A016061 -id:A016061 - cp's OEIS frontend

A135036 Sums of the products of n consecutive pairs of numbers.

0, 6, 26, 68, 140, 250, 406, 616, 888, 1230, 1650, 2156, 2756, 3458, 4270, 5200, 6256, 7446, 8778, 10260, 11900, 13706, 15686, 17848, 20200, 22750, 25506, 28476, 31668, 35090, 38750, 42656, 46816, 51238, 55930, 60900, 66156, 71706, 77558, 83720

Offset: 1

Cino Hilliard, Feb 10 2008

Number of integer solutions to 1-n <= x <= y <= z <= n-1 where x - 2*y + z is nonzero. - Michael Somos, Dec 27 2011
This sequence is related to A001105 by the transform a(n) = (n-1)*A001105(n)-Sum_{i=0..n-1} A001105(i). - Bruno Berselli, Mar 12 2012
a(n) is the maximum value obtainable by partitioning the set {x in the natural numbers | 0 <= x <= 2n+1} into pairs, taking the product of all such pairs, and taking the sum of all such products. - Thomas Anton, Oct 20 2020

			For n = 3, the sum of the first 3 pairs is 0*1+2*3+4*5 = 26, the 3rd entry in the sequence.
G.f.: 6*x^2 + 26*x^3 + 68*x^4 + 140*x^5 + 250*x^6 + 406*x^7 + 616*x^8 + 888*x^9 + ...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A016061, A001105.

[(n-1)*n*(4*n+1)/3: n in [1..40]]; // Bruno Berselli, Mar 12 2012

Accumulate[Times@@@Partition[Range[0,81],2]] (* or *) LinearRecurrence[ {4,-6,4,-1},{0,6,26,68},40] (* Harvey P. Dale, Jun 20 2013 *)
a[ n_] := n (n - 1) (4 n + 1)/3; (* Michael Somos, Oct 15 2015 *)
a[ n_] := If[ n >= 0, Length @ FindInstance[ 1 - n <= x <= y <= z <= n - 1 && x - 2 y + z != 0, {x, y, z}, Integers, 10^9], -(Length @ FindInstance[ n <= x < y <= z <= -n && x - 2 y + z != 0, {x, y, z}, Integers, 10^9] + n)]; (* Michael Somos, Oct 15 2015 *)

sumprod(n) = { local(x,s=0); forstep(x=0,n,2, s+=x*(x+1); print1(s",") ) }

{a(n) = n * (n - 1) * (4*n + 1) / 3}; /* Michael Somos, Dec 27 2011 */

a(n) = 0*1 + 2*3 + 4*5 + ... + 2*n*(2*n + 1).
a(n) = (4*n^3 - 3*n^2 - n)/3 = (n - 1)*n*(4*n + 1)/3.
From R. J. Mathar, Feb 14 2008: (Start)
O.g.f.: 2*x^2*(3 + x)/(1 - x)^4.
a(n) = 2*A016061(n-1). (End)
a(0)=0, a(1)=6, a(2)=26, a(3)=68, a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Jun 20 2013
From Amiram Eldar, Apr 30 2023: (Start)
Sum_{n>=2} 1/a(n) = 6*Pi/5 + 36*log(2)/5 - 213/25.
Sum_{n>=2} (-1)^n/a(n) = 6*sqrt(2)*Pi/5 + 6*(sqrt(2)+3)*log(2)/5 - 12*sqrt(2)*log(2-sqrt(2))/5 - 267/25. (End)
E.g.f.: exp(x)*x^2*(9 + 4*x)/3. - Stefano Spezia, Jun 14 2023

First formula corrected by Harvey P. Dale, Jun 20 2013

A210220 T(n, k) = -binomial(2n-k+2, k+1)hypergeom([2*n-k+3, 1], [k+2], 2). Triangle read by rows, T(n, k) for 1 <= k <= n.

Original entry on oeis.org

1, 2, 2, 3, 6, 3, 4, 12, 13, 4, 5, 20, 34, 24, 5, 6, 30, 70, 80, 40, 6, 7, 42, 125, 200, 166, 62, 7, 8, 56, 203, 420, 496, 314, 91, 8, 9, 72, 308, 784, 1211, 1106, 553, 128, 9, 10, 90, 444, 1344, 2576, 3108, 2269, 920, 174, 10, 11, 110, 615, 2160, 4956, 7476, 7274, 4352, 1461, 230, 11

Offset: 1

Clark Kimberling, Mar 19 2012

Previous name: Triangle of coefficients of polynomials v(n,x) jointly generated with A210217.

For a discussion and guide to related arrays, see A208510.

			First five rows:
  1
  2...2
  3...6....3
  4...12...13...4
  5...20...34...24...5
First three polynomials v(n,x): 1, 2 + 2x , 3 + 6x + 3x^2.

Cf. A210217, A210219, A208510, A016061, A112742.

T := (n,k) -> -binomial(2*n-k+2, k+1)*hypergeom([2*n-k+3, 1], [k+2], 2):
seq(seq(simplify(T(n,k)), k=1..n), n=1..10); # Peter Luschny, Oct 31 2019

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
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[%]     (* A210219 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]     (* A210220 *)
(* alternate program *)
T[n_,k_]:=Sum[Binomial[2*j+k-2,k-1],{j,1,n-k+1}];Flatten[Table[T[n,k],{n,1,11},{k,1,n}]] (* Detlef Meya, Dec 05 2023 *)

First and last term in row n: n.
Column 2: n*(n-1).
Column 3: A016061.
Column 4: A112742.
Row sums: -1+(even-indexed Fibonacci numbers).
Periodic alternating row sums: 1,0,0,1,0,0,1,0,0,...
u(n,x)=x*u(n-1,x)+v(n-1,x)+1,
v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
T(n,k) = Sum_{j=1..n-k+1} binomial(2*j+k-2,k-1). - Detlef Meya, Dec 05 2023

New name from Peter Luschny, Oct 31 2019

A240929 Number of 10-digit positive integers in base n where the sum of the first k digits equals the sum of the last k digits.

Original entry on oeis.org

126, 6046, 88428, 694360, 3705741, 15192604, 51418473, 150420187, 392406145, 933294637, 2056947827, 4253047045, 8329101326, 15566783605, 27934647638, 48371293570, 81155221112, 132379936520, 210555362990, 327359243694, 498565022483, 745175639274, 1094795785319

Offset: 2

Martin Renner, Aug 03 2014

These integers are sometimes called balanced numbers.

Cambridge Colleges Sixth Term Examination Papers (STEP) 2007, Paper I, Section A (Pure Mathematics), Nr. 1.

Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A016061, A197083, A240927, A240928.

def A240929(n): return n*(n*(n*(n*(n*(n*(n*(n*(156190*n-140571)+29400)-30870)+3990)-8379)-3100)-1620)-5040)//362880 # Chai Wah Wu, May 08 2024

a(n) = n*(n-1)*(156190*n^7 + 15619*n^6 + 45019*n^5 + 14149*n^4 + 18139*n^3 + 9760*n^2 + 6660*n + 5040)/362880
From Chai Wah Wu, May 08 2024: (Start)
a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4) + 252*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10) for n > 11.
G.f.: x^2*(x^7 + 326*x^6 + 7942*x^5 + 42341*x^4 + 67030*x^3 + 33638*x^2 + 4786*x + 126)/(x - 1)^10. (End)

A143814 Triangle T(n,m) read along rows: T(n,m) = n^2 - (m+1)^2 for 1<=m

Original entry on oeis.org

3, 5, 8, 12, 7, 15, 21, 16, 9, 24, 32, 27, 20, 11, 35, 45, 40, 33, 24, 13, 48, 60, 55, 48, 39, 28, 15, 63, 77, 72, 65, 56, 45, 32, 17, 80, 96, 91, 84, 75, 64, 51, 36, 19, 99, 117, 112, 105, 96, 85, 72, 57, 40, 21, 120, 140, 135, 128, 119, 108, 95, 80, 63, 44, 23, 143

Offset: 2

Paul Curtz, Sep 02 2008

The triangle appears taking the entries of A140978,
4;
9,9;
16,16,16;
25,25,25,25;
..
minus the entries of A133819 with the 1's moved to the end of the rows,
1;
4,1;
4,9,1;
4,9,16,1;
4,9,16,25,1;
The result T(n,m) is a variant of A120070, the first term in each row of A120070 transferred to the end of the row.

			3;
5,8;
12,7,15;
21,16,9,24;
32,27,20,11,35;

Cf. A016061 (row sums).

A143814 := proc(n,m) if mA143814(n,m),m=1..n-1),n=2..14) ; # R. J. Mathar, Jan 23 2011

A185958 Accumulation array of the array max{n,k}, by antidiagonals.

Original entry on oeis.org

1, 3, 3, 6, 7, 6, 10, 13, 13, 10, 15, 21, 22, 21, 15, 21, 31, 34, 34, 31, 21, 28, 43, 49, 50, 49, 43, 28, 36, 57, 67, 70, 70, 67, 57, 36, 45, 73, 88, 94, 95, 94, 88, 73, 45, 55, 91, 112, 122, 125, 125, 122, 112, 91, 55, 66, 111, 139, 154, 160, 161, 160, 154, 139, 111, 66, 78, 133, 169, 190, 200, 203, 203, 200, 190, 169, 133, 78, 91, 157, 202, 230, 245, 251, 252, 251, 245, 230, 202, 157, 91, 105, 183, 238, 274, 295, 305, 308, 308, 305, 295, 274, 238, 183, 105

Offset: 1

Clark Kimberling, Feb 07 2011

A member of the accumulation chain
... < A185917 < A051125 < A185958 < ...,
where A051125, written as a rectangular array M, is given by M(n,k)=max{n,k}.  See A144112 for the definition of accumulation array.
row 1: A000217
row 2: A002061
diag (1,7,...): A002412
diag (3,13,..): A016061
antidiagonal sums: A070893

			Northwest corner:
1....3....6....10....15
3....7....13...21....31
6....13...22...34....49
10...21...34...50....70

Cf. A144112, A051125.

A := proc(n,k) option remember; ## n >= 0 and k = 0 .. n
    if k < 0 or k > n then
        0
    elif n = 0 then
        1
    else
        A(n-1,k) + A(n-1,k-1) - A(n-2,k-1) + max(n-k+1,k+1)
    end if
end proc: # Yu-Sheng Chang, Jun 05 2020

From Yu-Sheng Chang, Jun 05 2020: (Start)
O.g.f.: F(z,v) = -(v^2*z^3+v*z^3-3*v*z^2+1)/((v*z^2-v*z-z+1)^2*(v*z^2-1)*(z-1)*(v*z-1)).
T(n,k) = [v^k] 1/2*n^2*(v^(n+2)+1)/(1-v)^2+1/2*n*(3*v^(n+3)-7*v^(n+2)+7*v-3)/(-1+v)^3-1/2*v*((1-v^(1/2))^4*(-1)^n+(1+v^(1/2))^4)*v^(1/2*n)/(1-v)^4+(6*v^2+6*v^(n+2)+v^(n+4)-3*v^(n+3)-3*v+1)/(1-v)^4.
(End)

A240928 Number of 8-digit positive integers in base n where the sum of the first k digits equals the sum of the last k digits.

Original entry on oeis.org

35, 750, 6174, 31025, 114961, 346193, 896876, 2072694, 4379055, 8606312, 15936426, 28073487, 47400509, 77164915, 121695128, 186650684, 279308283, 408886194, 586909430, 827618109, 1148421417, 1570399589, 2118856324, 2823924050, 3721224455, 4852586700

Offset: 2

Martin Renner, Aug 03 2014

These integers are sometimes called balanced numbers.

Cambridge Colleges Sixth Term Examination Papers (STEP) 2007, Paper I, Section A (Pure Mathematics), Nr. 1.

Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A016061, A197083, A240927, A240929.

Table[n(n-1)(1208n^5+151n^4+291n^3+116n^2+88n+60)/2520,{n,2,40}] (* Harvey P. Dale, Mar 18 2022 *)

a(n) = n*(n-1)*(1208*n^5+151*n^4+291*n^3+116*n^2+88*n+60)/2520.

G.f.: x^2*(x^5+83*x^4+673*x^3+1154*x^2+470*x+35)/(x-1)^8. - Alois P. Heinz, Mar 24 2022

A281381 a(n) = n(n + 1)(4*n + 5)/2.

Original entry on oeis.org

0, 9, 39, 102, 210, 375, 609, 924, 1332, 1845, 2475, 3234, 4134, 5187, 6405, 7800, 9384, 11169, 13167, 15390, 17850, 20559, 23529, 26772, 30300, 34125, 38259, 42714, 47502, 52635, 58125, 63984, 70224, 76857, 83895, 91350, 99234, 107559, 116337, 125580, 135300, 145509, 156219, 167442, 179190, 191475

Offset: 0

Peter M. Chema, Jan 21 2017

Shares its digital root, zero together with period 9: repeat [3, 3, 3, 6, 6, 6, 9, 9, 9] with A027480.

Final digits cycle a length period 20: repeat [0, 9, 9, 2, 0, 5, 9, 4, 2, 5, 5, 4, 4, 7, 5, 0, 4, 9, 7, 0].

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Partial sums of A195319.

Cf. A006002, A007742, A016061, A027480, A281258.

[n*(n+1)*(4*n+5)/2 : n in [0..50]]; // Wesley Ivan Hurt, Aug 30 2022

Table[n (n + 1) (4 n + 5)/2, {n, 0, 45}] (* or *)
CoefficientList[Series[3 x (3 + x)/(1 - x)^4, {x, 0, 45}], x] (* Michael De Vlieger, Jan 21 2017 *)

concat(0, Vec(3*x*(3 + x) / (1 - x)^4 + O(x^50))) \\ Colin Barker, Jan 21 2017

a(n) = n*(n + 1)*(4*n + 5)/2 \\ Charles R Greathouse IV, Feb 01 2017

a(n) = 2*n^3 + 9*n^2/2 + 5*n/2.
a(n) = 3*A016061(n).
a(n) = A006002(n+1)*(n) - A006002(n)*(n-1).
a(n) = A007742(n)*(n - 1)/2.
From Colin Barker, Jan 21 2017: (Start)
G.f.: 3*x*(3 + x) / (1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. (End)
From Stefano Spezia, Aug 30 2022: (Start)
E.g.f.: exp(x)*x*(18 + 21*x + 4*x^2)/2.
Sum_{n>0} 1/a(n) = 2*(20*log(8) + 10*Pi - 71)/25 = 0.1603805895595720759728288896228498341201... . (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*sqrt(2)*Pi/5 + 4*(3+sqrt(2))*log(2)/5 - 8*sqrt(2)*log(2-sqrt(2))/5 - 178/25. - Amiram Eldar, Sep 22 2022

A367964 Triangle of 2-parameter triangular numbers, read by rows. T(n, k) = (n(n + 1) + k(k + 1)) / 2.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 6, 7, 9, 12, 10, 11, 13, 16, 20, 15, 16, 18, 21, 25, 30, 21, 22, 24, 27, 31, 36, 42, 28, 29, 31, 34, 38, 43, 49, 56, 36, 37, 39, 42, 46, 51, 57, 64, 72, 45, 46, 48, 51, 55, 60, 66, 73, 81, 90, 55, 56, 58, 61, 65, 70, 76, 83, 91, 100, 110

Offset: 0

Peter Luschny, Dec 07 2023

If the rows of the triangle are extended for k > n, the array A144216 is created, which is symmetrical to the main diagonal and therefore contains no new information compared to this triangle.

			Triangle T(n, k) starts:
  0 |  0;
  1 |  1,  2;
  2 |  3,  4,  6;
  3 |  6,  7,  9, 12;
  4 | 10, 11, 13, 16, 20;
  5 | 15, 16, 18, 21, 25, 30;
  6 | 21, 22, 24, 27, 31, 36, 42;
  7 | 28, 29, 31, 34, 38, 43, 49, 56;
  8 | 36, 37, 39, 42, 46, 51, 57, 64, 72;
  9 | 45, 46, 48, 51, 55, 60, 66, 73, 81,  90;
 10 | 55, 56, 58, 61, 65, 70, 76, 83, 91, 100, 110;
.
Start at row 0, column 0 with 0. Go down by adding the column index in step n. At row n, restart the counting and go n steps right by adding the row index in step n, then change direction and go down again by adding the column index. After 3*n steps on this path you are at T(2*n, n) which is 2*triangular(n) + (triangular(2*n) - triangular(n)) = (5*n^2 + 3*n)/2. These are the sliced heptagonal numbers A147875 (see the illustration of Leo Tavares).
.
The equation T(n, k) = (n*(n + 1) + k*(k + 1))/2 can be extended to all n, k in ZZ.
  [n\k] ... -6  -5  -4  -3  -2  -1   0   1   2   3   4   5  ...
  -------------------------------------------------------------
  [-5] ..., 25, 20, 16, 13, 11, 10, 10, 11, 13, 16, 20, 25, ...
  [-4] ..., 21, 16, 12,  9,  7,  6,  6,  7,  9, 12, 16, 21, ...
  [-3] ..., 18, 13,  9,  6,  4,  3,  3,  4,  6,  9, 13, 18, ...
  [-2] ..., 16, 11,  7,  4,  2,  1,  1,  2,  4,  7, 11, 16, ...
  [-1] ..., 15, 10,  6,  3,  1,  0,  0,  1,  3,  6, 10, 15, ...
  [ 0] ..., 15, 10,  6,  3,  1,  0,  0,  1,  3,  6, 10, 15, ...
  [ 1] ..., 16, 11,  7,  4,  2,  1,  1,  2,  4,  7, 11, 16, ...
  [ 2] ..., 18, 13,  9,  6,  4,  3,  3,  4,  6,  9, 13, 18, ...
  [ 3] ..., 21, 16, 12,  9,  7,  6,  6,  7,  9, 12, 16, 21, ...
  [ 4] ..., 25, 20, 16, 13, 11, 10, 10, 11, 13, 16, 20, 25, ...

Columns: A000217, A000124, A152950, A166136, A121263.
Diagonals: A002378, A000290, A002061, A059100, A027689.
Cf. A147875 (T(2*n, n)), A016061 (row sums), A367965 (alternating row sums), A143216 (the multiplicative equivalent), A144216 (extended array).

T := (n, k) -> (n*(n + 1) + k*(k + 1)) / 2:
for n from 0 to 10 do seq(T(n, k), k = 0..n) od;

Module[{n=1},NestList[Append[#+n,n*++n]&,{0},10]] (* or *)
Table[(n(n+1)+k(k+1))/2,{n,0,10},{k,0,n}] (* Paolo Xausa, Dec 07 2023 *)

# A purely additive construction:
from functools import cache
@cache
def a_row(n: int) -> list[int]:
    if n == 0: return [0]
    row = a_row(n - 1) + [0]
    for k in range(n): row[k] += n
    row[n] = row[n - 1] + n
    return row

Recurrence: T(n, n) = n + T(n, n-1) starting with T(0, 0) = 0.
For k <> n: T(n, k) = n + T(n-1, k).
T(n, k) = t(n) + t(k), where t(n) are the triangular numbers A000217.
G.f.: (x + x*(2 - 5*x + x^2)*y + x^4*y^2)/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Dec 07 2023

A376923 T(n, k) = T(n - 1, k) + 2^(n - 1)*T(n - 2, k - 1), if k > 0 and T(n, 0) = 2^n.

Original entry on oeis.org

1, 2, 0, 4, 2, 0, 8, 10, 0, 0, 16, 42, 16, 0, 0, 32, 170, 176, 0, 0, 0, 64, 682, 1520, 512, 0, 0, 0, 128, 2730, 12400, 11776, 0, 0, 0, 0, 256, 10922, 99696, 206336, 65536, 0, 0, 0, 0, 512, 43690, 798576, 3380736, 3080192, 0, 0, 0, 0, 0, 1024, 174762, 6390640, 54425088, 108724224, 33554432, 0, 0, 0, 0, 0

Offset: 0

Thomas Scheuerle, Oct 17 2024

This is the case r = 2 of the more general recurrence: T(n, k, r) = T(n-1, k, r) + r^(n-1)*T(n - 2, k - 1, r), if k > 0 and T(n, 0, r) = 1 + (r^n - 1)/(r - 1) if r > 1. Consider the sequence b(n) = Sum_{k=0..n-1} b(n - k - 1)*T(n - 1, k, r)*(-1)^k, with b(0) = 1. The sequence b(n) will have an ordinary generating function which can be represented as the continued fraction expansion: 1/(1 - x/(1 - r^0*x/(1 - r^1*x/(1 - r^2*x/(1 - r^3*x/(...)))))). In short b(n) will have the ordinary generating function 1/(1-G(x)*x), where G(x) is the generating function of the Carlitz-Riordan q-Catalan numbers for q = r. The Hankel determinant of b(0)..b(2*n) will be r^A016061(n). The Hankel determinant of b(1)..b(2*n+1) will be r^A002412(n).

			Triangle begins:
n\k  0 |  1 |  2 | 3 | 4 | 5
[0]  1,
[1]  2,   0
[2]  4,   2,   0
[3]  8,  10,   0,  0
[4] 16,  42,  16,  0,  0
[5] 32, 170, 176,  0,  0,  0

Cf. A002412, A015083, A016061, A377132.

T(n, k) = if(n < 0, return(0), return(if(k == 0, return(2^n), T(n-1,k) + 2^(n-1)*T(n-2,k-1))))

Column k has o.g.f.: x^(2*k)*2^(k^2)/((1 - 2^(k+1)*x)*Product_{m=1..k}(1 - 2^(m-1)*x)).

A377132 The expansion of 1/(1 - G(x)*x), where G(x) is the ordinary generating function of the Carlitz-Riordan q-Catalan numbers for q = 2 (A015083).

Original entry on oeis.org

1, 1, 2, 6, 28, 228, 3592, 113880, 7267952, 929696784, 237968445472, 121835841547872, 124758916812038592, 255505766282965942848, 1046551115668335283290240, 8573345713494489300568753536, 140465691975467799273799959144192, 4602779760325164559879331800453222656, 301647773810532495378626041621616755442176, 39537576990478498231890121766124629197694682624

Offset: 0

Thomas Scheuerle, Oct 19 2024

nonn

Cf. A002412, A015083, A016061, A376923.

A376923(n, k) = if(n < 0, return(0),return(if(k == 0, return(2^n), T(n-1, k) + 2^(n-1)*T(n-2, k-1) )))
a(n) = if(n==0, 1, sum(k=0, n-1, a(n-k-1)*A376923(n-1, k)*(-1)^k))

O.g.f.: Continued fraction expansion 1/(1 - x/(1 - 2^0*x/(1 - 2^1*x/(1 - 2^2*x/(1 - 2^3*x/(...)))))).
a(n) = Sum_{k=0..n-1} a(n - k - 1)*A376923(n - 1, k)*(-1)^k, with a(0) = 1.
The Hankel determinant of a(0)..a(2*n) is 2^A016061(n). The Hankel determinant of a(1)..a(2*n+1) is 2^A002412(n).

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A135036 Sums of the products of n consecutive pairs of numbers.

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A210220 T(n, k) = -binomial(2*n-k+2, k+1)*hypergeom([2*n-k+3, 1], [k+2], 2). Triangle read by rows, T(n, k) for 1 <= k <= n.

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A240929 Number of 10-digit positive integers in base n where the sum of the first k digits equals the sum of the last k digits.

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A143814 Triangle T(n,m) read along rows: T(n,m) = n^2 - (m+1)^2 for 1<=m

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A185958 Accumulation array of the array max{n,k}, by antidiagonals.

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A240928 Number of 8-digit positive integers in base n where the sum of the first k digits equals the sum of the last k digits.

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A281381 a(n) = n*(n + 1)*(4*n + 5)/2.

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A210220 T(n, k) = -binomial(2n-k+2, k+1)hypergeom([2*n-k+3, 1], [k+2], 2). Triangle read by rows, T(n, k) for 1 <= k <= n.

A281381 a(n) = n(n + 1)(4*n + 5)/2.

A367964 Triangle of 2-parameter triangular numbers, read by rows. T(n, k) = (n(n + 1) + k(k + 1)) / 2.