A008459 -id:A008459 - cp's OEIS frontend

A113164 a(n) = binomial(6, n)^2.

Original entry on oeis.org

1, 36, 225, 400, 225, 36, 1

Offset: 0

Zerinvary Lajos, Jan 05 2006

The sequence sums to binomial(12, 6).

Cf. A008459, A113162.

Edited by Don Reble, Jan 26 2006

A113899 Number parallelogram based on Pascal's triangle (and special mirror of central and multiply of diagonal).

Original entry on oeis.org

252, 126, 126, 56, 140, 56, 21, 105, 105, 21, 6, 60, 120, 60, 6, 1, 25, 100, 100, 25, 1, 6, 60, 120, 60, 6, 21, 105, 105, 21, 56, 140, 56, 126, 126, 252

Offset: 0

Zerinvary Lajos, Jan 29 2006, May 28 2007

.............................C(0,0)*C(10,5)
......................C(1,0)*C(9,5)...C(1,1)*C(9,4)
...............C(2,0)*C(8,5)...C(2,1)*C(8,4)...C(2,2)*C(8,3)
........C(3,0)*C(7,5)...C(3,1)*C(7,4)...C(3,2)*C(7,3)...C(3,3)*C(7,2)
...C(4,0)*C(6,5)...C(4,1)*C(6,4)...C(4,2)*C(6,3)...C(4,3)*C(8,2)...C(4,4)*C(6,1)
C(5,0)*C(5,5)...C(5,1)*C(5,4)...C(5,2)*C(5,3)...C(5,3)*C(5,2)...C(5,4)*C(5,1)...C(5,5)*C(5,0)
...C(6,1)*C(4,4)...C(4,1)*C(6,4)...C(4,2)*C(6,3)...C(4,3)*C(8,2)...C(6,5)*C(4,0)
........C(7,2)*C(3,3)...C(7,3)*C(3,2)...C(7,4)*C(3,1)...C(7,5)*C(3,0)
...............C(8,3)*C(2,2)...C(8,4)*C(2,1)...C(8,5)*C(2,0)
......................C(9,4)*C(1,1)...C(9,5)*C(1,0)
.............................C(10,5)*C(0,0)
"m" matching: analog (permutations with exactly "m" fixed points.
if aaaaabbbbb (a 5 letters b 5 letters) permutations compared aaaaaaaaaa (a 10 times letters) or compared bbbbbbbbbb (b 10 times letters then 252 "5" matching. ("5" matching: analog (permutations with exactly 5 fixed points.)
If aaaaabbbbb (a 5 letters b 5 letters) permutations compared aaaaabbbbb (a 5 times letters b 5 times letters)then 1 "0" matching), 25 "2"matching 100 "4" matching, 100 "6" matching, 25 "8" matching and 1 "10" matching.(A008459 formatted as a triangular array: 6.rows)
If aaaaabbbbb (a 5 letters b 5 letters) permutations compared abbbbbbbbb (a 1 times letters b 9 times letters) or aaaaaaaaab (a 9 times letters b 1 times letters) then 126 "4" and 126 "6" matching.
etc...
matching equivalent "fixed-point"
example:
arrangement relevant!
compared
letters
times
matching:0.....1.....2.....3.....4.....5.....6.....7.....8.....9.....10
compared.
letters..
times....
.a..b
10..0.................................252..............................
.9..1...........................126.........126........................
.8..2......................56.........140..........56..................
.7..3................21.........105.........105..........21............
.6..4..........6...........60.........120..........60..........6.......
.5..5....1...........25.........100.........100..........25...........1
.4..6..........6...........60.........120..........60..........6.......
.3..7................21.........105.........105..........21............
.2..8......................56.........140..........56..................
.1..9...........................126.........126........................
0..10..................................252.............................
matching.0.....1.....2.....3.....4.....5.....6.....7.....8.....9.....10
The Maple code produces
252, 126, 56, 21, 6, 1
126, 140, 105, 60, 25, 6
56, 105, 120, 100, 60, 21
21, 60, 100, 120, 105, 56
6, 25, 60, 105, 140, 126
1, 6, 21, 56, 126, 252
which is the table rotated right by Pi/4.

Cf. A113162, A113163, A113164.

for n from 0 to 5 do seq(binomial(i,n)*binomial(10-i,5-n), i=0+n..10-5+n ); # Zerinvary Lajos, Mar 31 2009

A116647 Triangle of number of partitions that fit in an n X n box (but not in an (n-1) X (n-1) box) with Durfee square k.

Original entry on oeis.org

1, 3, 1, 5, 8, 1, 7, 27, 15, 1, 9, 64, 84, 24, 1, 11, 125, 300, 200, 35, 1, 13, 216, 825, 1000, 405, 48, 1, 15, 343, 1911, 3675, 2695, 735, 63, 1, 17, 512, 3920, 10976, 12740, 6272, 1232, 80, 1, 19, 729, 7344, 28224, 47628, 37044, 13104, 1944, 99, 1, 21, 1000, 12825

Offset: 1

Franklin T. Adams-Watters, Feb 20 2006

			Triangle begins
   1;
   3,   1;
   5,   8,   1;
   7,  27,  15,   1;
   9,  64,  84,  24,   1;
  11, 125, 300, 200,  35,   1;

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Eric Weisstein's World of Mathematics, Durfee Square.

Cf. A008459; row sums A051924.

Table[Binomial[n, k]^2 - Binomial[n - 1, k], {n, 1, 10}, {k, 1, n}] // Flatten (* G. C. Greubel, Nov 20 2017 *)

for(n=1,10, for(k=0,n, print1(binomial(n,k)^2 - binomial(n-1,k)^2, ", "))) \\ G. C. Greubel, Nov 20 2017

T(n,k) = binomial(n,k)^2 - binomial(n-1,k)^2.

A120406 Triangle read by rows: related to series expansion of the square root of 2 linear factors.

Original entry on oeis.org

1, 2, 2, 5, 6, 5, 14, 18, 18, 14, 42, 56, 60, 56, 42, 132, 180, 200, 200, 180, 132, 429, 594, 675, 700, 675, 594, 429, 1430, 2002, 2310, 2450, 2450, 2310, 2002, 1430, 4862, 6864, 8008, 8624, 8820, 8624, 8008, 6864, 4862

Offset: 0

David Callan, Jul 03 2006

The numbers T(n,k) arise in the expansion of the square root of 2 generic linear factors: 1 - sqrt((1-a*x)*(1-b*x)) = (a+b)*x/2 + (1/8)*(b-a)^2*x^2*Sum_{n>=0} (Sum_{k=0..n} T(n,k)*a^k*b^(n-k))*(x/4)^n. (The g.f. below simply reformulates this fact.) A combinatorial interpretation of T(n,k) would be very interesting.

			Table begins
  \ k..0....1....2....3....4....5....6
  n
  0 |..1
  1 |..2....2
  2 |..5....6....5
  3 |.14...18...18...14
  4 |.42...56...60...56...42
  5 |132..180..200..200..180..132
  6 |429..594..675..700..675..594..429

Column k=0 is the Catalan numbers A000108 (offset). The middle-of-row entries form A005566. Cf. A067804.

Table[2 Binomial[n,k]^2 Binomial[2n+2,n]/ Binomial[2n+2,2k+1],{n,0,9},{k,0,n}]

solve(A=x*(A^2*y^2-2*A^2*y-2*A*y+A^2-2*A+1),A); /* Vladimir Kruchinin, Oct 24 2020 */

T(n,k) = 2*binomial(n,k)^2*binomial(2n+2,n)/binomial(2n+2,2k+1). This shows that T(n,k) is positive and the rows are symmetric.
T(n,k) = (k+1)*CatalanNumber(n+1) - 2*Sum_{j=0..k-1} (k-j)*CatalanNumber(j)*CatalanNumber(n-j). This shows that T(n,k) is an integer.
G.f.: F(x,y):=Sum_{n>=0, k=0..n} T(n,k) x^n y^k is given by F(x,y) = ( 1-2x-2x*y-sqrt(1-4x)*sqrt(1-4x*y) )/( 2x^2*(1-y)^2 ). This shows that the row sums are the powers of 4 (A000302) because lim_{y->1} F(x,y) = 1/(1-4x).
1 + x*(d/dx)(log(F(x,y))) = 1 + (2 + 2*y)*x + (6 + 4*y + 6*y^2)*x^2 + ... is the o.g.f. for A067804. - Peter Bala, Jul 17 2015
G.f. A(x,y) = -G(-x,y), G(x,y) satisfies G(x,y) = x/A008459(G(x,y))^2. - Vladimir Kruchinin, Oct 24 2020

A144404 Triangle T(n,k) = 3*binomial(n, k)^2 - binomial(n, k) - 1, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 9, 1, 1, 23, 23, 1, 1, 43, 101, 43, 1, 1, 69, 289, 289, 69, 1, 1, 101, 659, 1179, 659, 101, 1, 1, 139, 1301, 3639, 3639, 1301, 139, 1, 1, 183, 2323, 9351, 14629, 9351, 2323, 183, 1, 1, 233, 3851, 21083, 47501, 47501, 21083, 3851, 233, 1, 1, 289, 6029, 43079, 132089, 190259, 132089, 43079, 6029, 289, 1

Offset: 0

Roger L. Bagula and Gary W. Adamson, Oct 03 2008

			Triangle begins as:
  1;
  1,   1;
  1,   9,    1;
  1,  23,   23,     1;
  1,  43,  101,    43,      1;
  1,  69,  289,   289,     69,      1;
  1, 101,  659,  1179,    659,    101,      1;
  1, 139, 1301,  3639,   3639,   1301,    139,     1;
  1, 183, 2323,  9351,  14629,   9351,   2323,   183,    1;
  1, 233, 3851, 21083,  47501,  47501,  21083,  3851,  233,   1;
  1, 289, 6029, 43079, 132089, 190259, 132089, 43079, 6029, 289, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000984, A007318, A008459, A144390, A144405.

[3*Binomial(n, k)^2 -Binomial(n, k) -1: k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 27 2021

T:= (n,m) -> 3*Binomial(n,m)^2 - Binomial(n,m)-1:
seq(seq(T(n,m),m=0..n),n=0..10); # Robert Israel, Jul 11 2016

Table[3*Binomial[n,k]^2 -Binomial[n,k] -1, {n,0,12}, {k,0,n}]//Flatten

flatten([[3*binomial(n, k)^2 -binomial(n, k) -1 for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 27 2021

From Robert Israel, Jul 11 2016: (Start)
T(n,m) = A144390(A007318(n,m)) = 3*A008459(n,m) - A007318(n,m).
Row sums: 3*binomial(2*n,n) - 2^n - n - 1.
G.f. as triangle: g(x,y) = 3/sqrt(1-2*x-2*x*y+x^2-2*x^2*y+x^2*y^2) - 1/(1-x-x*y)+1/((1-x)*(1-x*y)). (End)

Offset changed by Robert Israel, Jul 11 2016

A204621 Triangle read by rows: coordinator triangle for lattice A*_n.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 5, 5, 1, 1, 6, 16, 6, 1, 1, 7, 22, 22, 7, 1, 1, 8, 29, 64, 29, 8, 1, 1, 9, 37, 93, 93, 37, 9, 1, 1, 10, 46, 130, 256, 130, 46, 10, 1, 1, 11, 56, 176, 386, 386, 176, 56, 11, 1, 1, 12, 67, 232, 562, 1024, 562, 232, 67, 12, 1

Offset: 0

N. J. A. Sloane, Jan 17 2012

			Triangle begins:
                   1
                1    1
              1    4    1
            1    5    5    1
          1    6    16    6    1
        1    7    22    22    7    1
      1    8    29    64    29    8    1
    1    9    37    93    93    37    9    1
  1    10    46    130    256    130    46    10    1
1     11    56    176    386    386    176    56    11    1
...

Muniru A Asiru, Rows n=0..100 of triangle, flattened
J. H. Conway and N. J. A. Sloane, Low-dimensional lattices. VII Coordination sequences, Proc. R. Soc. Lond. A 453 (1997), 2369-2389.
Hidefumi Ohsugi, Akiyoshi Tsuchiya, The h*-polynomials of locally anti-blocking lattice polytopes and their gamma-positivity, arXiv:1906.04719 [math.CO], 2019.
Charles M. Wang, Josephine Yu, Toric h-vectors and Chow Betti Numbers of Dual Hypersimplices, arXiv:1707.04581 [math.CO], 2017.

The triangle for Z^n is A007318, A_n is A008459, D_n is A108558, D*_n is A008518.

T(2n,n) gives A000302.

Flat(List([0..10],n->List([0..n],k->Sum([0..Minimum(k,n-k)],i->Binomial(n+1,i))))); # Muniru A Asiru, Dec 14 2018

T[n_, k_] := Sum[Binomial[n+1, i] , {i, 0, Min[k, n-k]}]; Table[T[n,k], {n,0,10}, {k,0,n}] // Flatten (* Amiram Eldar, Dec 14 2018 *)

T(n, k) = Sum_{i=0..min(k,n-k)} binomial(n+1,i). [Wang and Yu, Theorem 4.1] - Eric M. Schmidt, Dec 07 2017

A306463 a(n) = Sum_{k=0..n} Sum_{m=0..floor(k/2)} binomial(k-m, m)*binomial(n-k, k-m)^2.

Original entry on oeis.org

1, 1, 2, 6, 15, 37, 98, 262, 699, 1883, 5110, 13918, 38045, 104355, 287028, 791320, 2186209, 6051113, 16776022, 46577806, 129491865, 360432855, 1004332322, 2801307498, 7820572153, 21851390549, 61101872126, 170977916730, 478755116117, 1341389394715, 3760507521800

Offset: 0

Vladimir Kruchinin, Feb 17 2019

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..2202

Cf. A008459, A306504.

a(n):=sum(sum(binomial(k-m,m)*binomial(n-k,k-m)^2,m,0,k/2),k,0,n);

a(n) = sum(k=0, n, sum(m=0, k\2, binomial(k-m, m)*binomial(n-k, k-m)^2)); \\ Michel Marcus, Feb 18 2019

N=66; x='x+O('x^N); Vec(1/sqrt(x^6+2*x^5-x^4-4*x^3-x^2-2*x+1)) \\ Seiichi Manyama, Feb 20 2019

G.f.: 1/sqrt(x^6 + 2*x^5 - x^4 - 4*x^3 - x^2 - 2*x + 1).

D-finite with recurrence: n*a(n) +(-2*n+1)*a(n-1) +(-n+1)*a(n-2) +2*(-2*n+3)*a(n-3) +(-n+2)*a(n-4) +(2*n-5)*a(n-5) +(n-3)*a(n-6)=0. - R. J. Mathar, Jan 16 2020

A142470 Triangle T(n, k) = ( (k+2)/(2binomial(k+2, 2)^2) )binomial(n, k)^2binomial(n+1, k)binomial(n+2, k), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 30, 30, 1, 1, 80, 300, 80, 1, 1, 175, 1750, 1750, 175, 1, 1, 336, 7350, 19600, 7350, 336, 1, 1, 588, 24696, 144060, 144060, 24696, 588, 1, 1, 960, 70560, 790272, 1728720, 790272, 70560, 960, 1, 1, 1485, 178200, 3492720, 14669424, 14669424, 3492720, 178200, 1485, 1

Offset: 0

Roger L. Bagula, Sep 20 2008

Row sums are 1, 2, 10, 62, 462, 3852, 34974, 338690, 3452306, 36683660, 403472368, ...
From Peter Bala, May 08 2012: (Start)
Define the action of the operator L on a sequence { a(i) }{0<=i<=n} by L{ a(i) }{0<=i<=n} = { a(i)^2 - a(i-1)*a(i+1) }_{0<=i<=n} with the conventions a(-1) = a(n+1) = 0. Extend the action of L to a lower triangular array T by letting L act on the rows of T. Then L acting on Pascal's triangle A007318 produces the triangle of Narayana numbers A001263 and L applied to A001263 produces the present triangle.
Since the Narayana polynomials are real-rooted it follows by a theorem of Branden that the row polynomials of this array are also real-rooted.
(End)

			The triangle begins as:
  1;
  1,    1;
  1,    8,      1;
  1,   30,     30,       1;
  1,   80,    300,      80,        1;
  1,  175,   1750,    1750,      175,        1;
  1,  336,   7350,   19600,     7350,      336,       1;
  1,  588,  24696,  144060,   144060,    24696,     588,      1;
  1,  960,  70560,  790272,  1728720,   790272,   70560,    960,    1;
  1, 1485, 178200, 3492720, 14669424, 14669424, 3492720, 178200, 1485, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Petter Brändén, Iterated sequences and the geometry of zeros, arXiv:0909.1927 [math.CO], 2009-2010; J. Reine Angew. Math. 658 (2011), 115-131.

Cf. A001263, A008459, A056939.

A142470:= func< n,k | ( (k+2)/(2*Binomial(k+2, 2)^2) )*Binomial(n, k)^2*Binomial(n+1, k)*Binomial(n+2, k) >;
[A142470(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 03 2021

f[n_, k_]:= f[n, k]= Binomial[n, k]*Product[j!*(n+j)!/((k+j)!*(n-k+j)!), {j,1,2}];
T[n_, k_]:= Binomial[n, k]*f[n, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* modified by G. C. Greubel, Apr 03 2021 *)

def A142470(n, k): return (2/((k+1)^2*(k+2)))*Binomial(n, k)^2*Binomial(n+1, k)*Binomial(n+2, k)
flatten([[A142470(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 03 2021

Let f(n, k) = binomial(n, k)*Product_{j=1.2} ( j!*(n+j)!/((k+j)!*(n-k+j)!) ), then T(n, k) = 2^(k-n)*f(n, k)*Sum_{j=k..n} binomial(n, j)*binomial(j, k) = binomial(n, k)*f(n, k).
From Peter Bala, May 08 2012: (Start)
T(n, k) = C(n, k)^2 * Product {i=1..2} i!*(n+i)!/((k+i)!*(n-k+i)!) = C(n, k)*C(n+2, k)*C(n+2, k+1)*C(n+2, k+2)/(C(n+2, 1)*C(n+2, 2)).
T(n, k) = 2/((n+1)*(n+2)*(n+3))*C(n, k)*C(n+1, k)*C(n+2, k+2)*C(n+3, k+1) = C(n, k)*A056939(n, k).
(End)
T(n, k) = ( (k+2)/(2*binomial(k+2, 2)^2) )*binomial(n, k)^2*binomial(n+1, k)*binomial(n+2, k). - G. C. Greubel, Apr 03 2021

Edited by G. C. Greubel, Apr 03 2021

A174689 Triangle T(n, k) = n! * binomial(n, k)^2 - n! + 1, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 49, 49, 1, 1, 361, 841, 361, 1, 1, 2881, 11881, 11881, 2881, 1, 1, 25201, 161281, 287281, 161281, 25201, 1, 1, 241921, 2217601, 6168961, 6168961, 2217601, 241921, 1, 1, 2540161, 31570561, 126403201, 197527681, 126403201, 31570561, 2540161, 1

Offset: 0

Roger L. Bagula, Mar 27 2010

			Triangle begins as:
  1;
  1,       1;
  1,       7,        1;
  1,      49,       49,         1;
  1,     361,      841,       361,         1;
  1,    2881,    11881,     11881,      2881,         1;
  1,   25201,   161281,    287281,    161281,     25201,        1;
  1,  241921,  2217601,   6168961,   6168961,   2217601,   241921,       1;
  1, 2540161, 31570561, 126403201, 197527681, 126403201, 31570561, 2540161, 1;

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A000108, A008459, A174690.

[Factorial(n)*(Binomial(n, k)^2 -1) + 1: k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 10 2021

T[n_, k_]:= n!*Binomial[n, k]^2 - n! + 1;
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten

flatten([[factorial(n)*(binomial(n, k)^2 -1) + 1 for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 10 2021

T(n, k) = n! * binomial(n, k)^2 - n! + 1.
From G. C. Greubel, Feb 10 2021: (Start)
T(n, k) = n! * ( A008459(n, k) - 1 ) + 1.
Sum_{k=0..n} T(n, k) = (n+1)*( n!*( C_{n} - 1 ) + 1 ) = (n+1)*( n!*( A000108(n) - 1 ) + 1). (End)

Edited by G. C. Greubel, Feb 10 2021

A303987 Triangle read by rows: T(n, k) = (binomial(n,k)*binomial(n+k,k))^2 = A063007(n, k)^2, for n >= 0, k = 0..n.

Original entry on oeis.org

1, 1, 4, 1, 36, 36, 1, 144, 900, 400, 1, 400, 8100, 19600, 4900, 1, 900, 44100, 313600, 396900, 63504, 1, 1764, 176400, 2822400, 9922500, 7683984, 853776, 1, 3136, 571536, 17640000, 133402500, 276623424, 144288144, 11778624, 1, 5184, 1587600, 85377600, 1200622500, 5194373184, 7070119056, 2650190400, 165636900

Offset: 0

Wolfdieter Lang, May 14 2018

The row sums of this triangle are b(n) = A005259(n), for n >= 0. This sequence b was used in R. Apéry's 1979 proof of the irrationality of Zeta(3). See A005259 for references and links.

Row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k = hypergeometric([-n, -n, n+1, n+1], [1, 1, 1], x), hence b(n) = hypergeometric([-n, -n, n+1, n+1], [1, 1, 1], 1) (see the formula in A005259 given by K. A. Penson. This is the solution to Exercise 2.14 of the Koepf reference given there, p. 29).

			The triangle T begins:
n\k  0    1       2        3          4          5          6          7 ...
0:   1
1:   1    4
2:   1   36      36
3:   1  144     900      400
4:   1  400    8100    19600       4900
5:   1  900   44100   313600     396900      63504
6:   1 1764  176400  2822400    9922500    7683984     853776
7:   1 3136  571536 17640000  133402500  276623424  144288144   11778624
----------------------------------------------------------------------------
row n = 8:   1 5184 1587600 85377600 1200622500 5194373184 7070119056 2650190400 165636900,
row n = 9: 1 8100 3920400 341510400 8116208100 63631071504 176752976400 169612185600 47869064100 2363904400,
row n = 10: 1 12100 8820900 1177862400 44188244100 572679643536 2828047622400 5446435737600 3877394192100 853369488400 34134779536.
...

Cf. A000984, A002894, A005259, A008459, A063007.

The column sequences (without zeros) are A000012, A035287(n+1) = 4*A000217(n)^2, 36*A288876, 400*A000579(n+6)^2, 4900*A000581(n+8)^2, 63504*A001287(n+10)^2, ...

Flat(List([0..10],n->List([0..n],k->(Binomial(n,k)*Binomial(n+k,k))^2))); # Muniru A Asiru, May 15 2018

T[n_, k_] := (Gamma[k + n + 1]/(Gamma[k + 1]^2*Gamma[-k + n + 1]))^2;
Flatten[Table[T[n, k], {n, 0, 8}, {k, 0, n}]] (* Peter Luschny, May 14 2018 *)

T(n, k) = (binomial(n,k)*binomial(n+k,k))^2 = A063007(n, k)^2, for n >= 0 and k = 0..n.
T(n, k) = (binomial(n+k, 2*k)*cbi(k))^2, with cbi(k) = A000984(k) = binomial(2*k, k), and cbi(k)^2 = A002894(k).
G.f. for column sequences (without leading zeros):
  cbi(k)^2*P2(2*k, x)/(1 - x)^(4*k+1), with the row polynomials of A008459 (Pascal entries squared) P2(2*k, x) = Sum_{j=0..2*k} A008459(2*k, j)*x^j. For a proof see the general comment in A288876 on the diagonals and columns of A008459.

cp's OEIS Frontend

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