sums of the elements), A229337 (sum of products of the elements).

Table[Times@@Times@@@Subsets[Divisors[n]],{n,20}] (* Harvey P. Dale, Mar 05 2015 *)

Conjecture: a(n) = n^s(n); where s(n) = A057711(tau(n)) = A057711(A000005(n)) = tau(n)*2^(tau(n)-2).

A086972 a(n) = n*3^(n-1) + (3^n + 1)/2.

Original entry on oeis.org

1, 3, 11, 41, 149, 527, 1823, 6197, 20777, 68891, 226355, 738113, 2391485, 7705895, 24712007, 78918989, 251105873, 796364339, 2518233179, 7942120025, 24988621541, 78452649023, 245818300271, 768835960421, 2400651060089

Offset: 0

Paul Barry, Jul 26 2003

Binomial transform of A057711 (without leading zero). Second binomial transform of (1,1,3,3,5,5,7,7,9,9,11,11,...).

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (7,-15,9).

Cf. A007051, A027471, A057711, A081038.

Partial sums of A199923.

[n*3^(n-1) + (3^n+1)/2: n in [0..30]]; // Vincenzo Librandi, Jun 09 2011

Table[((2*n+3)*3^(n-1) +1)/2, {n,0,30}] (* G. C. Greubel, Nov 24 2023 *)

Vec((1-4*x+5*x^2)/((1-x)*(1-3*x)^2) + O(x^40)) \\ Michel Marcus, Mar 08 2016

[((2*n+3)*3^(n-1) +1)//2 for n in range(31)] # G. C. Greubel, Nov 24 2023

a(n) = (1/2)*(A081038(n) + 1).
G.f.: (1-4*x+5*x^2)/((1-x)*(1-3*x)^2).
a(n) = A027471(n) + A007051(n).
E.g.f.: (1/2)*( exp(x) + (2*x+1)*exp(3*x) ). - G. C. Greubel, Nov 24 2023

A127952 Triangle read by rows, T(n,k) = (n+1)*binomial(n-1,k-1).

Original entry on oeis.org

1, 0, 2, 0, 3, 3, 0, 4, 8, 4, 0, 5, 15, 15, 5, 0, 6, 24, 36, 24, 6, 0, 7, 35, 70, 70, 35, 7, 0, 8, 48, 120, 160, 120, 48, 8, 0, 9, 63, 189, 315, 315, 189, 63, 9, 0, 10, 80, 280, 560, 700, 560, 280, 80, 10, 0, 11, 99, 396, 924, 1386, 1386, 924, 396, 99, 11

Offset: 0

Gary W. Adamson, Feb 09 2007

Row sums = A057711, starting (1, 2, 6, 16, 40, 96, ...).

T(2n,n) gives A033876(n-1) for n > 0. - Alois P. Heinz, Sep 04 2014

			First few rows of the triangle:
  1;
  0, 2;
  0, 3,  3;
  0, 4,  8,  4;
  0, 5, 15, 15,  5;
  0, 6, 24, 36, 24,  6;
  0, 7, 35, 70, 70, 35,  7;
  ...

G. C. Greubel, Rows n=0..100 of triangle, flattened

Cf. A007318, A126615, A057711.

[[n le 0 select 1 else (n+1)*Binomial(n-1,k-1): k in [0..n]]: n in [0..10]]; // G. C. Greubel, May 05 2018

T := (n,k) -> (n+1)*binomial(n-1, k-1);
seq(print(seq(T(n,k),k= 0..n)),n=0..6); # Peter Luschny, Sep 02 2014

Table[(n+1)*Binomial[n-1, k-1], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, May 05 2018 *)

for(n=0,10, for(k=0,n, print1(if(n==0, 1,(n+1)*binomial(n-1,k-1)), ", "))) \\ G. C. Greubel, May 05 2018

Name corrected after a suggestion of Joerg Arndt by Peter Luschny, Sep 02 2014

A129689 A007318 * A129688.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 7, 5, 3, 1, 15, 12, 8, 4, 1, 31, 27, 20, 12, 5, 1, 63, 58, 47, 32, 17, 6, 1, 127, 121, 105, 79, 49, 23, 7, 1, 255, 248, 226, 184, 128, 72, 30, 8, 1, 511, 503, 474, 410, 312, 200, 102, 38, 9, 1

Offset: 1

Gary W. Adamson, Apr 28 2007

Row sums = A057711: (1, 2, 6, 16, 40, 96, ...). A129690 = A129688 * A007318.

Riordan array ( (1-2*x+2*x^2)/((1-x)*(1-2*x)), x/(1-x) ). - Peter Bala, Mar 21 2018

			First few rows of the triangle are:
   1;
   1,  1;
   3,  2,  1;
   7,  5,  3,  1;
  15, 12,  8,  4,  1;
  31, 27, 20, 12,  5,  1;
  63, 58, 47, 32, 17,  6,  1;
  ...

Muniru A Asiru, Table of n, a(n) for n = 1..5151
Peter Bala, A note on the diagonals of a proper Riordan Array

Cf. A129688, A007318, A057711, A129690.

Flat(List([0..12],n->List([0..n],k->Binomial(n,k)+Sum([2..n],i->2^(i-1)*Binomial(n-i,n-k-i))))); # Muniru A Asiru, Mar 22 2018

C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if;
end proc:
for n from 0 to 12 do
   seq(C(n, n-k) + add(2^(i-1)*C(n-i, n-k-i), i = 2..n), k = 0..n)
end do; #  Peter Bala, Mar 21 2018

T[n_, k_] := Binomial[n, n-k] + Sum[2^(i-1) Binomial[n-i, n-k-i], {i, 2, n}]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 10 2019 *)

# uses[riordan_array from A256893]
# After Peter Bala.
riordan_array((1-2*x+2*x^2)/((1-x)*(1-2*x)), x/(1-x), 8) # Peter Luschny, Mar 21 2018

Binomial transform of A129688. A007318 * A129688 as infinite lower triangular matrices.
From Peter Bala, Mar 21 2018: (Start)
T(n,k) = C(n, n-k) + Sum_{i = 2..n} 2^(i-1)*C(n-i, n-k-i), where C(n,k) = n!/(k!*(n-k)!) for 0 <= k <= n, otherwise 0.
Exp(x) * the e.g.f. for row n = the e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(7 + 5*x + 3*x^2/2! + x^3/3!) = 7 + 12*x + 20*x^2/2! + 32*x^3/3! + 49*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1-x) ).
(End)

A192933 Triangle read by rows: T(n,k) = Sum_{i <= n, j <= k, (i,j) <> (n,k)} T(i,j), starting with T(1,1) = 1, for n >= 1 and 1 <= k <= n.

Original entry on oeis.org

1, 1, 2, 2, 6, 12, 4, 16, 44, 88, 8, 40, 136, 360, 720, 16, 96, 384, 1216, 3152, 6304, 32, 224, 1024, 3712, 11296, 28896, 57792, 64, 512, 2624, 10624, 36416, 108032, 273856, 547712, 128, 1152, 6528, 29056, 109696, 362624, 1056896, 2661504, 5323008, 256, 2560, 15872, 76800, 314880, 1135616, 3659776, 10528768, 26380544, 52761088

Offset: 1

Andrea Raffetti, Jul 13 2011

The outer diagonal is A059435.
The second outer diagonal is A090442.
The third outer diagonal is essentially 2*A068766.
The first column is A011782.
The second column is essentially A057711 (not considering its first two terms).
The second column is essentially A129952 (not considering its first two terms).
The second column is essentially 2*A001792.
The differences between the terms of the second column is essentially 2*A045623.
The third column is essentially 4*A084266.
The cumulative sums of the third column are essentially 4*A176027.
T(n,k) = 0 for n < k. If this overriding constraint is not applied, you get A059576. - Franklin T. Adams-Watters, Jul 24 2011
For n >= 2 and 1 <= k <= n, T(n,k) is the number of bimonotone subdivisions of a 2-row grid with n points on the first row and k points on the second row (with the lower left point of the grid being the origin). A bimonotone subdivision of a convex polygon (the convex hull of the grid) is one where the internal dividing lines have nonnegative (including infinite) slopes. See Robeva and Sun (2020). - Petros Hadjicostas and Michel Marcus, Jul 15 2020

			Triangle (with rows n >= 1 and columns k = 1..n) begins:
   1;
   1,   2;
   2,   6,   12;
   4,  16,   44,    88;
   8,  40,  136,   360,   720;
  16,  96,  384,  1216,  3152,   6304;
  32, 224, 1024,  3712, 11296,  28896,  57792;
  64, 512, 2624, 10624, 36416, 108032, 273856, 547712;
  ...
Example: T(4,3) = 44 = 1 + 1 + 2 + 2 + 6 + 12 + 4 + 16.
From _Petros Hadjicostas_, Jul 15 2020: (Start)
Consider the following 2-row grid with n = 3 points at the top and k = 2 points at the bottom:
   A  B  C
   *--*--*
   |    /
   |   /
   *--*
   D  E
The sets of the dividing internal lines of the T(3,2) = 6 bimonotone subdivisions of the above 2-row grid are as follows: { }, {DC}, {DB}, {EB}, {DB, DC}, and {DB, EB}. We exclude subdivisions {EA} and {EA, EB} because they have at least one dividing line with a negative slope. (End)

Andrea Raffetti, Rows n = 1..13 of triangle
Elina Robeva and Melinda Sun, Bimonotone Subdivisions of Point Configurations in the Plane, arXiv:2007.00877 [math.CO], 2020.

Cf. A001003, A001792, A006318, A011782, A045623, A057711, A059435, A059576, A068766, A084266, A090442, A129952, A166444, A176027, A336245.

lista(nn) = {my(T=matrix(nn, nn)); T[1,1] = 1; for (n=2, nn, for (k=1, n, T[n,k] = sum(i=1, n, sum(j=1, k, if ((i!=n) || (j!=k), T[i,j]))););); vector(nn, k, vector(k, i, T[k, i]));} \\ Michel Marcus, Mar 18 2020

T(n,1) = 2^(n-2) for n >= 2.
T(n,2) = n*2^(n-2) for n >= 2.
T(n,3) = 2^(n-2)*((n-k+1)^2 + 7*(n-k+1) + 4)/2 = 2^(n-3)*(n^2 + 3*n - 6) for k = 3 and n >= 3.
In general: For 1 <= k <= n with (n,k) <> 1,
T(n,k) = 2^(n-2)*Sum_{i=0..k-1} c(k,i)*(n-k+1)^(k-1-i)/(k-1)! and
T(n,k) = 2^(n-2)*Sum_{j=0..k-1} c(k,k-1-j)*(n-k+1)^j/(k-1)!
with c(k,i) being specific coefficients. Below are the first values for c(k,i):
  1;
  1,  1;
  1,  7,    4;
  1, 18,   77,   36;
  1, 34,  359, 1238,   528,
  1, 55, 1065, 8705, 26654, 10800;
... [Formula corrected by Petros Hadjicostas, Jul 15 2020]
The diagonal of this triangle for c(k,i) divided by (k-1)! (except for the first term) is equal to the Shroeder number sequence A006318(k-1).
From Petros Hadjicostas and Michel Marcus, Jul 15 2020: (Start)
T(n,1) = 2^(n-2) for n >= 2; T(n,k) = 2*(T(n,k-1) + T(n-1,k) - T(n-1,k-1)) for n > k >= 2; T(n,n) = 2*T(n,n-1) for n = k >= 2; and T(n,k) = 0 for 1 <= n < k. [Robeva and Sun (2020)] (They do not specify T(1,1) explicitly since they do not care about subdivisions of a degenerate polygon with only one side.)
T(n,k) = (2^(n-2)/(k-1)!) * P_k(n) = (2^(n-2)/(k-1)!) * Sum_{j=1..k} A336245(k,j)*n^(k-j) for n >= k >= 1 with (n,k) <> (1,1), where P_k(n) is some polynomial with integer coefficients of degree k-1. [Robeva and Sun (2020)]
A336245(k,j) = Sum_{s=0..j-1} c(k,s) * binomial(k-1-s, k-j) * (1-k)^(j-1-s) for 1 <= j <= k, in terms of the above coefficients c(k,i).
So c(k,s) = Sum_{j=1..s+1} A336245(k,j) * binomial(k-j, k-s-1) * (k-1)^(s+1-j) for k >= 1 and 0 <= s <= k-1, obtained by inverting the binomial transform.
Bivariate o.g.f.: x*y*(1 - x)*(1 - 2*y*g(2*x*y))/(1 - 2*x - 2*y + 2*x*y), where g(w) = 2/(1 + w + sqrt(1 - 6*w + w^2)) = g.f. of A001003.
Letting y = 1 in the above joint o.g.f., we get the o.g.f. of the row sums: x*(1-x)*(2*g(2*x) - 1). It can then be easily proved that
Sum_{k=1..n} T(n,k) = 2^n*A001003(n-1) - 2^(n-1)*A001003(n-2) for n >= 3. (End)

Offset changed by Andrew Howroyd, Dec 31 2017

Name edited by Petros Hadjicostas, Jul 15 2020

A119462 Triangle read by rows: T(n,k) is the number of circular binary words of length n having k occurrences of 01 (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 2, 2, 2, 2, 6, 2, 12, 2, 2, 20, 10, 2, 30, 30, 2, 2, 42, 70, 14, 2, 56, 140, 56, 2, 2, 72, 252, 168, 18, 2, 90, 420, 420, 90, 2, 2, 110, 660, 924, 330, 22, 2, 132, 990, 1848, 990, 132, 2, 2, 156, 1430, 3432, 2574, 572, 26, 2, 182, 2002, 6006, 6006, 2002, 182, 2, 2, 210, 2730, 10010, 12870, 6006, 910, 30

Offset: 0

Emeric Deutsch, May 21 2006

Row n contains 1 + floor(n/2) terms.
Sum of entries in row n is 2^n (A000079).
2*binomial(n-1,2k) is also the number of permutations avoiding both 123 and 132 with k valleys, i.e., positions with w[i]>w[i+1]Lara Pudwell, Dec 19 2018

			T(3,1) = 6 because we have 001, 010, 011, 100, 101 and 110.
Triangle starts:
  1;
  2;
  2,  2;
  2,  6;
  2, 12,  2;
  2, 20, 10;
  2, 30, 30, 2;
  ...

Muniru A Asiru, Rows n=0..150, flattened
M. Bukata, R. Kulwicki, N. Lewandowski, L. Pudwell, J. Roth, and T. Wheeland, Distributions of Statistics over Pattern-Avoiding Permutations, arXiv preprint arXiv:1812.07112 [math.CO], 2018.
L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.

Cf. A000079, A002378, A034827, A034839, A057711.

Concatenation([1],Flat(List([1..15],n->List([0..Int(n/2)],k->2*Binomial(n,2*k))))); # Muniru A Asiru, Dec 20 2018

T:=proc(n,k) if n=0 and k=0 then 1 else 2*binomial(n,2*k) fi end: for n from 0 to 15 do seq(T(n,k),k=0..floor(n/2)) od; # yields sequence in triangular form

T[0,0]:=1; T[n_,k_]:= 2*Binomial[n,2k]; Table[T[n,k],{n,0,15},{k,0,Floor[n/2]}]//Flatten (* Stefano Spezia, Apr 19 2025 *)

T(n,k) = 2*binomial(n,2k) for n >= 1; T(0,0) = 1.
T(n,k) = 2*T(n-1,k) - T(n-2,k) + T(n-2,k-1) for n >= 3.
G.f.: (1 - z^2 + t*z^2)/(1 - 2*z + z^2 - t*z^2).
T(n,0) = 2 for n >= 1.
T(n,1) = 2*binomial(n,2) = A002378(n-1).
T(n,2) = 2*binomial(n,4) = A034827(n).
T(n,k) = 2*A034839(n-1,k) for n >= 1. [Corrected by Georg Fischer, May 28 2023]
Sum_{k=0..floor(n/2)} k*T(n,k) = A057711(n).

A196389 Triangle T(n,k), read by rows, given by (0,1,-1,0,0,0,0,0,0,0,...) DELTA (1,0,0,1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 0, 2, 1, 0, 0, 0, 3, 1, 0, 0, 0, 0, 4, 1, 0, 0, 0, 0, 0, 5, 1, 0, 0, 0, 0, 0, 0, 6, 1, 0, 0, 0, 0, 0, 0, 0, 7, 1, 0, 0, 0, 0, 0, 0, 0, 0, 8, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 1

Offset: 0

Philippe Deléham, Oct 28 2011

Row sums are A028310; diagonal sums are A057979; column sums are A000027.

			Triangle begins:
  1;
  0, 1;
  0, 1, 1;
  0, 0, 2, 1;
  0, 0, 0, 3, 1;
  0, 0, 0, 0, 4, 1;
  0, 0, 0, 0, 0, 5, 1;
  0, 0, 0, 0, 0, 0, 6, 1;
  0, 0, 0, 0, 0, 0, 0, 7, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 8, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 1; ...

Cf. A084938.

T(n,n)=1, T(n+1,n)=n.
G.f.: (1-x*y+x^2*y)/(1-x*y)^2. - Philippe Deléham, Oct 31 2011
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A028310(n), A057711(n+1), A064017(n+1) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Oct 31 2011

A198792 Triangle T(n,k), read by rows, given by (0,1,1,0,0,0,0,0,0,0,...) DELTA (1,0,0,1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 4, 6, 3, 1, 0, 8, 16, 12, 4, 1, 0, 16, 40, 40, 20, 5, 1, 0, 32, 96, 120, 80, 30, 6, 1, 0, 64, 224, 336, 280, 140, 42, 7, 1, 0, 128, 512, 896, 896, 560, 224, 56, 8, 1, 0, 256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 1

Offset: 0

Philippe Deléham, Oct 30 2011

Row sums are A124302.

Variant of A119468.

			Triangle begins :
1
0, 1
0, 1, 1
0, 2, 2, 1
0, 4, 6, 3, 1
0, 8, 16, 12, 4, 1
0, 16, 40, 40, 20, 5, 1

Columns include A000007, A011782, A057711, A080929, A082138, A080951, A082139, A082140, A082141, A000012, A001477, A002378.

T(n,k) = A097805(n,k)*A011782(n-k).
Sum_{0<=k<=n} T(n,k)*2^k = A063376(n-1).
G.f.: (1-(y+2)*x+y*x^2)/((1-x*y)*(1-x*(y+2))).
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - 2*T(n-2,k-1) - T(n-2,k-2) for n>2, T(0,0) = T(1,1) = T(2,2) = T(2,1) = 1, T(1,0) = T(2,0) = 0, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Nov 10 2013

A303658 Decimal expansion of the alternating sum of the reciprocals of the triangular numbers.

Original entry on oeis.org

7, 7, 2, 5, 8, 8, 7, 2, 2, 2, 3, 9, 7, 8, 1, 2, 3, 7, 6, 6, 8, 9, 2, 8, 4, 8, 5, 8, 3, 2, 7, 0, 6, 2, 7, 2, 3, 0, 2, 0, 0, 0, 5, 3, 7, 4, 4, 1, 0, 2, 1, 0, 1, 6, 4, 8, 2, 7, 2, 0, 0, 3, 7, 9, 7, 3, 5, 7, 4, 4, 8, 7, 8, 7, 8, 7, 7, 8, 8, 6, 2, 4, 2, 3, 4, 5, 3

Offset: 0

Jon E. Schoenfield, Apr 28 2018

			1/1 - 1/3 + 1/6 - 1/10 + 1/15 - 1/21 + ... = 0.77258872223978123766892848583270627230200053744102...

Cf. A000217 (triangular numbers), A057711.

Apart from leading digit the same as A016639 (log(16)).

RealDigits[4*Log[2] - 2, 10, 100][[1]] (* Amiram Eldar, Aug 19 2020 *)
RealDigits[Log[16]-2,10,120][[1]] (* Harvey P. Dale, Apr 30 2022 *)

sumalt(n=1, (-1)^(n+1)*2/(n*(n+1))) \\ Michel Marcus, Apr 28 2018

PARI
```
log(16)-2 \\ Altug Alkan, May 07 2018
```

Equals log(16/e^2) = log(16) - 2.
Equals Sum_{k>=0} 1/((k+2)*2^k) = Sum_{k>=2} 1/A057711(k). - Amiram Eldar, Aug 19 2020
Equals 1 - Sum_{k>=1} 1/(k*(k+1)*(2*k+1)). - Davide Rotondo, May 24 2025

A109572 E.g.f.: x/[1-tan(x)].

Original entry on oeis.org

0, 1, 2, 6, 32, 200, 1536, 13664, 139264, 1595520, 20316160, 284534272, 4347396096, 71958415360, 1282685272064, 24497477345280, 499058556796928, 10802158189445120, 247567232110952448, 5989035387568652288

Offset: 0

Roger L. Bagula, Jun 27 2005

nonn

"Bernoulli numbers" for x/[1-tan(x)].

Cf. A057711.

G:=x/(1-tan(x)): Gser:=series(G,x=0,25): 0,seq(n!*coeff(Gser,x^n),n=1..22);

g[x_] = x/(-1 + Tan[x]) h[x_, n_] = Dt[g[x], {x, n}] a[x_] = Table[ -h[x, n], {n, 0, 50}]; b = a[0]

a(n) ~ n! * 2^(2*n-1)/Pi^n. - Vaclav Kotesovec, Jun 26 2013

cp's OEIS Frontend

A229338 Product of products of elements of subsets of divisors of n.

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A086972 a(n) = n*3^(n-1) + (3^n + 1)/2.

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A127952 Triangle read by rows, T(n,k) = (n+1)*binomial(n-1,k-1).

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A129689 A007318 * A129688.

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A192933 Triangle read by rows: T(n,k) = Sum_{i <= n, j <= k, (i,j) <> (n,k)} T(i,j), starting with T(1,1) = 1, for n >= 1 and 1 <= k <= n.

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A119462 Triangle read by rows: T(n,k) is the number of circular binary words of length n having k occurrences of 01 (0 <= k <= floor(n/2)).

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Formula

A196389 Triangle T(n,k), read by rows, given by (0,1,-1,0,0,0,0,0,0,0,...) DELTA (1,0,0,1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

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