A001792 -id:A001792 - cp's OEIS frontend

A027473 Second column of A027466.

1, 14, 147, 1372, 12005, 100842, 823543, 6588344, 51883209, 403536070, 3107227739, 23727920916, 179936733613, 1356446145698, 10173346092735, 75960984159088, 564959819683217, 4187349251769726, 30939858360298531, 227977903707462860, 1675637592249852021, 12288009009832248154

Offset: 1

Olivier Gérard

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 1..300
Frank Ellermann, Illustration of binomial transforms.
Index entries for linear recurrences with constant coefficients, signature (14,-49).

Cf. A001787, A003415, A053464, A053469.

[n*7^(n-1): n in [1..35]]; // Vincenzo Librandi, Jun 06 2011

Join[{a=1,b=14},Table[c=14*b-49*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)
LinearRecurrence[{14,-49},{1, 14},19] (* Stefano Spezia, May 05 2024 *)

a(n) = n*7^(n-1).
a(n) = 14*a(n-1) - 49*a(n-2) with a(1) = 1, a(2) = 14.
a(n) = A003415(7^n). - Bruno Berselli, Oct 22 2013
From Amiram Eldar, Oct 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 7*log(7/6).
Sum_{n>=1} (-1)^(n+1)/a(n) = 7*log(8/7). (End)
From Stefano Spezia, May 05 2024: (Start)
G.f.: x/(1 - 7*x)^2.
E.g.f.: x*exp(7*x). (End)

More terms from Larry Reeves (larryr(AT)acm.org), May 29 2001

Offset changed from 2 to 1 by Vincenzo Librandi, Jun 06 2011

A050143 A(n,k) = Sum_{h=0..n-1, m=0..k} A(h,m) for n >= 1 and k >= 1, with A(n,0) = 1 for n >= 0 and A(0,k) = 0 for k >= 1; square array A, read by descending antidiagonals.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 4, 7, 1, 0, 1, 5, 12, 15, 1, 0, 1, 6, 18, 32, 31, 1, 0, 1, 7, 25, 56, 80, 63, 1, 0, 1, 8, 33, 88, 160, 192, 127, 1, 0, 1, 9, 42, 129, 280, 432, 448, 255, 1, 0, 1, 10, 52, 180, 450, 832, 1120, 1024, 511, 1

Offset: 1

Clark Kimberling

The triangular version of this square array is defined by T(n,k) = A(k,n-k) for 0 <= k <= n. Conversely, A(n,k) = T(n+k,n) for n,k >= 0. We have [o.g.f of T](x,y) = [o.g.f. of A](x*y, x) and [o.g.f. of A](x,y) = [o.g.f. of T](y,x/y). - Petros Hadjicostas, Feb 11 2021
Formatted as a triangular array with offset (0,8), it is [0, 1, 0, -1, 1, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 1, 1, 0, 0, 0, 0, ...], where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 05 2006
The sum of the first two columns [of the rectangular array] gives the powers of 2; that is, Sum_{j=0..1} A(i,j) = 2^i, i >= 0. On the other hand, for i >= 1 and j >= 2, A(i,j) is the number of lattice paths of i-1 upsteps (1,1) and j-1 downsteps (1,-1) in which each downstep-free vertex is colored red or blue. A downstep-free vertex is one not incident with a downstep. For example, dots indicate the downstep-free vertices in the path .U.U.UDU.UDDU., and with i = j = 2, A(2,2) = 4 counts UD, *UD, DU, DU*, where asterisks indicate the red vertices. - David Callan, Aug 27 2011

			Square array A(n,k) (with rows n >= 0 and columns k >= 0) begins:
  1,   0,   0,    0,    0,    0,    0,     0,     0,     0, ...
  1,   1,   1,    1,    1,    1,    1,     1,     1,     1, ...
  1,   3,   4,    5,    6,    7,    8,     9,    10,    11, ...
  1,   7,  12,   18,   25,   33,   42,    52,    63,    75, ...
  1,  15,  32,   56,   88,  129,  180,   242,   316,   403, ...
  1,  31,  80,  160,  280,  450,  681,   985,  1375,  1865, ...
  1,  63, 192,  432,  832, 1452, 2364,  3653,  5418,  7773, ...
  1, 127, 448, 1120, 2352, 4424, 7700, 12642, 19825, 29953, ...
  ...
If we read the above square array by descending antidiagonals, we get the following triangular array T(n,k) (with rows n >= 0 and columns 0 <= k <= n):
   1;
   0, 1;
   0, 1, 1;
   0, 1, 3,  1;
   0, 1, 4,  7,   1;
   0, 1, 5, 12,  15,   1;
   0, 1, 6, 18,  32,  31,   1;
   0, 1, 7, 25,  56,  80,  63,   1;
   0, 1, 8, 33,  88, 160, 192, 127,   1;
   0, 1, 9, 42, 129, 280, 432, 448, 255, 1;
   ...

Yumin Cho, Jaehyun Kim, Jang Soo Kim, and Nakyung Lee, Enumeration of multiplex juggling card sequences using generalized q-derivatives, arXiv:2402.09903 [math.CO], 2024. See p. 9.
Clark Kimberling, Path-counting and Fibonacci numbers, Fib. Quart. 40(4) (2002), 328-338; see Example 3A and Eq. (8) on p. 333.

Antidiagonal sums are odd-indexed Fibonacci numbers (A001519).
Signed alternating antidiagonal sums are Fibonacci(n)-2, as in A001911.
Cf. A000225, A001792, A050147, A050148, A055807 (mirror array of triangle), A084938.

T[n_, k_] := If[n == k, 1, JacobiP[k - 1, 1, n - 2*k - 1, 3]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Peter Luschny, Nov 25 2021 *)

Formulas for the square array (A(n,k): n,k >= 0):
A(n,1) = -1 + 2^n = A000225(n) for n >= 1.
A(n+2,2) = 4*A001792(n) for n >= 0.
From Petros Hadjicostas, Feb 11 2021: (Start)
Recurrence: A(n,k) = 2*A(n-1,k) + A(n,k-1) - A(n-1,k-1) for n >= 1 and k >= 2; with A(n,0) = 1 for n >= 0, A(0,k) = 0 for k >= 1, and A(n,1) = -1 + 2^n for n >= 1.
Bivariate o.g.f.: Sum_{n,k>=0} A(n,k)*x^n*y^k = (1 - 2*x)*(1 - y)/((1 - x)*(1 - 2*x - y + x*y)).
A(n,k) = Sum_{s=1..n} binomial(n,s)*binomial(s+k-2,k-1) for n >= 0 and k >= 1. (It can be proved by using a partial fraction decomposition on the bivariate o.g.f. above.)
A(n,k) = n*hypergeom([-n + 1, k], [2], -1) for n >= 0 and k >= 1. (End)
Formulas for the triangular array (T(n,k): 0 <= k <= n):
Sum_{k=0..n} T(n,k) = Fibonacci(2*n-1) = A001519(n) with Fibonacci(-1) = 1.
Sum_{k=0..n} (-1)^(n+k-1)*T(n,k) = Fibonacci(n+1) - 2 = A001911(n-2) with A001911(-2) = A001911(-1) = -1.
T(n,k) = A055807(n,n-k) for 0 <= k <= n.
From Petros Hadjicostas, Feb 12 2021: (Start)
Recurrence: T(n,k) = 2*T(n-1,k-1) + T(n-1,k) - T(n-2,k-1) for n >= 3 and 1 <= k <= n-2; with T(n,n) = 1 for n >= 0, T(n,0) = 0 for n >= 1, and T(n+1, n) = 2^n - 1 for n >= 1.
Bivariate o.g.f: Sum_{n,k>=0} T(n,k)*x^n*y^k = (1 - x)*(1 - 2*x*y)/((1 - x*y)*(1 - x - 2*x*y + x^2*y)).
T(n,k) = Sum_{s=1..k} binomial(k,s)*binomial(s+n-k-2, s-1) = k*hypergeom([-k+1, n-k], [2], -1) for n >= 1 and 0 <= k <= n - 1. (End)
T(n, k) = JacobiP(k - 1, 1, n - 2*k - 1, 3) n >= 0 and 0 <= k < n. - Peter Luschny, Nov 25 2021

Various sections edited by Petros Hadjicostas, Feb 12 2021

A053540 a(n) = n*9^(n-1).

Original entry on oeis.org

1, 18, 243, 2916, 32805, 354294, 3720087, 38263752, 387420489, 3874204890, 38354628411, 376572715308, 3671583974253, 35586121596606, 343151886824415, 3294258113514384, 31501343210481297, 300189270593998242, 2851798070642983299, 27017034353459841780

Offset: 1

Barry E. Williams, Jan 15 2000

Vincenzo Librandi, Table of n, a(n) for n = 1..300
Frank Ellermann, Illustration of binomial transforms.
Index entries for linear recurrences with constant coefficients, signature (18,-81).

Related to computing A023052.

Cf. A001787, A053464, A053469.

List([1..20], n-> n*9^(n-1)); # G. C. Greubel, May 16 2019

[n*9^(n-1): n in [1..20]]; // Vincenzo Librandi, Jun 11 2011

f[n_]:=n*9^(n-1); f[Range[40]] (* Vladimir Joseph Stephan Orlovsky, Feb 09 2011*)

a(n)=n*9^(n-1) \\ Charles R Greathouse IV, Oct 07 2015

[n*9^(n-1) for n in (1..20)] # G. C. Greubel, May 16 2019

From Colin Barker, Oct 17 2012: (Start)
a(n) = 18*a(n-1) - 81*a(n-2).
G.f.: x/(1-9*x)^2. (End)
E.g.f.: x*exp(9*x). - G. C. Greubel, May 16 2019
From Amiram Eldar, Oct 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 9*log(9/8).
Sum_{n>=1} (-1)^(n+1)/a(n) = 9*log(10/9). (End)

More terms from Larry Reeves (larryr(AT)acm.org), May 29 2001

Edited by N. J. A. Sloane at the suggestion of Reinhard Zumkeller, Sep 16 2007

A055587 Triangle with columns built from row sums of the partial row sums triangles obtained from Pascal's triangle A007318. Essentially A049600 formatted differently.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 8, 8, 4, 1, 1, 16, 20, 13, 5, 1, 1, 32, 48, 38, 19, 6, 1, 1, 64, 112, 104, 63, 26, 7, 1, 1, 128, 256, 272, 192, 96, 34, 8, 1, 1, 256, 576, 688, 552, 321, 138, 43, 9, 1, 1, 512, 1280, 1696, 1520, 1002, 501, 190, 53, 10, 1, 1, 1024, 2816, 4096

Offset: 0

Wolfdieter Lang, May 30 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is 1/((1-z)*(1-x*z*(1-z)/(1-2*z))).

Column m (without leading zeros) is obtained from convolution of A000012 (powers of 1) with m-fold convoluted A011782.

			{1}; {1, 1}; {1, 2, 1}; {1, 4, 3, 1}; {1, 8, 8, 4, 1}; ...
Fourth row polynomial (n=3): p(3,x)= 1+4*x+3*x^2+x^3

Cf. A049600, column sequences are A000012 (powers of 1), A000079 (powers of 2), A001792, A049611, A049612, A055589, A055852-5 for m=0..9, row sums: A055588.

t[n_, k_] := Hypergeometric2F1[k, k-n, 1, -1]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 05 2014, after Paul D. Hanna *)

{T(n,k) = if( n<0 || k<0, 0, polcoeff( polcoeff( 1 / ((1 - z) * (1 - x*z * (1 - z) / (1 - 2*z) + z * O(z^n) + x * O(x^k))), k), n))}; /* Michael Somos, Sep 30 2003 */

{T(n,k)=if(k>n||n<0||k<0,0,if(k==0||k==n,1, sum(j=0,n-k,binomial(n-k,j)*binomial(k+j-1,k-1)););)} (Hanna)

a(n, m)= Am(n, 0) if n >= m >= 0 and a(n, m) := 0 if nA007318) with the lower triangular matrix A007318 (Pascal triangle) and prs^(m) is the partial row sums (prs) mapping for triangular matrices applied m times. See e.g. A055584 for m=4.
G.f. for column m: (1/(1-x))*(x*(1-x)/(1-2*x))^m, m >= 0.
T(n, k) = sum_{j=0..n-k} C(n-k, j)*C(k+j-1, k-1). - Paul D. Hanna, Jan 14 2004

A078836 a(n) = n*2^(n-6).

Original entry on oeis.org

6, 14, 32, 72, 160, 352, 768, 1664, 3584, 7680, 16384, 34816, 73728, 155648, 327680, 688128, 1441792, 3014656, 6291456, 13107200, 27262976, 56623104, 117440512, 243269632, 503316480, 1040187392, 2147483648, 4429185024, 9126805504, 18790481920, 38654705664

Offset: 6

Silvia Heubach (sheubac(AT)calstatela.edu), Jan 17 2003

a(n) is the number of occurrences of 5s in the palindromic compositions of 2n-1 = the number of occurrences of 6s in the palindromic compositions of 2n.
This sequence is part of a family of sequences, namely R(n,k), the number of ks in palindromic compositions of n. See also A057711, A001792, A079859, A079861 - A079863. General formula: R(n,k)=2^(floor(n/2) - k) * (2 + floor(n/2) - k) if n and k have different parity and R(n,k)=2^(floor(n/2) - k) * (2 + floor(n/2) - k + 2^(floor((k+1)/2 - 1)) otherwise, for n >= 2k.
Also the number of independent vertex sets and vertex covers in the (n-4)-sun graph. - Eric W. Weisstein, Sep 27 2017

			a(6) = 6 since the palindromic compositions of 11 that contain a 5 are 3+5+3, 1+2+5+2+1, 2+1+5+1+2, 1+1+1+5+1+1+1 and 5+1+5, for a total of 6 5s. The palindromic compositions of 12 that contain a 6 are 3+6+3, 1+2+6+2+1, 2+1+6+1+2, 1+1+1+6+1+1+1 and 6+6.

Vincenzo Librandi, Table of n, a(n) for n = 6..3000
Phyllis Chinn, Ralph Grimaldi and Silvia Heubach, The frequency of summands of a particular size in Palindromic Compositions, Ars Combin., Vol. 69 (2003), pp. 65-78.
Eric Weisstein's World of Mathematics, Independent Vertex Set.
Eric Weisstein's World of Mathematics, Sun Graph.
Eric Weisstein's World of Mathematics, Vertex Cover.
Index entries for linear recurrences with constant coefficients, signature (4,-4).

Cf. A057711, A001792, A079859, A079861, A079862, A079863.

[n*2^(n-6): n in [6..40]]; // Vincenzo Librandi, Oct 04 2011

Table[n 2^(n - 6), {m, 6, 50}]
LinearRecurrence[{4, -4}, {6, 14}, 20] (* Eric W. Weisstein, Sep 27 2017 *)
CoefficientList[Series[-2 (-3 + 5 x)/(-1 + 2 x)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Sep 27 2017 *)

a(n)=n<<(n-6) \\ Charles R Greathouse IV, Oct 03 2011

Vec(-2*x^6*(5*x-3)/(2*x-1)^2 + O(x^100)) \\ Colin Barker, Sep 29 2015

def a(n): return n << (n-6)
print([a(n) for n in range(6, 37)]) # Michael S. Branicky, Jun 14 2021

From Colin Barker, Sep 29 2015: (Start)
a(n) = 2*A045891(n-4).
a(n) = 4*a(n-1) - 4*a(n-2) for n > 7.
G.f.: -2*x^6*(5*x-3) / (2*x-1)^2.
(End)
From Amiram Eldar, Jan 12 2021: (Start)
Sum_{n>=6} 1/a(n) = 64*log(2) - 661/15.
Sum_{n>=6} (-1)^n/a(n) = 391/15 - 64*log(3/2). (End)

A080928 Triangle T(n,k) read by rows: T(n,k) = Sum_{i=0..n} C(n,2i)*C(2i,k).

Original entry on oeis.org

1, 1, 0, 2, 2, 1, 4, 6, 3, 0, 8, 16, 12, 4, 1, 16, 40, 40, 20, 5, 0, 32, 96, 120, 80, 30, 6, 1, 64, 224, 336, 280, 140, 42, 7, 0, 128, 512, 896, 896, 560, 224, 56, 8, 1, 256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 0, 512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10

Offset: 0

Paul Barry, Feb 26 2003

Gives the general solution to a(n) = 2*a(n-1) + k(k+2)*a(n-2), a(0) = a(1) = 1. The value k=1 gives the row sums of the triangle, or 1,1,5,13,... This is A046717, the solution to a(n) = 2*a(n-1) + 3*a(n-2), a(0)=a(1)=1.

Product of A007318 and A007318 with every odd-indexed row set to zero. - Paul Barry, Nov 08 2005

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

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identity 156.
J-L. Kim, Relation between weight distribution and combinatorial identities, Bulletin of the Institute of Combinatorics and its Applications, Canada, 31, 2001, pp. 69-79.

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150).
Saul Schleimer, Bert Wiest, On the conjugacy problem in braid groups: Garside theory and subsurfaces, arXiv:1807.01500 [math.GT], 2018.

Apart from k=n, T(n, k) equals (1/2)*A038207(n, k).

Columns include A011782, 2*A001792, A080929, 4*A080930. Row sums are in A046717.

Table[Sum[Binomial[n, 2 i] Binomial[2 i, k], {i, 0, n}], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Oct 11 2018 *)

T(n, n) = (n+1) mod 2, T(n, k) = C(n, k)*2^(n-k-1).
T(n, 0) = A011782(n), T(n, k)=0, k>n, T(2n, 2n)=1, T(2n-1, 2n-1)=0, T(n+1, n)=n+1. Otherwise T(n, k) = T(n-1, k-1) + 2T(n-1, k). Rows are the coefficients of the polynomials in the expansion of (1-x)/((1+k*x)*(1-(k+2)*x)). The main diagonal is 1, 0, 1, 0, 1, 0, ... with g.f. 1/(1-x^2). Subsequent subdiagonals are given by A011782(k)*C(n+k, k) with g.f. A011782(k)/(1-x)^k.
T(n, k) = Sum_{j=0..n} C(n, j)*C(j, k)*(1+(-1)^j)/2; T(n, k) = 2^(n-k-1)*(C(n, k) + (-1)^n*C(0, n-k)). - Paul Barry, Nov 08 2005

Edited by Ralf Stephan, Feb 04 2005

A103406 Triangle read by rows: n-th row = unsigned coefficients of the characteristic polynomials of an n X n matrix with 2's on the diagonal and 1's elsewhere.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 1, 6, 9, 4, 1, 8, 18, 16, 5, 1, 10, 30, 40, 25, 6, 1, 12, 45, 80, 75, 36, 7, 1, 14, 63, 140, 175, 126, 49, 8, 1, 16, 84, 224, 350, 336, 196, 64, 9, 1, 18, 108, 336, 630, 756, 588, 288, 81, 10, 1, 20, 135, 480, 1050, 1512, 1470, 960, 405, 100, 11, 1, 22, 165

Offset: 0

Gary W. Adamson, Feb 04 2005

This triangle * [1/1, 1/2, 1/3, ...] = (1, 2, 4, 8, 16, 32, ...). - Gary W. Adamson, Nov 15 2007

Triangle read by rows: T(n,k) = (k+1)*binomial(n,k), 0 <= k <= n. - Philippe Deléham, Apr 20 2009

			Characteristic polynomial of 3 X 3 matrix [2 1 1 / 1 2 1 / 1 1 2] = x^3 - 6x^2 + 9x - 4.
The first few characteristic polynomials are:
  1
  x - 2
  x^2 - 4x + 3
  x^3 - 6x^2 + 9x - 4
  x^4 - 8x^3 + 18x^2 - 16x + 5

Vincenzo Librandi, Rows n = 0..100, flattened

Row sums = A001792: 1, 3, 8, 20, 48, 112, ...
See A103283 for the mirror image.
Cf. A093375, A127648, A128064.

with(linalg): printf(`%d,`,1): for n from 1 to 15 do mymat:=array(1..n, 1..n): for i from 1 to n do for j from 1 to n do if i=j then mymat[i,j]:=2 else mymat[i,j]:=1 fi: od: od: temp:=charpoly(mymat,x): for j from n to 0 by -1 do printf(`%d,`,abs(coeff(temp, x, j))) od: od: # James Sellers, Apr 22 2005
p := (n,x) -> (x+1)^(n-1)+(x+1)^(n-2)*(n-1);
seq(seq(coeff(p(n,x),x,n-j-1),j=0..n-1),n=1..11); # Peter Luschny, Feb 25 2014

t[n_, k_] := (k+1)*Binomial[n, k]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 09 2012, after Philippe Deléham *)

Binomial transform of A127648. - Gary W. Adamson, Nov 15 2007
Equals A128064 * A007318. - Gary W. Adamson, Jan 03 2008
T(n,k) = (k+1)*A007318(n,k). - Philippe Deléham, Apr 20 2009
T(n,k) = Sum_{i=1..k+1} i*binomial(k+1,i)*binomial(n-k,k+1-i). - Mircea Merca, Apr 11 2012
O.g.f.: (1 - y)/(1 - y - x*y)^2 = 1 + (1 + 2*x)*y + (1 + 4*x + 3*x*2)*y^2 + .... - Peter Bala, Oct 18 2023

More terms from James Sellers, Apr 22 2005

A204235 Permanent of the n-th principal submatrix of A143182.

Original entry on oeis.org

1, 5, 42, 632, 14124, 449652, 19200336, 1063272704, 74068997888, 6344884818304, 655635015988864, 80447681129070080, 11565193558509497088, 1925787312858332101888, 367762470538537620457472, 79847718328265949957881856, 19560087897336150724249288704

Offset: 1

Clark Kimberling, Jan 13 2012

nonn

The n-th principal submatrix of A143182 is an n X n symmetric Toeplitz matrix whose first row consists of successive natural numbers 1, ..., n. - Stefano Spezia, Sep 23 2018

Conjecture: a(1) and a(2) are the only terms that are odd numbers. - Stefano Spezia, Oct 28 2018

Vaclav Kotesovec, Table of n, a(n) for n = 1..35 (terms 1..14 from Clark Kimberling, terms 15..22 from Stefano Spezia)

Cf. A143182.

f:= proc(n) uses LinearAlgebra;
Permanent(ToeplitzMatrix([seq(i, i=1 ..n)], n, symmetric))
end proc:
map(f, [$1..20]); # Stefano Spezia, Oct 28 2018

f[i_, j_] := Max[i - j + 1, j - i + 1];  (* A143182 *)
m[n_] := Table[f[i, j], {i, 1, n}, {j, 1, n}]
TableForm[m[4]] (* 4 X 4 principal submatrix *)
Table[Det[m[n]], {n, 1, 22}]  (* A001792 - signed *)
Permanent[m_] :=
  With[{a = Array[x, Length[m]]},
   Coefficient[Times @@ (m.a), Times @@ a]];
Table[Permanent[m[n]], {n, 1, 14}]  (* A204235 *)
b[i_]:=i; a[n_]:=Permanent[ToeplitzMatrix[Array[b, n], Array[b, n]]]; Array[a, 22] (* Stefano Spezia, Sep 23 2018 *)

{a(n) = matpermanent(matrix(n, n, i, j, max(i - j + 1, j - i + 1)))}
for(n=1, 20, print1(a(n), ", ")) \\ Vaclav Kotesovec, Apr 29 2019

Extended by Stefano Spezia, Oct 28 2018

A208341 Triangle read by rows, T(n,k) = hypergeometric_2F1([n-k+1, -k], [1], -1) for n>=0 and k>=0.

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 1, 4, 8, 8, 1, 5, 13, 20, 16, 1, 6, 19, 38, 48, 32, 1, 7, 26, 63, 104, 112, 64, 1, 8, 34, 96, 192, 272, 256, 128, 1, 9, 43, 138, 321, 552, 688, 576, 256, 1, 10, 53, 190, 501, 1002, 1520, 1696, 1280, 512, 1, 11, 64, 253, 743, 1683, 2972, 4048

Offset: 0

Clark Kimberling, Feb 25 2012

Previous name was: Triangle of coefficients of polynomials v(n,x) jointly generated with A160232; see the Formula section.
Row sums: (1,3,8,...), even-indexed Fibonacci numbers.
Alt. row sums: (1,-1,2,-3,...), signed Fibonacci numbers.
v(n,2) = A107839(n), v(n,n) = 2^(n-1), v(n+1,n) = A001792(n),
v(n+2,n) = A049611, v(n+3,n) = A049612.
Subtriangle of the triangle T(n,k) given by (1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 12 2012
Essentially triangle in A049600. - Philippe Deléham, Mar 23 2012

			First five rows:
  1;
  1, 2;
  1, 3,  4;
  1, 4,  8,  8;
  1, 5, 13, 20, 16;
First five polynomials v(n,x):
  1
  1 + 2x
  1 + 3x +  4x^2
  1 + 4x +  8x^2 +  8x^3
  1 + 5x + 13x^2 + 20x^3 + 16x^4
(1, 0, -1/2, 1/2, 0, 0, ...) DELTA (0, 2, 0, 0, 0, ...) begins:
  1;
  1, 0;
  1, 2,  0;
  1, 3,  4,  0;
  1, 4,  8,  8,  0;
  1, 5, 13, 20, 16,  0;
  1, 6, 19, 38, 48, 32, 0;
Triangle in A049600 begins:
  0;
  0, 1;
  0, 1, 2;
  0, 1, 3,  4;
  0, 1, 4,  8,  8;
  0, 1, 5, 13, 20, 16;
  0, 1, 6, 19, 38, 48, 32;
  ... - _Philippe Deléham_, Mar 23 2012

Reinhard Zumkeller, Rows n = 0..124 of triangle, flattened

Cf. A160232, A000045, A049600, A106195.

a208341 n k = a208341_tabl !! (n-1) !! (k-1)
a208341_row n = a208341_tabl !! (n-1)
a208341_tabl = map reverse a106195_tabl
-- Reinhard Zumkeller, Dec 16 2013

T := (n,k) -> hypergeom([n-k+1, -k],[1],-1):
seq(lprint(seq(simplify(T(n,k)),k=0..n)),n=0..7); # Peter Luschny, May 20 2015

u[1, x_] := 1; v[1, x_] := 1; z = 13;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
v[n_, x_] := u[n - 1, x] + 2*x*v[n - 1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A160232 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A208341 *)

T(n,k) = sum(i = 0, k, 2^(k-i)*binomial(n-k,i)*binomial(k,i));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print();); \\ Michel Marcus, Aug 14 2015

u(n,x) = u(n-1,x) + x*v(n-1,x), v(n,x) = u(n-1,x) + 2x*v(n-1,x), where u(1,x) = 1, v(1,x) = 1.
As DELTA-triangle with 0 <= k <= n: T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1), T(0,0) = T(1,0) = T(2,0) = 1, T(1,1) = T(2,2) = 0, T(2,1) = 2 and T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 12 2012
G.f.: (1-2*y*x+y*x^2)/(1-x-2*y*x+y*x^2). - Philippe Deléham, Mar 12 2012
T(n,k) = A106195(n-1,n-k), k = 1..n. - Reinhard Zumkeller, Dec 16 2013
From Peter Bala, Aug 11 2015: (Start)
The following remarks assume the row and column indexing start at 0.
T(n,k) = Sum_{i = 0..k} 2^(k-i)*binomial(n-k,i)*binomial(k,i) = Sum_{i = 0..k} binomial(n-k+i,i)*binomial(k,i).
Riordan array (1/(1 - x), x*(2 - x)/(1 - x)).
O.g.f. 1/(1 - (2*t + 1)*x + t*x^2) = 1 + (1 + 2*t)*x + (1 + 3*t + 4*t^2)*x^2 + ....
Read as a square array, this equals P * transpose(P^2), where P denotes Pascal's triangle A007318. (End)
For kGlen Whitney, Aug 17 2021

New name from Peter Luschny, May 20 2015

Offset corrected by Joerg Arndt, Aug 12 2015

A221876 T(n,k) is the number of order-preserving full contraction mappings (of an n-chain) with exactly k fixed points.

Original entry on oeis.org

1, 2, 1, 5, 2, 1, 12, 5, 2, 1, 28, 12, 5, 2, 1, 64, 28, 12, 5, 2, 1, 144, 64, 28, 12, 5, 2, 1, 320, 144, 64, 28, 12, 5, 2, 1, 704, 320, 144, 64, 28, 12, 5, 2, 1, 1536, 704, 320, 144, 64, 28, 12, 5, 2, 1, 3328, 1536, 704, 320, 144, 64, 28, 12, 5, 2, 1

Offset: 1

Abdullahi Umar, Feb 28 2013

Row sum is A001792(n-1).
The matrix inverse starts
1;
-2,1;
-1,-2,1;
0,-1,-2,1;
1,0,-1,-2,1;
2,1,0,-1,-2,1;
3,2,1,0,-1,-2,1;
4,3,2,1,0,-1,-2,1;
5,4,3,2,1,0,-1,-2,1;
6,5,4,3,2,1,0,-1,-2,1;
7,6,5,4,3,2,1,0,-1,-2,1; - R. J. Mathar, Apr 12 2013
...
T(n,k) is also the total number of occurrences of parts k in all compositions (ordered partitions) of n, see example. The equivalent sequence for partitions is A066633. Omar E. Pol, Aug 26 2013

			T(5,3) = 5 because there are exactly 5 order-preserving full contraction mappings (of a 5-chain) with exactly 3 fixed points, namely: (12333), (12334), (22344), (23345), (33345).
Triangle begins:
1,
2, 1,
5, 2, 1,
12, 5, 2, 1,
28, 12, 5, 2, 1,
64, 28, 12, 5, 2, 1,
144, 64, 28, 12, 5, 2, 1,
320, 144, 64, 28, 12, 5, 2, 1,
704, 320, 144, 64, 28, 12, 5, 2, 1,
1536, 704, 320, 144, 64, 28, 12, 5, 2, 1,
3328, 1536, 704, 320, 144, 64, 28, 12, 5, 2, 1,
...
Note that column k is column 1 shifted down by k positions.
Row 4 is [12, 5, 2, 1]: in the compositions of 4
[ 1]  [ 1 1 1 1 ]
[ 2]  [ 1 1 2 ]
[ 3]  [ 1 2 1 ]
[ 4]  [ 1 3 ]
[ 5]  [ 2 1 1 ]
[ 6]  [ 2 2 ]
[ 7]  [ 3 1 ]
[ 8]  [ 4 ]
there are 12 parts=1, 5 parts=2, 2 part=3, and 1 part=4.
- _Joerg Arndt_, Sep 01 2013

A. D. Adeshola, V. Maltcev and A. Umar, Combinatorial results for certain semigroups of order-preserving full contraction mappings of a finite chain, arXiv:1303.7428 [math.CO], 2013.
A. D. Adeshola, A. Umar, Combinatorial results for certain semigroups of order-preserving full contraction mappings of a finite chain, JMCC 106 (2017) 37-49

Cf. A001792, A221877, A221878, A221879, A221880, A221881, A221882.

T[n_, n_] = 1; T[n_, k_] := (n - k + 3)*2^(n - k - 2);
Table[T[n, k], {n, 1, 11}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 21 2018 *)

T(n,n) = 1, T(n,k) = (n-k+3)*2^(n-k-2) for n>=2 and n > k > 0.
T(2*n+1,n+1) = T(n+1,1) = A045623(n) for n>=0.
T(n,k) = A045623(n-k), n>=1, 1<=k<=n. - Omar E. Pol, Sep 01 2013

cp's OEIS Frontend

A027473 Second column of A027466.

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