A034856 -id:A034856 - cp's OEIS frontend

A323255 The permanent of an n X n Toeplitz matrix M(n) whose first row consists of successive positive integer numbers 2n - 1, n - 1, ..., 1 and whose first column consists of 2n - 1, 2*n - 2, ..., n.

1, 1, 11, 248, 9968, 638772, 60061657, 7798036000, 1336715859150, 292406145227392, 79483340339739367, 26280500564448081664, 10386012861097225139356, 4834639222955142417477888, 2618110215141486526589786501, 1631888040186649673361825042432, 1159983453675106278249250918734938

Offset: 0

Stefano Spezia, Jan 09 2019

nonn

The trace of the matrix M(n) is A000384(n).
The sum of the first row of the matrix M(n) is A034856(n).
The sum of the first column of the matrix M(n) is A000326(n).
For n > 1, the sum of the superdiagonal of the matrix M(n) is A000290(n-1).
For n > 1, the sum of the subdiagonal of the matrix M(n) is A001105(n-1).

			For n = 1 the matrix M(1) is
   1
with permanent a(1) = 1.
For n = 2 the matrix M(2) is
   3, 1
   2, 3
with permanent a(2) = 11.
For n = 3 the matrix M(3) is
   5, 2, 1
   4, 5, 2
   3, 4, 5
with permanent a(3) = 248.

Stefano Spezia, Table of n, a(n) for n = 0..35
Wikipedia, Toeplitz matrix

Cf. A000290, A000326, A000384, A001105, A001792.
Cf. A034856, A204235, A318173, A322908, A322909.
Cf. A323254 (determinant of matrix M(n)).

b[i_]:=i; a[n_]:=If[n==0, 1, Permanent[ToeplitzMatrix[Join[{b[2*n-1]}, Array[b, n-1, {2*n-2,n }]], Join[{b[2*n-1]},Array[b, n-1, {n-1,1}]]]]]; Array[a, 16, 0]

tm(n) = {my(m = matrix(n, n, i, j, if (j==1, 2*n-i, n-j+1))); for (i=2, n, for (j=2, n, m[i, j] = m[i-1, j-1]; ); ); m;}
a(n) = matpermanent(tm(n)); \\ Stefano Spezia, Dec 11 2019

a(0) = 1 prepended by Stefano Spezia, Dec 08 2019

A349736 Binomial coefficients C(m,k) such that C(m,k), C(m,k+1), C(m,k+2) with 0 <= k <= m-2 form an increasing arithmetic progression.

Original entry on oeis.org

7, 1001, 490314, 927983760, 6973199770790, 209769429934732479, 25331521183260952835630, 12289694242827235919344118592, 23955991473971122736214778043009679, 187581456720371323313917970237305876898550, 5898404991626652623457605084827693331568853294440, 744569299056744628602691379013860201165514803170616390880

Offset: 1

Bernard Schott, Nov 28 2021

Exercise A1 of 33rd Putnam Exam in 1972 asked one to prove that there are no four consecutive binomial coefficients C(m,k), C(m,k+1), C(m,k+2), C(m,k+3) in arithmetic progression (see link and reference).
However, as there exist such progressions with 3 terms, this sequence lists the smallest term of these arithmetic progressions.
Three consecutive binomial coefficients form an arithmetic progression iff m = n^2+4n+2, n >= 1 (2nd comment of A008865), and then, corresponding k = (n^2+3n-2)/2. Successive pairs (m,k) are (7,1), (14,4), (23,8), (34,13), (47,19), ...
By symmetry of Pascal's triangle with C(m,k) = C(m,m-k), there exists another decreasing arithmetic progression with the same 3 terms in the same row.
Corresponding common differences are in A349737.

			For n = 1, row 7 of Pascal's triangle is 1, 7, 21, 35, 35, 21, 7, 1; C(7,1) = 7, C(7,2) = 21 and C(7,3) = 35 form an arithmetic progression with common difference = 14, hence a(3) = 7 = C(7,1).
For n = 2, row 14 is 1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1; C(14,4) = 1001 , C(14,5) = 2002 and C(14,6) = 3003 form an arithmetic progression with common difference = 1001, hence a(4) = 1001 = C(14,4).

G. L. Alexanderson, L. F. Klosinski and L. C. Larson, The William Lowell Putnam Mathematical Competition, Problems and Solutions 1965-1984, The Mathematical Association of America, 1985, page 17.

Andrew Howroyd, Table of n, a(n) for n = 1..50
John Scholes, 33rd Putnam 1972, Problem A1.
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A007318, A008865, A034856, A349737.

Sequence = seq(binomial(n^2+4*n+2,(n^2+3*n-2)/2), n=1..16);

nterms=15; Table[Binomial[n^2+4n+2, (n^2+3n-2)/2], {n, nterms}] (* Paolo Xausa, Nov 29 2021 *)

a(n) = binomial(n^2+4*n+2,(n^2+3*n-2)/2) \\ Andrew Howroyd, Oct 29 2023

a(n) = C(n^2+4n+2,(n^2+3n-2)/2) = C(A008865(n+2),A034856(n)), for n >= 1.

a(n) ~ c*2^(n^2+4*n)/n, where c = 4*sqrt(2/(Pi*e)). - Stefano Spezia, Nov 29 2021

Missing a(9) = 23955...79 inserted by Georg Fischer, Oct 29 2023

A101919 Triangle read by rows: T(n,k) is the number of Schroeder paths of length 2n and having k up steps starting at even heights.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 12, 8, 1, 1, 33, 42, 13, 1, 1, 88, 183, 102, 19, 1, 1, 232, 717, 624, 205, 26, 1, 1, 609, 2622, 3275, 1650, 366, 34, 1, 1, 1596, 9134, 15473, 11020, 3716, 602, 43, 1, 1, 4180, 30691, 67684, 64553, 30520, 7483, 932, 53, 1, 1, 10945, 100284, 279106

Offset: 0

Emeric Deutsch, Dec 20 2004

A Schroeder path of length 2n is a lattice path starting from (0,0), ending at (2n,0), consisting only of steps U=(1,1) (up steps), D=(1,-1) (down steps) and H=(2,0) (level steps) and never going below the x-axis (Schroeder paths are counted by the large Schroeder numbers, A006318). Also number of Schroeder paths of length 2n and having k humps. A hump is an up step U followed by 0 or more level steps H followed by a down step D. The T(3,2)=8 Schroeder paths of length 6 and having 2 humps are: H(UD)(UD), (UD)H(UD), (UD)(UD)H, (UD)(UHD), (UD)(UUDD), (UHD)(UD), (UUDD)(UD) and U(UD)(UD)D, the humps being shown between parentheses. Row sums are the large Schroeder numbers (A006318). Column 1 yields the odd-indexed Fibonacci numbers minus 1 (A027941). T(n,n-1)=A034856(n)=binomial(n + 1, 2) + n - 1.

Product A085478*A090181 (Morgan-Voyce times Narayana). [From Paul Barry, Jan 29 2009]

			T(3,2)=8 because we have HU'DU'D, U'DHU'D, U'DU'DH, U'DU'HD, U'DU'UDD, U'HDU'D, U'UDDU'D and U'UU'DDD, the up steps starting at an even height being shown with a prime sign.
Triangle begins:
1;
1,1;
1,4,1;
1,12,8,1;
1,33,42,13,1;

Cf. A006318, A027941, A034856, A101920.

G:=1/2/(-z+z^2)*(-1+z+t*z-z^2+sqrt(1-6*z-2*t*z+11*z^2+2*t*z^2-6*z^3+t^2*z^2-2*t*z^3+z^4)): Gser:=simplify(series(G,z=0,13)): P[0]:=1: for n from 1 to 11 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 11 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields the sequence in triangular form

G.f.=G=G(t, z) satisfies z(1-z)G^2-(1-z-tz+z^2)G+1-z=0.

G.f.: 1/(1-x-xy/(1-x-x/(1-x-xy/(1-x-xy/(1-x-x/(1-x-xy/(1-.... (continued fraction). [From Paul Barry, Jan 29 2009]

A128219 A000012 * A127701. a(1) = 1, a(2) = 2, a(3) = 2; by rows, n-1 terms of 2, 3, 4, ... followed by "n".

Original entry on oeis.org

1, 2, 2, 2, 3, 3, 2, 3, 4, 4, 2, 3, 4, 5, 5, 2, 3, 4, 5, 6, 6, 2, 3, 4, 5, 6, 7, 7, 2, 3, 4, 5, 6, 7, 8, 8, 2, 3, 4, 5, 6, 7, 8, 9, 9, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 12, 2, 3, 4, 5, 6

Offset: 1

Gary W. Adamson, Feb 19 2007

			First few rows of the triangle:
  1;
  2, 2;
  2, 3, 3;
  2, 3, 4, 4;
  2, 3, 4, 5, 5;
  2, 3, 4, 5, 6, 6;
  2, 3, 4, 5, 6, 7, 7;
  ...

Cf. A000012, A127701, A034856 (row sums), A128220.

trm[i_]:=Join[Range[2,i],{i}]; Flatten[Table[trm[n],{n,13}]] (* Harvey P. Dale, Nov 14 2012 *)

A000012 * A127701 as infinite lower triangular matrices.

A131819 A131818 * A000012 as infinite lower triangular matrices. Triangle read by rows, partial sums starting from the right of A131818.

Original entry on oeis.org

1, 4, 2, 8, 5, 3, 13, 9, 7, 4, 19, 14, 12, 9, 5, 26, 20, 18, 15, 11, 6, 34, 27, 25, 22, 18, 13, 7, 43, 35, 33, 30, 26, 21, 15, 8, 53, 44, 42, 39, 35, 30, 24, 17, 9, 64, 54, 52, 49, 45, 40, 34, 27, 19, 10, 76, 65, 63, 60, 56, 51, 45, 38, 30, 21, 11

Offset: 1

Gary W. Adamson, Jul 18 2007

Left column = A034856: (1, 4, 8, 13, 19, 26, 34, ...). Row sums = A131820: (1, 6, 16, 33, 59, 96, ...).

			First few rows of the triangle:
   1;
   4,  2;
   8,  5,  3;
  13,  9,  7,  4;
  19, 14, 12,  9,  5;
  26, 20, 18, 15, 11,  6;
  34, 27, 25, 22, 18, 13,  7;
  ...

Cf. A131818, A000012, A131820, A034856.

tabl(nn) = {m131818 = matrix(nn, nn, n, k, if (k==1, n, if (k <= n, k, 0))); m000012 = matrix(nn, nn, n, k, (k<=n)); m131819 = m131818 * m000012; for (n = 1, nn, for (k = 1, n, print1(m131819[n, k], ", ");););} \\ Michel Marcus, Feb 12 2014

More terms from Michel Marcus, Feb 12 2014

A134225 A007436 + A134082 - A000012 as infinite lower triangular matrices; where A000012 = (1; 1,1; 1,1,1; ...).

Original entry on oeis.org

1, 3, 1, 2, 5, 1, 3, 2, 7, 1, 4, 3, 2, 9, 1, 5, 4, 3, 2, 11, 1, 6, 5, 4, 3, 2, 13, 1, 7, 6, 5, 4, 3, 2, 15, 1, 8, 7, 6, 5, 4, 3, 2, 17, 1, 9, 8, 7, 6, 5, 4, 3, 2, 19, 1

Offset: 1

Gary W. Adamson, Oct 14 2007

Row sums = A034856: (1, 4, 8, 13, 19, 26, ...).

			First few rows of the triangle:
  1;
  3, 1;
  2, 5, 1;
  3, 2, 7, 1;
  4, 3, 2, 9,  1;
  5, 4, 3, 2, 11,  1;
  6, 5, 4, 3,  2, 13,  1;
  7, 6, 5, 4,  3,  2, 15, 1;
  ...

Cf. A007436, A134082, A034856.

A237124 Triangle of numbers related to Catalan numbers (A000108).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 4, 3, 1, 9, 11, 8, 4, 1, 28, 33, 24, 13, 5, 1, 90, 104, 76, 43, 19, 6, 1, 297, 339, 249, 145, 69, 26, 7, 1, 1001, 1133, 836, 497, 248, 103, 34, 8, 1, 3432, 3861, 2860, 1727, 891, 394, 146, 43, 9, 1, 11934, 13364, 9932, 6071, 3211, 1484, 593, 199, 53, 10, 1

Offset: 0

Philippe Deléham, Feb 03 2014

Riordan array (1 +x +x^2*C(x)^3, x*C(x)) where C(x) is the g.f. of A000108.
Diagonal sums are A000108(n).
Row sums are T(n+1,1).
T(n,0) = A071724(n-1).
T(n,1) = A220902(n), n>=2.
T(n,2) = A228404(n-2), n>=4.
T(n+3,3) = A033434(n).
T(n,n) = 1.
T(n+1,n) = n+1.
T(n+2,n) = A034856(n+1).

			Triangle begins:
      1;
      1,     1;
      1,     2,    1;
      3,     4,    3,    1;
      9,    11,    8,    4,    1;
     28,    33,   24,   13,    5,    1;
     90,   104,   76,   43,   19,    6,   1;
    297,   339,  249,  145,   69,   26,   7,   1;
   1001,  1133,  836,  497,  248,  103,  34,   8,  1;
   3432,  3861, 2860, 1727,  891,  394, 146,  43,  9,  1;
  11934, 13364, 9932, 6071, 3211, 1484, 593, 199, 53, 10, 1;
  ...

G. C. Greubel, Rows n = 0..30 of the triangle, flattened

Cf. A000108.

b[n_, k_]:= Binomial[2*n-k+1, n-k];
T[n_, k_]:= If[n<3, Binomial[n, k], b[n, k] -2*b[n, k+1] -b[n, k+2] +3*b[n, k+3] - 2*b[n, k+4]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, May 08 2021 *)

def b(n,k): return binomial(2*n-k+1, n-k)
def T(n,k): return binomial(n,k) if (n<3) else b(n,k) -2*b(n, k+1) -b(n, k+2) +3*b(n, k+3) -2*b(n, k+4)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 08 2021

From Peter Bala, Feb 18 2018: (Start)
T(n,k) = C(2*n+1-k, n-k) - 2*C(2*n-k, n-k-1) - C(2*n-1-k, n-k-2) + 3*C(2*n-2-k, n-k-3) - 2*C(2*n-3-k, n-k-4), for n > 2, otherwise C(n, k).
The n-th row polynomial of the row reverse triangle equals the n-th degree Taylor polynomial of the function (1 - x^2 + x^3)*(1 - 2*x)/(1 - x)^2 * 1/(1 - x)^n about 0. For example, for n = 4, (1 - x^2 + x^3)*(1 - 2*x)/(1 - x)^2 * 1/(1 - x)^4 = 1 + 4*x + 8*x^2 + 11*x^3 + 9*x^4 + O(x^5), giving (9, 11, 8, 4, 1) as row 4. (End)

A237519 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the partial sums of the column k of A237273 starting in row k^2.

Original entry on oeis.org

1, 4, 8, 13, 2, 19, 2, 26, 7, 34, 7, 43, 13, 53, 13, 3, 64, 20, 3, 76, 20, 3, 89, 28, 10, 103, 28, 10, 118, 37, 10, 134, 37, 18, 151, 47, 18, 4, 169, 47, 18, 4, 188, 58, 27, 4, 208, 58, 27, 4, 229, 70, 27, 13, 251, 70, 37, 13, 274, 83, 37, 13, 298, 83, 37, 13, 323, 97, 48, 23, 349, 97, 48, 23, 5

Offset: 1

Omar E. Pol, Feb 08 2014

The sum of row n is A024916(n), the sum of all divisors of all positive integers <= n.
Row n has length A000196(n).
Column 1 is A034856.
Column 2 lists the elements of A155212 repeated.
Column 3 lists the elements of A051938 repeated with three copies of every element.
Column k contains k copies of every element.

			Triangle begins:
1;
4;
8;
13,   2;
19,   2;
26,   7;
34,   7;
43,  13;
53,  13,  3;
64,  20,  3;
76,  20,  3;
89,  28, 10;
103, 28, 10;
118, 37, 10;
134, 37, 18;
151, 47, 18,  4;
169, 47, 18,  4;
188, 58, 27,  4;
208, 58, 27,  4;
229, 70, 27, 13;
251, 70, 37, 13;
274, 83, 37, 13;
298, 83, 37, 13;
323, 97, 48, 23;
349, 97, 48, 23,  5;
...

Cf. A000203, A000196, A024916, A034856, A051938, A155212, A236104, A235791, A237273.

A283054 Triangle read by rows: T(n,k) = T(n,k-1) + T(n-1,k), T(n,0)=1, T(n,n) = T(n,n-1) + 1.

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 1, 4, 8, 9, 1, 5, 13, 22, 23, 1, 6, 19, 41, 64, 65, 1, 7, 26, 67, 131, 196, 197, 1, 8, 34, 101, 232, 428, 625, 626, 1, 9, 43, 144, 376, 804, 1429, 2055, 2056, 1, 10, 53, 197, 573, 1377, 2806, 4861, 6917, 6918, 1, 11, 64, 261, 834, 2211, 5017, 9878, 16795, 23713, 23714, 1, 12, 76, 337, 1171, 3382, 8399, 18277, 35072, 58785, 82499, 82500

Offset: 0

Ely Golden, Feb 27 2017

The left diagonals form polynomial sequences. This is due to the observation that diagonal 0 D_0(x) = 1, and D_n(x) = D_n(x-1)+D_(n-1)(x+1), with D_n(-1) = 1 which is a recurrence that can be solved.
These polynomials begin 1, x+2, (x(x+7)+8)/2, (x(x(x+15)+62)+54)/6, (x(x(x(x+26)+227)+730)+552)/24, etc., the first 3 of which correspond to A000012(n), A000027(n+2), and A034856(n+2), respectively.
The rightmost diagonal appears to follow A014137(n). The second rightmost appears to follow A014138(n+1), the third appears to follow A001453(n+2), the fourth appears to follow A114277(n), and the fifth appears to follow A143955(n+3).
A closed-form formula for T(n,k) would be very desirable.

			First 7 rows:
  1;
  1,   2;
  1,   3,   4;
  1,   4,   8,   9;
  1,   5,  13,  22,  23;
  1,   6,  19,  41,  64,  65;
  1,   7,  26,  67, 131, 196, 197;

Ely Golden, Rows n = 0..124 of triangle, flattened
Ely Golden, Java program for generating the triangle
Ely Golden, Sage program for computing the polynomial of the n-th left diagonal

T[0, 0] = 1; T[n_, k_] := T[n, k] = Which[k == 0, 1, k == n, T[n, n - 1] + 1, True, T[n, k - 1] + T[n - 1, k]]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 27 2017 *)

T(n,k)=if(k==0,return(1));if(k==n,return(T(n,n-1)+1));T(n,k-1)+T(n-1,k)
for(n=0,10,for(k=0,n,print1(T(n,k),", "))) \\ Derek Orr, Feb 28 2017

def sideTriangleAt(a,b):
    if(b==0): return 1
    elif(b==a): return sideTriangleAt(a,b-1)+1
    else: return sideTriangleAt(a,b-1)+sideTriangleAt(a-1,b)
def sideTriangle(size):
    li=[]
    for c in range(size):
        for d in range(c+1):
            if(d==0): li.append([1])
            elif(d==c): li[c].append(li[c][d-1]+1)
            else: li[c].append(li[c][d-1]+li[c-1][d])
    return li
trig=sideTriangle(125)
for c in range(len(trig)):
    print(str(trig[c])[1:-1].replace(",",""))

A303273 Array T(n,k) = binomial(n, 2) + k*n + 1 read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 4, 4, 1, 4, 6, 7, 7, 1, 5, 8, 10, 11, 11, 1, 6, 10, 13, 15, 16, 16, 1, 7, 12, 16, 19, 21, 22, 22, 1, 8, 14, 19, 23, 26, 28, 29, 29, 1, 9, 16, 22, 27, 31, 34, 36, 37, 37, 1, 10, 18, 25, 31, 36, 40, 43, 45, 46, 46, 1, 11, 20, 28, 35, 41

Offset: 0

Franck Maminirina Ramaharo, Apr 20 2018

Columns are linear recurrence sequences with signature (3,-3,1).
8*T(n,k) + A166147(k-1) are squares.
Columns k are binomial transforms of [1, k, 1, 0, 0, 0, ...].
Antidiagonals sums yield A116731.

			The array T(n,k) begins
1    1    1    1    1    1    1    1    1    1    1    1    1  ...  A000012
1    2    3    4    5    6    7    8    9   10   11   12   13  ...  A000027
2    4    6    8   10   12   14   16   18   20   22   24   26  ...  A005843
4    7   10   13   16   19   22   25   28   31   34   37   40  ...  A016777
7   11   15   19   23   27   31   35   39   43   47   51   55  ...  A004767
11  16   21   26   31   36   41   46   51   56   61   66   71  ...  A016861
16  22   28   34   40   46   52   58   64   70   76   82   88  ...  A016957
22  29   36   43   50   57   64   71   78   85   92   99  106  ...  A016993
29  37   45   53   61   69   77   85   93  101  109  117  125  ...  A004770
37  46   55   64   73   82   91  100  109  118  127  136  145  ...  A017173
46  56   66   76   86   96  106  116  126  136  146  156  166  ...  A017341
56  67   78   89  100  111  122  133  144  155  166  177  188  ...  A017401
67  79   91  103  115  127  139  151  163  175  187  199  211  ...  A017605
79  92  105  118  131  144  157  170  183  196  209  222  235  ...  A190991
...
The inverse binomial transforms of the columns are
1    1    1    1    1    1    1    1    1    1    1    1    1  ...
0    1    2    3    4    5    6    7    8    9   10   11   12  ...
1    1    1    1    1    1    1    1    1    1    1    1    1  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
0    0    0    0    0    0    0    0    0    0    0    0    0  ...
...
T(k,n-k) = A087401(n,k) + 1 as triangle
1
1   1
1   2   2
1   3   4   4
1   4   6   7   7
1   5   8  10  11  11
1   6  10  13  15  16  16
1   7  12  16  19  21  22  22
1   8  14  19  23  26  28  29  29
1   9  16  22  27  31  34  36  37  37
1  10  18  25  31  36  40  43  45  46  46
...

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, 1994.

Cf. A085475, A086270, A086271, A086272, A086273, A130154, A159798, A162609, A162610, A300401.

T := (n, k) -> binomial(n, 2) + k*n + 1;
for n from 0 to 20 do seq(T(n, k), k = 0 .. 20) od;

Table[With[{n = m - k}, Binomial[n, 2] + k n + 1], {m, 0, 11}, {k, m, 0, -1}] // Flatten (* Michael De Vlieger, Apr 21 2018 *)

T(n, k) := binomial(n, 2)+ k*n + 1$
for n:0 thru 20 do
    print(makelist(T(n, k), k, 0, 20));

T(n,k) = binomial(n, 2) + k*n + 1;
tabl(nn) = for (n=0, nn, for (k=0, nn, print1(T(n, k), ", ")); print); \\ Michel Marcus, May 17 2018

G.f.: (3*x^2*y - 3*x*y + y - 2*x^2 + 2*x - 1)/((x - 1)^3*(y - 1)^2).
E.g.f.: (1/2)*(2*x*y + x^2 + 2)*exp(y + x).
T(n,k) = 3*T(n-1,k) - 3*T(n-2,k) + T(n-3,k), with T(0,k) = 1, T(1,k) = k + 1 and T(2,k) = 2*k + 2.
T(n,k) = T(n-1,k) + n + k - 1.
T(n,k) = T(n,k-1) + n, with T(n,0) = 1.
T(n,0) = A152947(n+1).
T(n,1) = A000124(n).
T(n,2) = A000217(n).
T(n,3) = A034856(n+1).
T(n,4) = A052905(n).
T(n,5) = A051936(n+4).
T(n,6) = A246172(n+1).
T(n,7) = A302537(n).
T(n,8) = A056121(n+1) + 1.
T(n,9) = A056126(n+1) + 1.
T(n,10) = A051942(n+10) + 1, n > 0.
T(n,11) = A101859(n) + 1.
T(n,12) = A132754(n+1) + 1.
T(n,13) = A132755(n+1) + 1.
T(n,14) = A132756(n+1) + 1.
T(n,15) = A132757(n+1) + 1.
T(n,16) = A132758(n+1) + 1.
T(n,17) = A212427(n+1) + 1.
T(n,18) = A212428(n+1) + 1.
T(n,n) = A143689(n) = A300192(n,2).
T(n,n+1) = A104249(n).
T(n,n+2) = T(n+1,n) = A005448(n+1).
T(n,n+3) = A000326(n+1).
T(n,n+4) = A095794(n+1).
T(n,n+5) = A133694(n+1).
T(n+2,n) = A005449(n+1).
T(n+3,n) = A115067(n+2).
T(n+4,n) = A133694(n+2).
T(2*n,n) = A054556(n+1).
T(2*n,n+1) = A054567(n+1).
T(2*n,n+2) = A033951(n).
T(2*n,n+3) = A001107(n+1).
T(2*n,n+4) = A186353(4*n+1) (conjectured).
T(2*n,n+5) = A184103(8*n+1) (conjectured).
T(2*n,n+6) = A250657(n-1) = A250656(3,n-1), n > 1.
T(n,2*n) = A140066(n+1).
T(n+1,2*n) = A005891(n).
T(n+2,2*n) = A249013(5*n+4) (conjectured).
T(n+3,2*n) = A186384(5*n+3) = A186386(5*n+3) (conjectured).
T(2*n,2*n) = A143689(2*n).
T(2*n+1,2*n+1) = A143689(2*n+1) (= A030503(3*n+3) (conjectured)).
T(2*n,2*n+1) = A104249(2*n) = A093918(2*n+2) = A131355(4*n+1) (= A030503(3*n+5) (conjectured)).
T(2*n+1,2*n) = A085473(n).
a(n+1,5*n+1)=A051865(n+1) + 1.
a(n,2*n+1) = A116668(n).
a(2*n+1,n) = A054569(n+1).
T(3*n,n) = A025742(3*n-1), n > 1 (conjectured).
T(n,3*n) = A140063(n+1).
T(n+1,3*n) = A069099(n+1).
T(n,4*n) = A276819(n).
T(4*n,n) = A154106(n-1), n > 0.
T(2^n,2) = A028401(n+2).
T(1,n)*T(n,1) = A006000(n).
T(n*(n+1),n) = A211905(n+1), n > 0 (conjectured).
T(n*(n+1)+1,n) = A294259(n+1).
T(n,n^2+1) = A081423(n).
T(n,A000217(n)) = A158842(n), n > 0.
T(n,A152947(n+1)) = A060354(n+1).
floor(T(n,n/2)) = A267682(n) (conjectured).
floor(T(n,n/3)) = A025742(n-1), n > 0 (conjectured).
floor(T(n,n/4)) = A263807(n-1), n > 0 (conjectured).
ceiling(T(n,2^n)/n) = A134522(n), n > 0 (conjectured).
ceiling(T(n,n/2+n)/n) = A051755(n+1) (conjectured).
floor(T(n,n)/n) = A133223(n), n > 0 (conjectured).
ceiling(T(n,n)/n) = A007494(n), n > 0.
ceiling(T(n,n^2)/n) = A171769(n), n > 0.
ceiling(T(2*n,n^2)/n) = A046092(n), n > 0.
ceiling(T(2*n,2^n)/n) = A131520(n+2), n > 0.

cp's OEIS Frontend

A323255 The permanent of an n X n Toeplitz matrix M(n) whose first row consists of successive positive integer numbers 2*n - 1, n - 1, ..., 1 and whose first column consists of 2*n - 1, 2*n - 2, ..., n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Extensions

A349736 Binomial coefficients C(m,k) such that C(m,k), C(m,k+1), C(m,k+2) with 0 <= k <= m-2 form an increasing arithmetic progression.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A101919 Triangle read by rows: T(n,k) is the number of Schroeder paths of length 2n and having k up steps starting at even heights.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Formula

A128219 A000012 * A127701. a(1) = 1, a(2) = 2, a(3) = 2; by rows, n-1 terms of 2, 3, 4, ... followed by "n".

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

A131819 A131818 * A000012 as infinite lower triangular matrices. Triangle read by rows, partial sums starting from the right of A131818.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Extensions

A134225 A007436 + A134082 - A000012 as infinite lower triangular matrices; where A000012 = (1; 1,1; 1,1,1; ...).

Views

Author

Keywords

Comments

Examples

Crossrefs

A237124 Triangle of numbers related to Catalan numbers (A000108).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Sage

Formula

A237519 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the partial sums of the column k of A237273 starting in row k^2.

Views

Author

Keywords

A323255 The permanent of an n X n Toeplitz matrix M(n) whose first row consists of successive positive integer numbers 2n - 1, n - 1, ..., 1 and whose first column consists of 2n - 1, 2*n - 2, ..., n.