A228196 -id:A228196 - cp's OEIS frontend

A137688 2^A003056: 2^n appears n+1 times.

1, 2, 2, 4, 4, 4, 8, 8, 8, 8, 16, 16, 16, 16, 16, 32, 32, 32, 32, 32, 32, 64, 64, 64, 64, 64, 64, 64, 128, 128, 128, 128, 128, 128, 128, 128, 256, 256, 256, 256, 256, 256, 256, 256, 256, 512, 512, 512, 512, 512, 512, 512, 512, 512, 512, 1024, 1024, 1024, 1024, 1024, 1024

Offset: 0

M. F. Hasler, Feb 06 2008

First differences of A007664.
Viewed as a triangle, it is computed like Pascal's triangle, but with 2^n on the triangle edges. - T. D. Noe, Jul 31 2013
From Paul Curtz, Oct 23 2018: (Start)
Oresme numbers O(n) = n/2^n are an autosequence of the first kind. The corresponding sequence of the second kind is 1/2^n. The difference table is
   1     1/2   1/4   1/8 ...
  -1/2  -1/4  -1/8  -1/16 ...
   1/4   1/8   1/16  1/32 ...
  -1/8  -1/16 -1/32 -1/64 ...
etc.
The denominators on the antidiagonals are a(n). (End)

			Triangle T(n,k) begins:
1
2, 2
4, 4, 4
8, 8, 8, 8
16, 16, 16, 16, 16
32, 32, 32, 32, 32, 32
64, 64, 64, 64, 64, 64, 64
- _Philippe Deléham_, Dec 26 2013

Vincenzo Librandi, Rows n = 0..100, flattened

Cf. A003056, A007664 (gives partial sums).

Flat(List([0..10],n->List([1..n+1],k->2^n))); # Muniru A Asiru, Oct 23 2018

a137688 n = a137688_list !! n
a137688_list = concat $ zipWith ($) (map replicate [1..]) (map (2^) [0..])
-- Reinhard Zumkeller, Mar 18 2011

seq(seq(2^n,k=1..n+1),n=0..10); # Muniru A Asiru, Oct 23 2018

t = {}; Do[r = {}; Do[If[k == 0||k == n, m = 2^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 9}]; t = Flatten[t] (* Vincenzo Librandi, Aug 01 2013 *)

PARI
```
A137688(n)= 1<
				
```

from math import isqrt
def A137688(n): return 1<<(isqrt((n<<3)+1)-1>>1) # Chai Wah Wu, Feb 10 2023

a(n) = 2^[sqrt(2n+2)-.5] = 2^A003056(n) = A007664(n+1) - A007664(n).
Closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013
Viewed as a triangle T(n,k) : T(n,k)=2*T(n-1,k)+2*T(n-1,k-1)-4*T(n-2,k-1), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 26 2013
Sum_{n>=0} 1/a(n) = 4. - Amiram Eldar, Aug 16 2022

A106516 A Pascal-like triangle based on 3^n.

Original entry on oeis.org

1, 3, 1, 9, 4, 1, 27, 13, 5, 1, 81, 40, 18, 6, 1, 243, 121, 58, 24, 7, 1, 729, 364, 179, 82, 31, 8, 1, 2187, 1093, 543, 261, 113, 39, 9, 1, 6561, 3280, 1636, 804, 374, 152, 48, 10, 1, 19683, 9841, 4916, 2440, 1178, 526, 200, 58, 11, 1, 59049, 29524, 14757, 7356, 3618, 1704, 726, 258, 69, 12, 1

Offset: 0

Paul Barry, May 05 2005

Row sums are A027649. Antidiagonal sums are A106517.
From Wolfdieter Lang, Jan 09 2015: (Start)
Alternating row sums give A025192. The A-sequence of this Riordan lower triangular matrix is [1, 1, repeat(0, )] (leading to the Pascal recurrence for T(n,k) for n >= k >= 1. The Z-sequence is [3, repeat(0, )] (leading to the recurrence T(n,0) = 3*T(n-1,0), n >= 1. For A- and Z-sequences see the W. Lang link under A006232.
The inverse of this Riordan matrix is Tinv = ((1 - 2*x)/(1 + x), x/(1 + x)) given as a signed version of A093560: Tinv(n,m) = (-1)^(n-m)*A093560(n,m). (End)

			The triangle T(n,k) begins:
n\k     0     1     2    3    4    5   6   7  8  9 10 ...
0:      1
1:      3     1
2:      9     4     1
3:     27    13     5    1
4:     81    40    18    6    1
5:    243   121    58   24    7    1
6:    729   364   179   82   31    8   1
7:   2187  1093   543  261  113   39   9   1
8:   6561  3280  1636  804  374  152  48  10  1
9:  19683  9841  4916 2440 1178  526 200  58 11  1
10: 59049 29524 14757 7356 3618 1704 726 258 69 12  1
... reformatted and extended. - _Wolfdieter Lang_, Jan 06 2015
----------------------------------------------------------
With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
/ 1        \/1           \/1        \       /1         \
| 3  1     ||0  1        ||0 1      |      | 3  1      |
| 9  4 1   ||0  3  1     ||0 0 1    |... = | 9  7  1   |
|27 13 5 1 ||0  9  4 1   ||0 0 3 1  |      |27 37 12 1 |
|...       ||0 27 13 5 1 ||0 0 9 4 1|      |...        |
|...       ||...         ||...      |      |...        |
= A143495. - _Peter Bala_, Dec 23 2014

Columns 1, 2, 3, 4, 5: A003462, A000340, A052150, A097786, A097787.

Cf. A055248, A143495.

a106516[n_] := Block[{a, k},
a[x_] := Flatten@ Last@ Reap[For[k = -1, k < x, Sow[Binomial[x, k] +
2 Sum[3^(i - 1)*Binomial[x - i, k], {i, 1, x}]], k++]]; Flatten@Array[a, n, 0]]; a106516[11] (* Michael De Vlieger, Dec 23 2014 *)

Riordan array (1/(1-3x), x/(1-x)); Number triangle T(n, 0)=A000244(n), T(n, k)=T(n-1, k-1)+T(n-1, k); T(n, k)=sum{j=0..n, binomial(n, k+j)2^j}.
From Peter Bala, Jul 16 2013: (Start)
T(n,k) = binomial(n,k) + 2*sum {i = 1..n} 3^(i-1)*binomial(n-i,k).
O.g.f.: (1 - t)/( (1 - 3*t)*(1 - (1 + x)*t) ) = 1 + (3 + x)*t + (9 + 4*x + x^2)*t^2 + ....
The n-th row polynomial R(n,x) = 1/(x - 2)*( x*(x + 1)^n - 2*3^n ). (End)
Closed-form formula for arbitrary left and right borders of Pascal-like triangle see A228196. - Boris Putievskiy, Aug 19 2013
T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 3*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, Dec 26 2013
From Peter Bala, Dec 23 2014: (Start)
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(27 + 13*x + 5*x^2/2! + x^3/3!) = 27 + 40*x + 58*x^2/2! + 82*x^3/3! + 113*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
Let M denote the present triangle. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A143495 (but with a different offset). See the Example section. Cf. A055248. (End)
n-th row polynomial R(n, x) = (2*3^n - x*(1 + x)^n)/(2 - x). - Peter Bala, Mar 05 2025

A164844 Generalized Pascal Triangle - satisfying the same recurrence as Pascal's triangle, but with a(n,0)=1 and a(n,n)=10^n (instead of both being 1).

Original entry on oeis.org

1, 1, 10, 1, 11, 100, 1, 12, 111, 1000, 1, 13, 123, 1111, 10000, 1, 14, 136, 1234, 11111, 100000, 1, 15, 150, 1370, 12345, 111111, 1000000, 1, 16, 165, 1520, 13715, 123456, 1111111, 10000000, 1, 17, 181, 1685, 15235, 137171, 1234567, 11111111, 100000000, 1, 18, 198, 1866, 16920, 152406, 1371738, 12345678, 111111111, 1000000000, 1, 19, 216, 2064, 18786, 169326, 1524144, 13717416, 123456789, 1111111111, 10000000000

Offset: 0

Mark Dols, Aug 28 2009

Like with Pascal's triangle, the columns grown polynomially. For example, a(n,1)=10+n, a(n,2)=(1/2)*(180+19n+n^2), a(n,3)=(1/6)*(5400 + 569n + 30n^2 + n^3). Likewise, diagonals grow exponentially: a(n,n)=10^n, a(n,n-1) = (10^n-1) / 9. - Kellen Myers, Jan 24 2010

			Triangle begins:
  1
  1,10
  1,11,100
  1,12,111,1000
  1,13,123,1111,10000
  1,14,136,1234,11111,100000

Robert Israel, Table of n, a(n) for n = 0..10152 (rows 0 to 141, flattened)

Cf. A007318, A093645, A011557, A228196.

f:= proc(n,k) option remember;
if k=n then 10^n elif k=0 then 1 else procname(n-1,k-1)+procname(n-1,k) fi
end proc:
seq(seq(f(n,k),k=0..n),n=0..10); # Robert Israel, Jul 01 2016

f[r_, k_] := Sum[10^i*Binomial[r - i - 1, r - k - 1], {i, 0, k}]; TableForm[Table[f[n, k], {n, 0, 15}, {k, 0, n}]] (* Alex Meiburg, Aug 21 2010 *)
a[n_, k_] := a[n, k] = Piecewise[{{0, k > n || k < 0}, {1, k == 0}, {10^n, k == n}}, a[n - 1, k - 1] + a[n - 1, k]]; TableForm[Table[a[n, k], {n, 0, 10}, {k, 0, n}]] (* Kellen Myers, Jan 24 2010 *)

From  Kellen Myers, Jan 24 2010: (Start)
a(n,k) = Sum_{i = 0..k} 10^i * binomial(n-i-1, n-k-1), for 0<=k<=n.
a(n,0) = 1, a(n,n) = 10^n, a(n,k) = a(n-1,k-1)+a(n-1,k). (End)
T(n,k) = T(n-1,k)+11*T(n-1,k-1)-10*T(n-2,k-1)-10*T(n-2,k-2), T(0,0)=1, T(1,0)=1, T(1,1)=10, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
G.f. of triangle: g(x,y) = (1-xy)/((1-10xy)(1-x-xy)). - Robert Israel, Jul 01 2016

Definition clarified, more terms, and revision of Meiburg's Mathematica code by Kellen Myers, Jan 24 2010

A134636 Triangle formed by Pascal's rule given borders = 2n+1.

Original entry on oeis.org

1, 3, 3, 5, 6, 5, 7, 11, 11, 7, 9, 18, 22, 18, 9, 11, 27, 40, 40, 27, 11, 13, 38, 67, 80, 67, 38, 13, 15, 51, 105, 147, 147, 105, 51, 15, 17, 66, 156, 252, 294, 252, 156, 66, 17, 19, 83, 222, 408, 546, 546, 408, 222, 83, 19, 21, 102, 305, 630, 954, 1092, 954, 630, 305, 102, 21

Offset: 0

Gary W. Adamson, Nov 04 2007

Row sums = A048487: (1, 6, 16, 36, 76, 156, ...).

			First few rows of the triangle:
   1;
   3,  3;
   5,  6,  5;
   7, 11, 11,  7;
   9, 18, 22, 18,  9;
  11, 27, 40, 40, 27, 11;
  13, 38, 67, 80, 67, 38, 13;
  ...

Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A048487, A051601, A051597.

a134636 n k = a134636_tabl !! n !! k
a134636_row n = a134636_tabl !! n
a134636_tabl = iterate (\row -> zipWith (+) ([2] ++ row) (row ++ [2])) [1]
-- Reinhard Zumkeller, Nov 23 2012

T:= proc(n,k) option remember;
      `if`(k<0 or k>n, 0,
      `if`(k=0 or k=n, 2*n+1,
         T(n-1, k-1) + T(n-1, k) ))
    end:
seq(seq(T(n, k), k=0..n), n=0..14);  # Alois P. Heinz, May 26 2013

NestList[Append[Prepend[Map[Apply[Plus, #] &, Partition[#, 2, 1]], #[[1]] + 2], #[[1]] + 2] &, {1}, 10] // Grid  (* Geoffrey Critzer, May 26 2013 *)
T[n_, k_] := Binomial[n, k-1] + Binomial[n, k] + 2 Binomial[n, k+1] + Binomial[n, n-k+1];
Table[T[n, k], {n, 0, 14}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 07 2021 *)

Triangle, given borders = (1, 3, 5, 7, 9, ...); apply Pascal's rule T(n,k) = T(n-1,k) P T(n-1,k-1).
T(n,k) = A051601(n,k) + A051597(n,k); T(n,k) mod 2 = A047999(n,k). - Reinhard Zumkeller, Nov 23 2012
Closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013

Offset changed by Reinhard Zumkeller, Nov 23 2012

A051666 Rows of triangle formed using Pascal's rule except begin and end n-th row with n^2.

Original entry on oeis.org

0, 1, 1, 4, 2, 4, 9, 6, 6, 9, 16, 15, 12, 15, 16, 25, 31, 27, 27, 31, 25, 36, 56, 58, 54, 58, 56, 36, 49, 92, 114, 112, 112, 114, 92, 49, 64, 141, 206, 226, 224, 226, 206, 141, 64, 81, 205, 347, 432, 450, 450, 432, 347, 205, 81, 100, 286, 552, 779, 882, 900, 882, 779

Offset: 0

Asher Auel

Row sums give 6*2^n - 4*n - 6 (A051667).
Central terms: T(2*n,n) = 2 * A220101(n). - Reinhard Zumkeller, Aug 05 2013
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

			Triangle begins:
   0;
   1,  1;
   4,  2,  4;
   9,  6,  6,  9;
  16, 15, 12, 15, 16;
  ...

Reinhard Zumkeller, Rows n = 0..100 of table, flattened

Cf. A007318, A000290, A005408, A228196, A228576.

a051666 n k = a051666_tabl !! n !! k
a051666_row n = a051666_tabl !! n
a051666_tabl = map fst $ iterate
   (\(vs, w:ws) -> (zipWith (+) ([w] ++ vs) (vs ++ [w]), ws))
   ([0], [1, 3 ..])
-- Reinhard Zumkeller, Aug 05 2013

T[n_, 0] := n^2; T[n_, n_] := n^2;
T[n_, k_] := T[n, k] = T[n-1, k-1] + T[n-1, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 13 2018 *)

More terms from James Sellers

A164847 (100^n,1) Pascal triangle.

Original entry on oeis.org

1, 100, 1, 10000, 101, 1, 1000000, 10101, 102, 1, 100000000, 1010101, 10203, 103, 1, 10000000000, 101010101, 1020304, 10306, 104, 1, 1000000000000, 10101010101, 102030405, 1030610, 10410, 105, 1, 100000000000000, 1010101010101

Offset: 0

Mark Dols, Aug 28 2009

(For row n=0,1,2,3,4,5 : Sum of digits = Pascal triangle)

			Triangle begins:
1
100,1
10000,101,1
1000000,10101,102,1
100000000,1010101,10203,103,1
10000000000,101010101,1020304,10306,104,1
1000000000000,10101010101,102030405,1030610,10410,105,1
100000000000000,1010101010101,10203040506,103061015,1041020,10515,106,1
10000000000000000,101010101010101,1020304050607,10306101521,104102035,1051535,10621,107,1,

Cf. A228196

T(n,0)=100^n, T(n,n)=1, T(n,k)=T(n-1,k-1)+T(n-1,k) for 0Philippe Deléham, Dec 27 2013

T(n,k)=101*T(n-1,k)+T(n-1,k-1)-100*T(n-2,k)-100*T(n-2,k-1), T(0,0)=1, T(1,0)=100, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013

A164851 Generalized Lucas-Pascal triangle; (11*10^n, 1).

Original entry on oeis.org

1, 11, 1, 110, 12, 1, 1100, 122, 13, 1, 11000, 1222, 135, 14, 1, 110000, 12222, 1357, 149, 15, 1, 1100000, 122222, 13579, 1506, 164, 16, 1, 11000000, 1222222, 135801, 15085, 1670, 180, 17, 1

Offset: 0

Mark Dols, Aug 28 2009

			Triangle begins:
         1;
        11,      1;
       110,     12,      1;
      1100,    122,     13,     1;
     11000,   1222,    135,    14,    1;
    110000,  12222,   1357,   149,   15,   1;
   1100000, 122222,  13579,  1506,  164,  16,  1;
  11000000,1222222, 135801, 15085, 1670, 180, 17, 1;
  ...

Robert Israel, Table of n, a(n) for n = 0..10010(rows 0 to 140, flattened)

Cf. A029635, A164844, A228196.

G[0]:= 1;
G[1]:= 11+x;
G[2]:= 110+12*x+x^2;
for nn from 3 to 20 do
  G[nn]:= expand((x+11)*G[nn-1]-10*(x+1)*G[nn-2]);
od:
seq(seq(coeff(G[n],x,j),j=0..n),n=0..20); # Robert Israel, Jul 17 2017

T[0, 0] := 1; T[n_, n_] := 1; T[n_, 0] := 11*10^(n - 1); T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k];  Table[T[n, k], {n, 0, 10}, {k, 0, n}] //Flatten (* G. C. Greubel, Dec 22 2017 *)

T(0,0)=1, T(n+1,0)=11*10^n, T(n,n)=1, T(n,k)=T(n-1,k-1)+T(n-1,k) for 0Philippe Deléham, Dec 27 2013

G.f. as triangle: (1-x^2)/((1-10*x)*(1-x-x*y)). - Robert Israel, Jul 17 2017

Initial 1 added by Philippe Deléham, Dec 27 2013

A164855 Generalized Lucas-Pascal triangle: (101*100^n,1).

Original entry on oeis.org

1, 101, 1, 10100, 102, 1, 1010000, 10202, 103, 1, 101000000, 1020202, 10305, 104, 1, 10100000000, 102020202, 1030507, 10409, 105, 1, 1010000000000, 10202020202, 103050709, 1040916, 10514, 106, 1, 101000000000000, 1020202020202

Offset: 0

Mark Dols, Aug 28 2009

			Triangle begins:
1
101,1
10100,102,1
1010000,10202,103,1
101000000,1020202,10305,104,1
10100000000,102020202,1030507,10409,105,1
1010000000000,10202020202,103050709,1040916,10514,106,1
101000000000000,1020202020202,10305070911,104091625,1051430,10620,107,1

Cf. A164847, A164851, A029635, A228196.

A164855 := proc(n,k)
    option remember;
    if n=k then
        1;
    elif k>n or k<0 then
        0;
    elif k = 0 then
        101*100^(n-1) ;
    else
        procname(n-1,k-1)+procname(n-1,k) ;
    end if;
end proc:
seq(seq(A164855(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Nov 03 2016

T(0,0)=1, T(n+1,0)=101*100^n, T(n,n)=1, T(n,k)=T(n-1,k-1)+T(n-1,k) for 0Philippe Deléham, Dec 27 2013

Initial 1 added by Philippe Deléham, Dec 27 2013

A160760 Triangle read by rows, binomial transform of an infinite lower triangular Toeplitz matrix with A078008 in every column.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 9, 5, 3, 1, 27, 14, 8, 4, 1, 81, 41, 22, 12, 5, 1, 243, 122, 63, 34, 17, 6, 1, 729, 365, 185, 97, 51, 23, 7, 1, 2187, 1094, 550, 282, 148, 74, 30, 8, 1, 6561, 3281, 1644, 832, 430, 222, 104, 38, 9, 1

Offset: 0

Gary W. Adamson, May 25 2009

Row sums = A025192: (1, 2, 6, 18, 54, 162, 486, 1458,...).

A triangle formed like Pascal's triangle, but with 3^n for n>=0 on the left border instead of 1. - Boris Putievskiy, Aug 19 2013

			First few rows of the triangle =
     1;
     1,    1;
     3,    2,    1;
     9,    5,    3,   1;
    27,   14,    8,   4,   1;
    81,   41,   22,  12,   5,   1;
   243,  122,   63,  34,  17,   6,   1;
   729,  365,  185,  97,  51,  23,   7,  1;
  2187, 1094,  550, 282, 148,  74,  30,  8, 1;
  6561, 3281, 1644, 832, 430, 222, 104, 38, 9, 1;
...

Cf. A078008, A025192.

A007318 * an infinite lower triangular Toeplitz matrix with A078008 in every column: (1, 0, 2, 2, 6, 10, 22, 42, 86,...).

Closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013

T(7,4) corrected by Georg Fischer, Oct 08 2021

A336014 Irregular triangle read by rows: T(n,1) = T(n,2) = T(n,3n-2) = T(n,3n-1) = n for n >= 1 and T(n,k) = T(n-1,k-2) + T(n-1,k-1) for n > 1, 3 <= k <= 3*(n-1).

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 3, 3, 4, 4, 6, 7, 8, 8, 8, 7, 6, 4, 4, 5, 5, 8, 10, 13, 15, 16, 16, 15, 13, 10, 8, 5, 5, 6, 6, 10, 13, 18, 23, 28, 31, 32, 31, 28, 23, 18, 13, 10, 6, 6, 7, 7, 12, 16, 23, 31, 41, 51, 59, 63, 63, 59, 51, 41, 31, 23, 16, 12, 7, 7

Offset: 1

Lechoslaw Ratajczak, Jul 04 2020

The number of terms in row n is 3*n-1 = A016789(n-1).
The sum of row n is equal to 2*A094002(n-1) = 2*A188589(n).
Fibonacci(n) = T(n+k,n) - T(n+k-1,n) for n >= 1, k = 1,2,3,...
The elements b(k) of the main diagonal, superdiagonal 1 and all subdiagonals have the recursive formula: b(k) = 2*b(k-1) + b(k-2) - 2*b(k-3) - b(k-4) for k > 4.

			Triangle begins:
n\k 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20...
1   1  1
2   2  2  2  2  2
3   3  3  4  4  4  4  3  3
4   4  4  6  7  8  8  8  7  6  4  4
5   5  5  8 10 13 15 16 16 15 13 10  8  5  5
6   6  6 10 13 18 23 28 31 32 31 28 23 18 13 10  6  6
7   7  7 12 16 23 31 41 51 59 63 63 59 51 41 31 23 16 12  7  7
...

Superdiagonal 1 is A029907 for n >= 1.
The main diagonal is A208354 for n >= 1.
Subdiagonal 1 is A102702(n-1) for n >= 1.
Subdiagonal 2 is A206268(n+2) for n >= 1 (conjectured).
Subdiagonal 3 is A191830(n+3) for n >= 1.
Cf. A007318, A051597, A228196.

T(n,k) = T(n,3*k-n) for 1 <= k <= 3*n-1.
T(n,k) = Sum_{u=2*(n-k)+3..2*n-k+1} ceiling(u/2)*A065941(k-2,u-2*(n-k)-3) for n >= 3, 3 <= k <= n.
T(n,k) = Sum_{m1=1..k-n} A208354(m1)*binomial(n-m1-1, k-n-m1) + Sum_{m2=1..2*n-k} A208354(m2)*binomial(n-m2-1, 2*n-k-m2) for n >= 2, n+1 <= k <= 2*n-1.
T(n,k) = Sum_{u=2*(k-2*n)+3..k-n+1} ceiling(u/2)*A065941(3*n-k-2,u-2*(k-2*n)-3) for n>= 3, 2*n <= k <= 3*(n-1).
T(n,k) = A208354(k) + (n-k)*Fibonacci(k) for n >= 3, 3 <= k <= n.
T(n,k) = A029907(k-1) + (n-k+1)*Fibonacci(k) for n >= 2, 3 <= k <= n+1.

cp's OEIS Frontend

A137688 2^A003056: 2^n appears n+1 times.

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A106516 A Pascal-like triangle based on 3^n.

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A164844 Generalized Pascal Triangle - satisfying the same recurrence as Pascal's triangle, but with a(n,0)=1 and a(n,n)=10^n (instead of both being 1).

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A134636 Triangle formed by Pascal's rule given borders = 2n+1.

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A051666 Rows of triangle formed using Pascal's rule except begin and end n-th row with n^2.

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A164847 (100^n,1) Pascal triangle.

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A164851 Generalized Lucas-Pascal triangle; (11*10^n, 1).

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A336014 Irregular triangle read by rows: T(n,1) = T(n,2) = T(n,3n-2) = T(n,3n-1) = n for n >= 1 and T(n,k) = T(n-1,k-2) + T(n-1,k-1) for n > 1, 3 <= k <= 3*(n-1).