A045618 -id:A045618 - cp's OEIS frontend

A055251 Eighth column of triangle A055249.

1, 10, 57, 244, 874, 2772, 8054, 21920, 56751, 141326, 341303, 804276, 1858080, 4223784, 9474444, 21018144, 46195149, 100734354, 218190469, 469866964, 1006759110, 2147634364, 4563581746, 9663887808, 20401343003, 42949963286, 90194651043, 188978952404

Offset: 0

Wolfdieter Lang, May 26 2000

A045618 Partial sums of A000337(n+4),n>=0,
A045889 Partial sums of A045618,
A034009 Partial sums of A045889,
(A055250 Seventh column of triangle A055249) Partial sums of A034009,
(A055251 Eighth column of triangle A055249) Partial sums of A055250. - Vladimir Joseph Stephan Orlovsky, Jul 09 2011

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-43,104,-155,146,-85,28,-4).

Cf. A055249, A035039, partial sums of A055250.

a:= n-> (Matrix(8, (i,j)-> if (i=j-1) then 1 elif j=1 then [10,-43,104,-155, 146,-85,28,-4][i] else 0 fi)^(n))[1,1]: seq(a(n), n=0..25); # Alois P. Heinz, Aug 05 2008

Table[Sum[(-1)^(n - k) k (-1)^(n - k) Binomial[n + 6, k + 6], {k, 0, n}], {n, 1, 26}] (* Zerinvary Lajos, Jul 08 2009 *)

Vec(1 / ((1 - x)^6*(1 - 2*x)^2) + O(x^30)) \\ Colin Barker, Sep 20 2017

G.f.: 1 / (((1-2*x)^2)*(1-x)^6).
a(n) = A055249(n+7, 7).
For n >= 1, a(n) = A035039(n+7) + Sum_{j=0..n-1} a(j).
a(n) = Sum_{k=0..n+6} Sum_{i=0..n+6} (i-k) * C(n-k+6,i+4). - Wesley Ivan Hurt, Sep 19 2017
a(n) = (1/120)*(38520 - 75*2^(9+n) + 2*(9637 + 15*2^(8+n))*n + 4285*n^2 + 525*n^3 + 35*n^4 + n^5). - Colin Barker, Sep 20 2017

A213747 Rectangular array: (row n) = bc, where b(h) = -1 + 2^h, c(h) = b(n-1+h), n>=1, h>=1, and = convolution.

Original entry on oeis.org

1, 6, 3, 23, 16, 7, 72, 57, 36, 15, 201, 170, 125, 76, 31, 522, 459, 366, 261, 156, 63, 1291, 1164, 975, 758, 533, 316, 127, 3084, 2829, 2448, 2007, 1542, 1077, 636, 255, 7181, 6670, 5905, 5016, 4071, 3110, 2165, 1276, 511, 16398, 15375, 13842

Offset: 1

Clark Kimberling, Jun 19 2012

Principal diagonal: A213748.
Antidiagonal sums: A213749.
Row 1,  (1,3,7,15,31,...)**(1,3,7,15,31,...): A045618.
Row 2,  (1,3,7,15,31,...)**(3,7,15,31,...).
Row 3,  (1,3,7,15,31,...)**(7,15,31,...).
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1....6.....23....72.....201
3....16....57....170....459
7....36....125...366....975
15...76....261...758....1007
31...156...533...1542...4071

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A213500.

b[n_] := -1 + 2^n; c[n_] := -1 + 2^n;
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213747 *)
Table[t[n, n], {n, 1, 40}] (* A213748 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A213749 *)

T(n,k) = 6*T(n,k-1)-13*T(n,k-2)+12*T(n,k-3)-4*T(n,k-4).

G.f. for row n: f(x)/g(x), where f(x) = -1 + 2^n - (-2 - 2^n)*x and g(x) = (1 - 3*x + 2*x^2 )^2.

A053219 Reverse of triangle A053218, read by rows.

Original entry on oeis.org

1, 3, 2, 8, 5, 3, 20, 12, 7, 4, 48, 28, 16, 9, 5, 112, 64, 36, 20, 11, 6, 256, 144, 80, 44, 24, 13, 7, 576, 320, 176, 96, 52, 28, 15, 8, 1280, 704, 384, 208, 112, 60, 32, 17, 9, 2816, 1536, 832, 448, 240, 128, 68, 36, 19, 10, 6144, 3328, 1792, 960, 512, 272, 144, 76, 40

Offset: 1

Asher Auel, Jan 01 2000

First element in each row gives A001792. Difference between center element of row 2n-1 and row sum of row n (A053220(n+4) - A053221(n+4)) gives A045618(n).
Subtriangle of triangle in A062111. - Philippe Deléham, Nov 21 2011
Can be seen as the transform of 1, 2, 3, 4, 5, ... by a variant of the boustrophedon algorithm (see the Sage implementation). - Peter Luschny, Oct 30 2014

			Triangle begins:
1
3, 2
8, 5, 3
20, 12, 7, 4
48, 28, 16, 9, 5 ...

Cf. A053218 (reverse of this triangle), A053220 (center elements), A053221 (row sums), A001792, A045618, A062111.

Map[Reverse,NestList[FoldList[Plus,#[[1]]+1,#]&,{1},10]]//Grid (* Geoffrey Critzer, Jun 27 2013 *)

def u():
    for n in PositiveIntegers():
        yield n
def bous_variant(f):
    k = 0
    am = next(f)
    a = [am]
    while True:
        yield list(a)
        am = next(f)
        a.append(am)
        for m in range(k,-1,-1):
            am += a[m]
            a[m] = am
        k += 1
b = bous_variant(u())
[next(b) for  in range(8)] # _Peter Luschny, Oct 30 2014

A058394 A square array based on natural numbers (A000027) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 2, 1, 1, 0, 2, 2, 1, 3, 2, 3, 3, 1, 0, 3, 4, 5, 4, 1, 4, 3, 5, 7, 8, 5, 1, 0, 4, 6, 9, 12, 12, 6, 1, 5, 4, 7, 11, 16, 20, 17, 7, 1, 0, 5, 8, 13, 20, 28, 32, 23, 8, 1, 6, 5, 9, 15, 24, 36, 48, 49, 30, 9, 1, 0, 6, 10, 17, 28, 44, 64, 80, 72, 38, 10, 1, 7, 6, 11, 19, 32, 52, 80, 112, 129

Offset: 0

Henry Bottomley, Nov 24 2000

Changing the formula by replacing T(2n,0)=T(n,2) by T(2n,0)=T(n,m) for some other value of m, would make the generating function change to coefficient of x^n in expansion of (1+x)^k/(1-x^2)^m. This would produce A058393, A058395, A057884 (and effectively A007318).

			Rows are (1,0,2,0,3,0,4,...), (1,1,2,2,3,3,...), (1,2,3,4,5,6,...), (1,3,5,7,9,11,...), etc.

Rows are A027656 (A000027 with zeros), A008619, A000027, A005408, A008574 etc. Columns are A000012, A001477, A022856 etc. Diagonals include A034007, A045891, A045623, A001792, A001787, A000337, A045618, A045889, A034009, A055250, A055251 etc. The triangle A055249 also appears in half of the array.

T(n, k)=T(n-1, k-1)+T(n, k-1) with T(0, k)=1, T(2n, 0)=T(n, 2) and T(2n+1, 0)=0. Coefficient of x^n in expansion of (1+x)^k/(1-x^2)^2.

A086804 a(0)=0; for n > 0, a(n) = (n+1)^(n-2)*2^(n^2).

Original entry on oeis.org

0, 1, 16, 2048, 1638400, 7247757312, 164995463643136, 18446744073709551616, 9803356117276277820358656, 24178516392292583494123520000000, 271732164163901599116133024293512544256

Offset: 0

Yuval Dekel (dekelyuval(AT)hotmail.com), Aug 05 2003

Discriminant of Chebyshev polynomial U_n (x) of second kind.
Chebyshev second kind polynomials are defined by U(0)=0, U(1)=1 and U(n) = 2xU(n-1) - U(n-2) for n > 1.
The absolute value of the discriminant of Pell polynomials is a(n-1).
Pell polynomials are defined by P(0)=0, P(1)=1 and P(n) = 2x P(n-1) + P(n-2) if n > 1. - Rigoberto Florez, Sep 01 2018

Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990; p. 219, 5.1.2.

Rigoberto Flórez, Robinson Higuita, and Alexander Ramírez, The resultant, the discriminant, and the derivative of generalized Fibonacci polynomials, arXiv:1808.01264 [math.NT], 2018.
Rigoberto Flórez, Robinson Higuita, and Antara Mukherjee, Star of David and other patterns in the Hosoya-like polynomials triangles, Journal of Integer Sequences, Vol. 21 (2018), Article 18.4.6.
R. Flórez, N. McAnally, and A. Mukherjees, Identities for the generalized Fibonacci polynomial, Integers, 18B (2018), Paper No. A2.
R. Flórez, R. Higuita and A. Mukherjees, Characterization of the strong divisibility property for generalized Fibonacci polynomials, Integers, 18 (2018), Paper No. A14.
Eric Weisstein's World of Mathematics, Discriminant
Eric Weisstein's World of Mathematics, Chebyshev Polynomial of the Second Kind
Eric Weisstein's World of Mathematics, Pell Polynomial
Index entries for sequences related to Chebyshev polynomials.

Cf. A007701, A127670 (discriminant for S-polynomials), A006645, A001629, A001871, A006645, A007701, A045618, A045925, A093967, A193678, A317404, A317405, A317408, A317451, A318184, A318197.

[0] cat [(n+1)^(n-2)*2^(n^2): n in [1..10]]; // G. C. Greubel, Nov 11 2018

Join[{0},Table[(n+1)^(n-2) 2^n^2,{n,10}]] (* Harvey P. Dale, May 01 2015 *)

PARI
```
a(n)=if(n<1,0,(n+1)^(n-2)*2^(n^2))
```

a(n)=if(n<1,0,n++; poldisc(poltchebi(n)'/n))

a(n) = ((n+1)^(n-2))*2^(n^2), n >= 1, a(0):=0.
a(n) = ((2^(2*(n-1)))*Det(Vn(xn[1],...,xn[n])))^2, n >= 1, with the determinant of the Vandermonde matrix Vn with elements (Vn)i,j:= xn[i]^j, i=1..n, j=0..n-1 and xn[i]:=cos(Pi*i/(n+1)), i=1..n, are the zeros of the Chebyshev U(n,x) polynomials.
a(n) = ((-1)^(n*(n-1)/2))*(2^(n*(n-2)))*Product_{i=1..n}((d/dx)U(n,x)|_{x=xn[i]}), n >= 1, with the zeros xn[i], i=1..n, given above.

Formula and more terms from Vladeta Jovovic, Aug 07 2003

A193605 Triangle: (row n) = partial sums of partial sums of row n of Pascal's triangle.

Original entry on oeis.org

1, 1, 3, 1, 4, 8, 1, 5, 12, 20, 1, 6, 17, 32, 48, 1, 7, 23, 49, 80, 112, 1, 8, 30, 72, 129, 192, 256, 1, 9, 38, 102, 201, 321, 448, 576, 1, 10, 47, 140, 303, 522, 769, 1024, 1280, 1, 11, 57, 187, 443, 825, 1291, 1793, 2304, 2816, 1, 12, 68, 244, 630, 1268, 2116, 3084, 4097, 5120, 6144

Offset: 0

Clark Kimberling, Jul 31 2011

The n-th row is contains the partial sums of the n-th row of the array interpretation of A052509. - R. J. Mathar, Apr 22 2013

			First 5 rows of A193605:
1
1....3
1....4....8
1....5....12....20
1....6....17....32....48

Denis Neiter and Amsha Proag, Links Between Sums Over Paths in Bernoulli's Triangles and the Fibonacci Numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.8.3.

Cf. A193606.

A052509 := proc(n,k)
    if k = 0 then
        1;
    else
        procname(n,k-1)+binomial(n,k) ;
    end if;
end proc:
A193605 := proc(n,k)
    if k = 0 then
        1;
    else
        procname(n,k-1)+A052509(n,k) ;
    end if;
end proc: # R. J. Mathar, Apr 22 2013
# Alternative after Vladimir Kruchinin:
gf := ((x*y-1)/(1-2*x*y))^2/(1-x*y-x): ser := series(gf, x, 12):
p := n -> coeff(ser,x,n): row := n -> seq(coeff(p(n),y,k), k=0..n):
seq(row(n), n=0..10); # Peter Luschny, Aug 19 2019

u[n_, k_] := Sum[Binomial[n, h], {h, 0, k}]
p[n_, k_] := Sum[u[n, h], {h, 0, k}]
Table[p[n, k], {n, 0, 12}, {k, 0, n}]
Flatten[%]   (* A193605 as a sequence *)
TableForm[Table[p[n, k], {n, 0, 12}, {k, 0, n}]]  (* A193605 as a triangle *)

T(n,k):=sum(((i+3)*2^(i-2))*binomial(n-i,k-i),i,1,min(n,k))+binomial(n,k);
/* Vladimir Kruchinin, Aug 20 2019 */

Writing the general term as T(n,k), for 0<=k<=n:
T(n,n)=A001792, T(n,n-1)=A001787, T(n,n-2)=A000337, T(n,n-3)=A045618.
T(n-1,k-1) + T(n-1,k) = T(n,k). - David A. Corneth, Oct 18 2016
G.f.: -(1-x*y)^2/(4*x^3*y^3+(4*x^3-8*x^2)*y^2+(5*x-4*x^2)*y+x-1). - Vladimir Kruchinin, Aug 19 2019
T(n,k) = C(n,k)+Sum_{i=1..n} (i+3)*2^(i-2)*C(n-i,k-i), - Vladimir Kruchinin, Aug 20 2019

More terms from David A. Corneth, Oct 18 2016

A246118 T(n,k), for n,k >= 1, is the number of partitions of the set [n] into k blocks, where, if the blocks are arranged in order of their minimal element, the odd-indexed blocks are all singletons.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 1, 0, 1, 4, 11, 6, 1, 0, 1, 5, 26, 23, 9, 1, 0, 1, 6, 57, 72, 50, 12, 1, 0, 1, 7, 120, 201, 222, 86, 16, 1, 0, 1, 8, 247, 522, 867, 480, 150, 20, 1, 0, 1, 9, 502, 1291, 3123, 2307, 1080, 230, 25, 1, 0, 1, 10, 1013, 3084, 10660, 10044, 6627, 2000, 355, 30, 1

Offset: 1

Peter Bala, Aug 14 2014

Unsigned matrix inverse of A246117. Analog of the Stirling numbers of the second kind, A048993.
This is the triangle of connection constants between the monomial polynomials x^n and the polynomial sequence [x, x^2, x^2*(x - 1), x^2*(x - 1)^2, x^2*(x - 1)^2*(x - 2), x^2*(x - 1)^2*(x - 2)^2, ...]. An example is given below.
Except for differences in offset, this triangle is the Galton array G(floor(k/2),1) in the notation of Neuwirth with inverse array G(-floor(n/2),1).
Essentially the same as A256161. - Peter Bala, Apr 14 2018
From Peter Bala, Feb 10 2020: (Start)
The sums S(n):= Sum_{k >= 0} k^n*(x^k/k!)^2, n = 2,3,4,..., can be expressed as a linear combination of the sums S(0) and S(1) with polynomial coefficients, namely, S(n) = E(n,x)*S(0) + (1/x)*O(n,x)* S(1,x), where E(n,x) = Sum_{k >= 1} T(n,2*k)*x^(2*k) and O(n,x) = Sum_{k >= 0} T(n,2*k+1)*x^(2*k+1) are the even and odd parts of the n-th row polynomial of this array. This result is the analog of the Dobinski formula Sum_{k >= 0} (k^n)*x^k/k! = exp(x)*Bell(n,x), where Bell(n,x) is the n-th row polynomial of A048993.
For example, for n = 6 we have S(6) = Sum_{k >= 1} k^6*(x^k/k!)^2 = (x^2 + 11*x^4 + x^6) * Sum_{k >= 0} (x^k/k!)^2 + (1/x)*(4*x^3 + 6*x^5) * Sum_{k >= 1} k*(x^k/k!)^2.
Setting x = 1 in the above result gives Sum_{k >= 0} k^n*/k!^2 = A000994(n)*Sum_{k >= 0} 1/k!^2 + A000995(n)*Sum_{k >= 1} k/k!^2. See A086880. (End)

			Triangle begins
n\k| 1    2    3    4    5    6    7    8
1  | 1
2  | 0    1
3  | 0    1    1
4  | 0    1    2    1
5  | 0    1    3    4    1
6  | 0    1    4   11    6    1
7  | 0    1    5   26   23    9    1
8  | 0    1    6   57   72   50   12    1
...
Connection constants: Row 6 = (0, 1, 4, 11, 6, 1) so
x^6 = x^2 + 4*x^2*(x - 1) + 11*x^2*(x - 1)^2 + 6*x^2*(x - 1)^2*(x - 2) + x^2*(x - 1)^2*(x - 2)^2.
Row 5 = [0, 1, 3, 4, 1]. There are 9 set partitions of {1,2,3,4,5} of the type described in the Name section:
= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Number of      Set partitions                Count
blocks
= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
2                {1}{2,3,4,5}                   1
3           {1}{2,4,5}{3}, {1}{2,3,5}{4},
            {1}{2,3,4}{5}                       3
4          {1}{2,3}{4}{5}, {1}{2,4}{3}{5},
           {1}{2,5}{3}{4}, {1}{2}{3}{4,5}       4
5          {1}{2}{3}{4}{5}                      1

Yue Cai and Margaret Readdy, Negative q-Stirling numbers, arXiv:1506.03249 [math.CO], 2015.
Emrah Kiliç and Helmut Prodinger, Identities with Squares of Binomial Coefficients: an Elementary and Explicit Approach, Publications de l'Institut Mathématique (Beograd) (N.S.), Vol.99(113) (2016), 243-248. See p. 248.
Erich Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Tech Report TR 99-05, 1999.
E. Neuwirth, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 (2001) 33-51.

Cf. A000295 (column 4), A007476 (row sums), A008277, A045618 (column 5), A048993, A246117 (unsigned matrix inverse), A256161, A000994, A000995, A086880.

Flatten[Table[Table[Sum[StirlingS2[j,Floor[k/2]] * StirlingS2[n-j-1,Floor[(k-1)/2]],{j,0,n-1}],{k,1,n}],{n,1,12}]] (* Vaclav Kotesovec, Feb 09 2015 *)

T(n,k) = Sum_{i = 0..n-1} Stirling2(i, floor(k/2))*Stirling2(n-i-1, floor((k - 1)/2)) for n,k >= 1.
Recurrence equation: T(1,1) = 1, T(n,1) = 0 for n >= 2; T(n,k) = 0 for k > n; otherwise T(n,k) = floor(k/2)*T(n-1,k) + T(n-1,k-1).
O.g.f. (with an extra 1): A(z) = 1 + Sum_{k >= 1} (x*z)^k/( ( Product_{i = 1..floor((k-1)/2)} (1 - i*z) ) * ( Product_{i = 1..floor(k/2)} (1 - i*z) ) ) = 1 + x*z + x^2*z^2 + (x^2 + x^3)*z^3 + (x^2 + 2*x^3 + x^4)*z^4 + .... satisfies A(z) = 1 + x*z + x^2*z^2/(1 - z)*A(z/(1 - z)).
k-th column generating function z^k/( ( Product_{i = 1..floor((k-1)/2)} (1 - i*z) ) * ( Product_{i = 1..floor(k/2)} (1 - i*z) ) ).
Recurrence for row polynomials: R(n,x) = x^2*Sum_{k = 0..n-2} binomial(n-2,k)*R(k,x) with initial conditions R(0,x) = 1 and R(1,x) = x. Compare with the recurrence satisfied by the Bell polynomials: Bell(n,x) = x*Sum_{k = 0..n-1} binomial(n-1,k) * Bell(k,x).
Row sums are A007476.

A383843 Square array A(n,k), n>=0, k>=0, read by antidiagonals downwards, where column k is the expansion of 1/Product_{j=0..k} (1 - j*x)^2.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 6, 3, 0, 1, 12, 23, 4, 0, 1, 20, 86, 72, 5, 0, 1, 30, 230, 480, 201, 6, 0, 1, 42, 505, 2000, 2307, 522, 7, 0, 1, 56, 973, 6300, 14627, 10044, 1291, 8, 0, 1, 72, 1708, 16464, 65002, 95060, 40792, 3084, 9, 0, 1, 90, 2796, 37632, 227542, 587580, 567240, 157440, 7181, 10, 0

Offset: 0

Seiichi Manyama, May 12 2025

			Square array begins:
  1, 1,    1,     1,      1,       1,        1, ...
  0, 2,    6,    12,     20,      30,       42, ...
  0, 3,   23,    86,    230,     505,      973, ...
  0, 4,   72,   480,   2000,    6300,    16464, ...
  0, 5,  201,  2307,  14627,   65002,   227542, ...
  0, 6,  522, 10044,  95060,  587580,  2725380, ...
  0, 7, 1291, 40792, 567240, 4817990, 29331038, ...

Columns k=0..4 give A000007, A000027(n+1), A045618, A383841, A383842.
Main diagonal gives A350376.
A(n,n-1) gives A383880.
Cf. A106800, A287532.

a(n, k) = sum(j=0, n, stirling(j+k, k, 2)*stirling(n-j+k, k, 2));

A(n,k) = Sum_{j=0..n} Stirling2(j+k,k) * Stirling2(n-j+k,k).

A110952 Triangle read by rows: T(n,k) = number of permutations of [n] where the first increasing run has length k and the last increasing run has length n-k-1, 0

Original entry on oeis.org

1, 3, 3, 6, 11, 6, 10, 26, 26, 10, 15, 50, 71, 50, 15, 21, 85, 155, 155, 85, 21, 28, 133, 295, 379, 295, 133, 28, 36, 196, 511, 799, 799, 511, 196, 36, 45, 276, 826, 1519, 1849, 1519, 826, 276, 45, 55, 375, 1266, 2674, 3829, 3829, 2674, 1266, 375, 55, 66, 495, 1860

Offset: 3

David Scambler, Nov 22 2006

Permutations of [n] with exactly 2 descents and the descents are adjacent. Adjusting for initial index: row sums are A045618; first diagonal is A000217, the triangular numbers; 2nd diagonal is A051925; and 3rd diagonal is A001701, generalized Stirling numbers.

			Triangle (beginning with n=3, k=1) is:
   1
   3  3
   6 11  6
  10 26 26 10
  15 50 71 50 15
  ...
For n=5, k = 2: T(5,2) = 11 = permutations of [5] with first run 2 long and last run 5-2-1 = 2 long, namely {14325, 15324, 15423, 24315, 25314, 25413, 34215, 35214, 35412, 45213, 45312}.

Joseph Adams, Table of n, a(n) for n = 3..4755

Cf. A045618, A000217, A051925, A001701, A112858.

T(n,k) = k*C(n,k+1) - C(n,k) + 1.

A143088 Triangle T(n,m) = (2^(m+1) - 1) * (2^(n-m+1) - 1), read by rows, 0 <= m <= n.

Original entry on oeis.org

1, 3, 3, 7, 9, 7, 15, 21, 21, 15, 31, 45, 49, 45, 31, 63, 93, 105, 105, 93, 63, 127, 189, 217, 225, 217, 189, 127, 255, 381, 441, 465, 465, 441, 381, 255, 511, 765, 889, 945, 961, 945, 889, 765, 511, 1023, 1533, 1785, 1905, 1953, 1953, 1905, 1785, 1533, 1023, 2047

Offset: 0

Roger L. Bagula and Gary W. Adamson, Oct 16 2008

Row sums are A045618.

Considered as a square array A(m,n) = (2^m - 1)(2^n - 1), (m, n >= 1), read by rising antidiagonals, this gives the number of m X n matrices of rank 1 over the field F_2. For a different field F_q, that number would be A(m,n) = (q^m - 1)(q^n - 1)/(q - 1). It satisfies the recurrence relation A(m,n) = A(m,n-1)*q + A(m,1). - M. F. Hasler, Sep 12 2024

			1;
3, 3;
7, 9, 7;
15, 21, 21, 15;
31, 45, 49, 45, 31;
63, 93, 105, 105, 93, 63;
127, 189, 217, 225, 217, 189, 127;
255, 381, 441, 465, 465, 441, 381, 255;
511, 765, 889, 945, 961, 945, 889, 765, 511;
1023, 1533, 1785, 1905, 1953, 1953, 1905, 1785, 1533, 1023;
2047, 3069, 3577, 3825, 3937, 3969, 3937, 3825, 3577, 3069, 2047;
...
From _M. F. Hasler_, Sep 12 2024: (Start)
Considered as a square array A(m,n), read by antidiagonals, with m, n >= 1, this represents the following matrix A:
    m \ n: 1  |  2  |  3  |  4  |  5  | ...
  -----+------+-----+-----+-----+-----+-----
    1  |   1  |  3  |   7 |  15 |  31 | ...
    2  |   3  |  9  |  21 |  45 |  93 | ...
    3  |   7  | 21  |  49 | 105 | 217 | ...
    4  |  15  | 45  | 105 | 225 | 465 | ...
   ...
Here each row equals twice the previous row plus the first row, and likewise for columns. See my comment relating this to rank 1 matrices over F_2. (End)

Cf. A000225 (first column), A068156 (second column).

Table[Table[(2^(m + 1) - 1)*(2^(n - m + 1) - 1), {m, 0, n}], {n, 0, 10}]; Flatten[%]

T(n,m) = (2^(m+1) - 1) * (2^(n-m+1) - 1) \\ M. F. Hasler, Sep 12 2024

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

T(n,0) = T(n,n) = 2^(n+1) - 1. - M. F. Hasler, Sep 12 2024

cp's OEIS Frontend

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A213747 Rectangular array: (row n) = bc, where b(h) = -1 + 2^h, c(h) = b(n-1+h), n>=1, h>=1, and = convolution.