A000337 -id:A000337 - cp's OEIS frontend

A127984 a(n) = (n/3 + 7/9)*2^(n - 1) + (-1)^n/9.

1, 3, 7, 17, 39, 89, 199, 441, 967, 2105, 4551, 9785, 20935, 44601, 94663, 200249, 422343, 888377, 1864135, 3903033, 8155591, 17010233, 35418567, 73633337, 152859079, 316902969, 656175559, 1357090361, 2803659207, 5786275385, 11930464711, 24576757305

Offset: 1

Artur Jasinski, Feb 09 2007

a(n) is the number of runs of strictly increasing parts in all compositions of n. a(3) = 7: (1)(1)(1), (12), (2)(1), (3). - Alois P. Heinz, Apr 30 2017
From Hugo Pfoertner, Feb 19 2020: (Start)
a(n)/2^(n-2) apparently is the expected number of flips of a fair coin to completion of a game where the player advances by 1 for heads and by 2 for tails, starting at position 0 and repeating to flip until the target n+1 is exactly reached. If the position n (1 below the target) is reached, the player stays at this position and continues to flip the coin and count the flips until he can advance by 1.
The expected number of flips for targets 1, 2, 3,... , found by inversion of the corresponding Markov matrices, is 2, 2, 3, 7/2, 17/4, 39/8, 89/16, 199/32, 441/64, ...
Target 1 needs an expected number of 2 flips and would require a(0) = 1/2.
  n=1, target n+1 = 2: 1 / 2^(1-2) = 2;
  n=2, target n+1 = 3: 3 / 2^(2-2) = 3;
  n=3, target n+1 = 4: 7 / 2^(3-2) = 7/2.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
W. Bosma, Signed bits and fast exponentiation, J. Th. des Nombres de Bordeaux Vol.13, Fasc. 1, 2001.
Aruna Gabhe, Problem 11623, Am. Math. Monthly 119 (2012) 161.
Index entries for linear recurrences with constant coefficients, signature (3,0,-4).

Cf. A059570, A073371, A127976, A127978, A127979, A127980, A127981, A127982, A127983, A073371, A000337, A172481.

[(n/3+7/9)*2^(n-1)+(-1)^n/9: n in [1..35]]; // Vincenzo Librandi, Jun 15 2017

A127984:=n->(n/3 + 7/9)*2^(n - 1) + (-1)^n/9; seq(A127984(n), n=1..50); # Wesley Ivan Hurt, Mar 14 2014

Table[(n/3 + 7/9)2^(n - 1) + (-1)^n/9, {n, 50}] (* Artur Jasinski *)
CoefficientList[Series[(1 - 2 x^2) / ((-1 + 2 x)^2 (1 + x)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 15 2017 *)

a(n) = (n/3 + 7/9)*2^(n - 1) + (-1)^n/9.
From R. J. Mathar, Apr 04 2008: (Start)
O.g.f.: -x*(-1+2x^2)/((-1+2x)^2*(1+x)).
a(n) = 3*a(n-1) - 4*a(n-3). (End)
a(n) + a(n+1) = A087447(n+1). - R. J. Mathar, Feb 21 2009
A172481(n) = a(n) + 2^(n-1). Application: Problem 11623, AMM 119 (2012) 161. - Stephen J. Herschkorn, Feb 11 2012
From Wolfdieter Lang, Jun 14 2017: (Start)
a(n) = f(n+1)*2^(n-1), where f(n) is a rational Fibonacci type sequence based on fuse(a,b) = (a+b+1)/2 with f(0) = 0, f(1) = 1 and f(n) = fuse(f(n-1),f(n-2)), for n >= 2. For fuse(a,b) see the Jeff Erickson link under A188545. Proof: f(n) = (3*n+4 - (-1)^n/2^(n-2))/9, n >= 0, by induction.
a(n) = a(n-1) + a(n-2) + 2^(n-2), n >= 1, with inputs a(-1) = 0, a(0) = 1/2.
(End)
E.g.f.: (2*exp(-x) + exp(2*x)*(7 + 6*x) - 9)/18. - Stefano Spezia, Feb 19 2020

A193844 Triangular array: the fission of ((x+1)^n) by ((x+1)^n); i.e., the self-fission of Pascal's triangle.

Original entry on oeis.org

1, 1, 3, 1, 5, 7, 1, 7, 17, 15, 1, 9, 31, 49, 31, 1, 11, 49, 111, 129, 63, 1, 13, 71, 209, 351, 321, 127, 1, 15, 97, 351, 769, 1023, 769, 255, 1, 17, 127, 545, 1471, 2561, 2815, 1793, 511, 1, 19, 161, 799, 2561, 5503, 7937, 7423, 4097, 1023

Offset: 0

Clark Kimberling, Aug 07 2011

See A193842 for the definition of fission of two sequences of polynomials or triangular arrays.
A193844 is also the fission of (p1(n,x)) by (q1(n,x)), where p1(n,x)=x^n+x^(n-1)+...+x+1 and q1(n,x)=(x+2)^n.
Essentially A119258 but without the main diagonal. - Peter Bala, Jul 16 2013
From Robert Coquereaux, Oct 02 2014: (Start)
This is also a rectangular array A(n,p) read down the antidiagonals:
1 1 1 1 1 1 1 1 1
3 5 7 9 11 13 15 17 19
7 17 31 49 71 97 127 161 199
15 49 111 209 351 545 799 1121 1519
31 129 351 769 1471 2561 4159 6401 9439
...
Calling Gr(n) the Grassmann algebra with n generators, A(n,p) is the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence. If p is odd A(n,p) is the dimension of the cyclic cohomology group of order p of the Z2 graded algebra Gr(n). If p is even, the dimension of this cohomology group is A(n,p)+1. A(n,p) = 2^n*A059260(p,n-1)-(-1)^p.
(End)
The n-th row are also the coefficients of the polynomial P=sum_{k=0..n} (X+2)^k (in falling order, i.e., that of X^n first). - M. F. Hasler, Oct 15 2014

			First six rows:
1
1....3
1....5....7
1....7....17....15
1....9....31....49....31
1....11...49....111...129...63

Jean-François Chamayou, A Random Difference Equation with Dufresne Variables revisited, arXiv:1410.1708 [math.PR], 2014.
R. Coquereaux and E. Ragoucy, Currents on Grassmann algebras, J. of Geometry and Physics, 1995, Vol 15, pp 333-352.
R. Coquereaux and E. Ragoucy, Currents on Grassmann algebras, arXiv:hep-th/9310147, 1993.
C. Kassel, A Künneth formula for the cyclic cohomology of Z2-graded algebras, Math. Ann. 275 (1986) 683.

Cf. A193842, A193845, A119258.
Columns, diagonals: A000225, A000337, A055580, A027608, A211386, A211388, A000012, A005408, A056220, A199899.
A145661 is an essentially identical triangle.

A193844 := (n,k) -> 2^k*binomial(n+1,k)*hypergeom([1,-k],[-k+n+2],1/2);
for n from 0 to 5 do seq(round(evalf(A193844(n,k))),k=0..n) od; # Peter Luschny, Jul 23 2014
# Alternatively
p := (n,x) -> add(x^k*(1+2*x)^(n-k), k=0..n): for n from 0 to 7 do [n], PolynomialTools:-CoefficientList(p(n,x), x) od; # Peter Luschny, Jun 18 2017

z = 10;
p[n_, x_] := (x + 1)^n;
q[n_, x_] := (x + 1)^n
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A193844 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]]  (* A193845 *)

# uses[fission from A193842]
p = lambda n,x: (x+1)^n
A193844_row = lambda n: fission(p, p, n)
for n in range(7): print(A193844_row(n)) # Peter Luschny, Jul 23 2014

From Peter Bala, Jul 16 2013: (Start)
T(n,k) = sum {i = 0..k} (-1)^i*binomial(n+1,k-i)*2^(k-i).
O.g.f.: 1/( (1 - x*t)*(1 - (2*x + 1)*t) ) = 1 + (1 + 3*x)*t + (1 + 5*x + 7*x^2)*t^2 + ....
The n-th row polynomial R(n,x) = 1/(x+1)*( (2*x+1)^(n+1) - x^(n+1) ). (End)
T(n,k) = T(n-1,k) + 3*T(n-1,k-1) - T(n-2,k-1) - 2*T(n-2,k-2), T(0,0) = 1, T(1,0) = 1, T(1,1) = 3, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 17 2014
T(n,k) = 2^k*binomial(n+1,k)*hyper2F1(1,-k,-k+n+2, 1/2). - Peter Luschny, Jul 23 2014

A066185 Sum of the first moments of all partitions of n with weights starting at 0.

Original entry on oeis.org

0, 0, 1, 4, 12, 26, 57, 103, 191, 320, 537, 843, 1342, 2015, 3048, 4457, 6509, 9250, 13170, 18316, 25483, 34853, 47556, 64017, 86063, 114285, 151462, 198871, 260426, 338275, 438437, 564131, 724202, 924108, 1176201, 1489237, 1881273, 2365079, 2966620, 3705799

Offset: 0

Wouter Meeussen, Dec 15 2001

The first element of each partition is given weight 0.
Consider the partitions of n, e.g., n=5. For each partition sum T(e-1) and sum all these. E.g., 5 -> T(4)=10, 41 -> T(3)+T(0)=6, 32 -> T(2)+T(1)=4, 311 -> T(2)+T(0)+T(0)=3, 221 -> T(1)+T(1)+T(0)=2, 21111 ->1 and 11111 ->0. Summing, 10+6+4+3+2+1+0 = 26 as desired. - Jon Perry, Dec 12 2003
Also equals the sum of f(p) over the partitions p of n, where f(p) is obtained by replacing each part p_i of partition p by p_i*(p_i-1)/2. See I. G. Macdonald: Symmetric functions and Hall polynomials 2nd edition, p. 3, eqn (1.5) and (1.6). - Wouter Meeussen, Sep 25 2014

			a(3)=4 because the first moments of all partitions of 3 are {3}.{0},{2,1}.{0,1} and {1,1,1}.{0,1,2}, resulting in 0,1,3; summing to 4.

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A000337, A001788, A066183, A066184, A066186, A161680, A264034.

b:= proc(n, i) option remember; `if`(n=0 or i=1, [1, 0],
      b(n, i-1)+(h-> h+[0, h[1]*i*(i-1)/2])(b(n-i, min(n-i, i))))
    end:
a:= n-> b(n$2)[2]:
seq(a(n), n=0..50);  # Alois P. Heinz, Jan 29 2014

Table[ Plus@@ Map[ #.Range[ 0, -1+Length[ # ] ]&, IntegerPartitions[ n ] ], {n, 40} ]
b[n_, i_] := b[n, i] = If[n==0, {1, 0}, If[i<1, {0, 0}, If[i>n, b[n, i-1], b[n, i-1] + Function[h, h+{0, h[[1]]*i*(i-1)/2}][b[n-i, i]]]]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Oct 26 2015, after Alois P. Heinz *)

a(n) = 1/2*(A066183(n) - A066186(n)). - Vladeta Jovovic, Mar 23 2003
G.f.: Sum_{k>=1} x^(2*k)/(1 - x^k)^3 / Product_{j>=1} (1 - x^j). - Ilya Gutkovskiy, Mar 05 2021
a(n) = Sum_{k=0..A161680(n)} k * A264034(n,k). - Alois P. Heinz, Jan 20 2023
a(n) ~ 3 * zeta(3) * sqrt(n) * exp(Pi*sqrt(2*n/3)) / (sqrt(2) * Pi^3). - Vaclav Kotesovec, Jul 06 2025

A102301 a(n) = ((3n + 1)2^(n+3) + 9 + (-1)^n)/18.

Original entry on oeis.org

1, 4, 13, 36, 93, 228, 541, 1252, 2845, 6372, 14109, 30948, 67357, 145636, 313117, 669924, 1427229, 3029220, 6407965, 13514980, 28428061, 59652324, 124897053, 260978916, 544327453, 1133394148, 2356266781, 4891490532, 10140895005, 20997617892, 43426891549

Offset: 0

Creighton Dement, Feb 20 2005

A floretion-generated sequence resulting from particular transform of A000975.

Floretion Algebra Multiplication Program, FAMP Code: 2jesforseq[ + .5'i + 'kk' + .5'jk' ], 1vesforseq(n) = A000975(n+2)*(-1)^(n+1), ForType: 1A, LoopType: tes (2nd iteration)

G. C. Greubel, Table of n, a(n) for n = 0..1000
T. Etzion, On the stopping redundancy of Reed-Muller codes, IEEE Trans. Information Theory 52 (11) (2006) 4867-4879, also, arXiv:cs/0511056 [cs.IT], 2005.
Toufik Mansour and Armend Sh. Shabani, Bargraphs in bargraphs, Turkish Journal of Mathematics (2018) Vol. 42, Issue 5, 2763-2773.
Index entries for linear recurrences with constant coefficients, signature (4,-3,-4,4).

Cf. A000975, A000337, A045883, A001787, A059570, A137266, A008574, A059260.

[((3*n+1)*2^(n+3)+9+(-1)^n)/18: n in [0..40]]; // Vincenzo Librandi, Nov 21 2018

Table[((3n+1)*2^(n+3) + 9 + (-1)^n)/18, {n,0,50}] (* G. C. Greubel, Sep 27 2017 *)
LinearRecurrence[{4, -3, -4, 4}, {1, 4, 13, 36}, 50] (* Vincenzo Librandi, Nov 21 2018 *)

a(n)=((3*n+1)*2^(n+3)+9+(-1)^n)/18 \\ Charles R Greathouse IV, Oct 16 2015

G.f.: 1/((1-x^2)*(1-2*x)^2).
a(n+1) - 2*a(n) = A000975(n+2) (n-th number without consecutive equal binary digits)
a(n) + a(n+1) = A000337(n+2);
a(n+1) - a(n) = A045883(n+2);
a(n+2) - a(n) = A001787(n+3) ( Number of edges in n-dimensional hypercube );
a(n+2) - 2*a(n+1) + a(n) = A059570(n+3);
Convolution of "Number of fixed points in all 231-avoiding involutions in S_n" (A059570) with the natural numbers (A000027), treating the result as if offset=0. - Graeme McRae, Jul 12 2006
Equals triangle A059260 * A008574 as a vector, where A008574 = [1, 4, 8, 12, 16, 20, ...]. - Gary W. Adamson, Mar 06 2012

A130328 Triangle of differences between powers of 2, read by rows.

Original entry on oeis.org

1, 3, 2, 7, 6, 4, 15, 14, 12, 8, 31, 30, 28, 24, 16, 63, 62, 60, 56, 48, 32, 127, 126, 124, 120, 112, 96, 64, 255, 254, 252, 248, 240, 224, 192, 128, 511, 510, 508, 504, 496, 480, 448, 384, 256

Offset: 0

Gary W. Adamson, May 24 2007

A130321 * A059268 as infinite lower triangular matrices.
Row sums = A000337: (1, 5, 17, 49, 129, 321, ...). A130329 = A059268 * A130321.
From Alonso del Arte, Mar 13 2008: (Start)
Column 0 contains the Mersenne numbers A000225.
Column 1 is A000918.
An even perfect number (A000396) is found in the triangle by reference to its matching exponent for the Mersenne prime p (A000043) thus: go to row 2p - 1 and then column p - 1 (remembering that the first position is column 0).
Likewise divisors of multiply perfect numbers, if not the multiply perfect numbers themselves, can also be found in this triangle. (End)

			First few rows of the triangle are;
   1;
   3,  2;
   7,  6,  4;
  15, 14, 12,  8;
  31, 30, 28, 24, 16;
  63, 62, 60, 56, 48, 32;
  ...
a(5, 2) = 28 because 2^5 = 32, 2^2 = 4 and 32 - 4 = 28.

Cf. A130321, A059268, A000337, A130329.

ColumnForm[Table[2^n - 2^k, {n, 15}, {k, 0, n - 1}], Center] (* Alonso del Arte, Mar 13 2008 *)

t(n, k) = 2^n - 2^k, where n is the row number and k is the column number, running from 0 to n - 1. (If k is allowed to reach n, then the triangle would have an extra diagonal filled with zeros) - Alonso del Arte, Mar 13 2008

Better definition from Alonso del Arte, Mar 13 2008

A055251 Eighth column of triangle A055249.

Original entry on oeis.org

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

A059045 Square array T(n,k) read by antidiagonals where T(0,k) = 0 and T(n,k) = 1 + 2k + 3k^2 + ... + n*k^(n-1).

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 5, 1, 0, 1, 10, 17, 7, 1, 0, 1, 15, 49, 34, 9, 1, 0, 1, 21, 129, 142, 57, 11, 1, 0, 1, 28, 321, 547, 313, 86, 13, 1, 0, 1, 36, 769, 2005, 1593, 586, 121, 15, 1, 0, 1, 45, 1793, 7108, 7737, 3711, 985, 162, 17, 1, 0, 1, 55, 4097, 24604, 36409

Offset: 0

Henry Bottomley, Dec 18 2000

			   0,   0,   0,    0,     0,      0,      0,      0,       0, ...
   1,   1,   1,    1,     1,      1,      1,      1,       1, ...
   1,   3,   5,    7,     9,     11,     13,     15,      17, ...
   1,   6,  17,   34,    57,     86,    121,    162,     209, ...
   1,  10,  49,  142,   313,    586,    985,   1534,    2257, ...
   1,  15, 129,  547,  1593,   3711,   7465,  13539,   22737, ...
   1,  21, 321, 2005,  7737,  22461,  54121, 114381,  219345, ...
   1,  28, 769, 7108, 36409, 131836, 380713, 937924, 2054353, ...

Cf. A000004, A000012, A005408, A056109, A056578, A056579 (rows 1-6).

Cf. A057427, A000217, A000337, A014915, A014916, A014917, A014918, A014920 (columns 1-7).

A059045 := proc(n,k)
    if k = 1 then
        n*(n+1) /2 ;
    else
        (1+n*k^(n+1)-k^n*(n+1))/(k-1)^2 ;
    end if;
end proc: # R. J. Mathar, Mar 29 2013

T(n,k) = n*k^(n-1)+T(n-1, k) = (n*k^(n+1)-(n+1)*k^n+1)/(k-1)^2.

A127717 Triangle read by rows. T(n, k) = k * binomial(n + 1, k + 1), for 1 <= k <= n.

Original entry on oeis.org

1, 3, 2, 6, 8, 3, 10, 20, 15, 4, 15, 40, 45, 24, 5, 21, 70, 105, 84, 35, 6, 28, 112, 210, 224, 140, 48, 7, 36, 168, 378, 504, 420, 216, 63, 8, 45, 240, 630, 1008, 1050, 720, 315, 80, 9, 55, 330, 990, 1848, 2310, 1980, 1155, 440, 99, 10

Offset: 1

Gary W. Adamson, Jan 25 2007

T(n,k) is the sum of the greatest element in each size k subset of {1,2,...,n}. - Geoffrey Critzer, Oct 17 2009
Reversed unsigned rows of A055137 with the diagonal and first subdiagonal removed. - Tom Copeland, Nov 04 2012
Consider the transformation 1 + 2x + 3x^2 + 4x^3 + ... + (n+1)*x^n = A_0*(x-1)^0 + A_1*(x-1)^1 + A_2*(x-1)^2 + ... + A_n*(x-1)^n. This sequence gives A_0, ..., A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 30 2014

			First few rows of the triangle:
    [1    2    3     4     5    6    7     8    9]
[1]  1;
[2]  3,   2;
[3]  6,   8,   3;
[4] 10,  20,  15,    4;
[5] 15,  40,  45,   24,    5;
[6] 21,  70, 105,   84,   35,    6;
[7] 28, 112, 210,  224,  140,   48,    7;
[8] 36, 168, 378,  504,  420,  216,   63,   8;
[9] 45, 240, 630, 1008, 1050,  720,  315,  80,  9;
  ...
T(4, 3) = 15 because the size 3 subsets of {1, 2, 3, 4} are {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}. Adding the largest element from each subset we get 3 + 4 + 4 + 4 = 15. - _Geoffrey Critzer_, Oct 17 2009

Cyann Donnot, Antoine Genitrini, Yassine Herida, Unranking Combinations Lexicographically: an efficient new strategy compared with others, hal-02462764 [cs] / [cs.DS] / [math] / [math.CO], 2020.
Antoine Genitrini and Martin Pépin, Lexicographic unranking of combinations revisited, hal-03040740v2 [cs.DM] [cs.DS] [math.CO], 2020.

Cf. A002260, A007318, A000337, A127718.

# Assuming (1,1)-based triangle:
T := (n, k) -> k*binomial(n+1, k+1):
seq(seq(T(n, k), k = 1..n), n = 1..9);
# Assuming (0,0)-based triangle:
gf := 1/((1 - x)*(1 - x - x*y)^2): ser := series(gf, x, 11):
seq(seq(coeff(coeff(ser, x, n), y, k), k=0..n), n=0..9); # Peter Luschny, Jan 07 2023

Table[Table[Sum[Binomial[i - 1, k - 1]*i, {i, k, n}], {k, 1, n}], {n, 1, 10}] // Grid (* Geoffrey Critzer, Oct 17 2009 *)

T(n,k) = k*sum(i=0,n-k,binomial(i+k,k))
for(n=1,15,for(k=1,n,print1(T(n,k),", "))) \\ Derek Orr, Oct 30 2014

A002260 * A007318 (Pascal's Triangle), where A002260 = the matrix [1; 1,2; 1,2,3,...].
T(n,k) = Sum_{i=k..n} binomial(i-1, k-1)*i. - Geoffrey Critzer, Oct 17 2009
Row sums = A000337: (1, 5, 17, 49, 129, ...) A007318 * A002260 = A127718.
From Geoffrey Critzer, Oct 18 2009: (Start)
T(n,k) = k*binomial(n+1, k+1).
Recurrence for column k: a(n) = a(n-1) + n*binomial(n-1, k-1) = a(n-1) + k*binomial(n, k).
O.g.f. for column k: k*x^k/(1-x)^(k+2). (End)
T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n+2-k, k+2-i). - Mircea Merca, Apr 11 2012
G.f.: 1/((1 - x)*(1 - x - x*y)^2), assuming the triangle (0,0)-based. - Vladimir Kruchinin, Jan 07 2023

a(8) = 20, corrected by Geoffrey Critzer, Oct 17 2009
More terms from Derek Orr, Oct 30 2014
Offset set to 1 and new name using a formula of Geoffrey Critzer by Peter Luschny, Jan 07 2023

A127982 a(n) = (n - 1/3)*2^n - n/2 + 1/4 + (-1)^n/12.

Original entry on oeis.org

1, 6, 20, 57, 147, 360, 850, 1959, 4433, 9894, 21840, 47781, 103759, 223908, 480590, 1026723, 2184525, 4631202, 9786700, 20621985, 43341131, 90876576, 190141770, 397060767, 827675977, 1722460830, 3579139400, 7426714269, 15390299463

Offset: 1

Artur Jasinski, Feb 09 2007

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000
W. Bosma, Signed bits and fast exponentiation, J. Th. des Nombres de Bordeaux Vol.13, Fasc. 1, 2001.
Index entries for linear recurrences with constant coefficients, signature (5,-7,-1,8,-4).

Cf. A073371, A127976, A127978, A127979, A127980, A073371, A000337.

[(n-1/3)*2^n -n/2 +1/4 +(-1)^n/12: n in [1..50]]; // G. C. Greubel, May 08 2018

Table[(n-1/3)*2^n -n/2 +1/4 +(-1)^n/12, {n, 1, 50}]
LinearRecurrence[{5,-7,-1,8,-4}, {1,6,20,57,147}, 50] (* G. C. Greubel, May 08 2018 *)

a(n) = (n-1/3)*2^n -n/2 +1/4 +(-1)^n/12 \\ G. C. Greubel, May 08 2018

a(n) = (n - 1/3)*2^n - n/2 + 1/4 + (-1)^n/12.

G.f.: -x*(3*x^2-x-1)/((1+x)*(2*x-1)^2*(x-1)^2). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 14 2009 [checked and corrected by R. J. Mathar, Sep 16 2009]

A193845 Mirror of the triangle A193844.

Original entry on oeis.org

1, 3, 1, 7, 5, 1, 15, 17, 7, 1, 31, 49, 31, 9, 1, 63, 129, 111, 49, 11, 1, 127, 321, 351, 209, 71, 13, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1

Offset: 0

Clark Kimberling, Aug 07 2011

This triangle is obtained by reversing the rows of the triangle A193844.
From Philippe Deléham, Jan 17 2014: (Start)
Subtriangle of the triangle in A112857.
T(n,0)   = A000225(n+1).
T(n,1)   = A000337(n).
T(n+2,2) = A055580(n).
T(n+3,3) = A027608(n).
T(n+4,4) = A211386(n).
T(n+5,5) = A211388(n).
T(n,n)   = A000012(n).
T(n+1,n) = A005408(n).
T(n+2,n) = A056220(n+2).
T(n+3,n) = A199899(n+1).
Row sums are A003462(n+1).
Diagonal sums are A048739(n).
Riordan array (1/((1-2*x)*(1-x)), x/(1-2*x)). (End)
Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-2)^0 + A_1*(x-2)^1 + A_2*(x-2)^2 + ... + A_n*(x-2)^n. This sequence gives A_0, ... A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 14 2014
The n-th row lists the coefficients of the polynomial sum_{k=0..n} (X+2)^k, in order of increasing powers. - M. F. Hasler, Oct 15 2014

			First six rows:
1
3....1
7....5....1
15...17...7....1
31...49...31...9...1
63...129..111..49..11..1

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened)
Russell Jay Hendel, A Method for Uniformly Proving a Family of Identities, arXiv:2107.03549 [math.CO], 2021.

Cf. A193844.
Cf. Columns: A000225, A000337, A055580, A027608, A211386, A211388
Cf. Diagonals: A000012, A005408, A056220, A199899

z = 10;
p[n_, x_] := (x + 1)^n;
q[n_, x_] := (x + 1)^n
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A193844 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]]  (* A193845 *)
Table[2^k*Binomial[n + 1, k]*Hypergeometric2F1[1, -k, -k + n + 2, 1/2], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Nov 09 2021 *)

for(n=0,20,for(k=0,n,print1(1/k!*sum(i=0,n,(2^(i-k)*prod(j=0,k-1,i-j))),", "))) \\ Derek Orr, Oct 14 2014

T(n,k) = A193844(n,n-k).

T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) - T(n-2,k-1), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 17 2014

cp's OEIS Frontend

A127984 a(n) = (n/3 + 7/9)*2^(n - 1) + (-1)^n/9.

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A193844 Triangular array: the fission of ((x+1)^n) by ((x+1)^n); i.e., the self-fission of Pascal's triangle.

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A066185 Sum of the first moments of all partitions of n with weights starting at 0.

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A102301 a(n) = ((3*n + 1)*2^(n+3) + 9 + (-1)^n)/18.

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A130328 Triangle of differences between powers of 2, read by rows.

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A055251 Eighth column of triangle A055249.

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A059045 Square array T(n,k) read by antidiagonals where T(0,k) = 0 and T(n,k) = 1 + 2k + 3k^2 + ... + n*k^(n-1).

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A102301 a(n) = ((3n + 1)2^(n+3) + 9 + (-1)^n)/18.