A228053 -id:A228053 - cp's OEIS frontend

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024

Offset: 1

Boris Putievskiy, Aug 15 2013

The third row is (n^4 - n^2 + 24*n + 24)/12.

For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Rely Pellicer and David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4
Index entries for triangles and arrays related to Pascal's triangle

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).

Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

T:= function(n,k)
    if k=0 then return n^2;
    elif k=n then return 2^n;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019

T:= proc(n, k) option remember;
      if k=0 then n^2
    elif k=n then 2^k
    else T(n-1, k-1) + T(n-1, k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019

T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)

T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019

def funcL(n):
   q = n**2
   return q
def funcR(n):
   q = 2**n
   return q
for n in range (1,9871):
   t=int((math.sqrt(8*n-7) - 1)/ 2)
   i=n-t*(t+1)/2-1
   j=(t*t+3*t+4)/2-n-1
   sum1=0
   sum2=0
   for m1 in range (1,i+1):
      sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
   for m2 in range (1,j+1):
      sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
   sum=sum1+sum2

@CachedFunction
def T(n, k):
    if (k==0): return n^2
    elif (k==n): return 2^n
    else: return T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A026637 Triangular array T read by rows: T(n,0) = T(n,n) = 1 for n >= 0, T(n,1) = T(n,n-1) = floor((3*n-1)/2) for n >= 1, otherwise T(n,k) = T(n-1,k-1) + T(n-1,k) for 2 <= k <= n-2, n >= 4.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 5, 8, 5, 1, 1, 7, 13, 13, 7, 1, 1, 8, 20, 26, 20, 8, 1, 1, 10, 28, 46, 46, 28, 10, 1, 1, 11, 38, 74, 92, 74, 38, 11, 1, 1, 13, 49, 112, 166, 166, 112, 49, 13, 1, 1, 14, 62, 161, 278, 332, 278, 161, 62, 14, 1, 1, 16, 76, 223, 439, 610, 610, 439, 223, 76, 16, 1

Offset: 0

Clark Kimberling

T(n, k) = number of paths from (0, 0) to (n-k, k) in directed graph having vertices (i, j) and edges (i, j)-to-(i+1, j) and (i, j)-to-(i, j+1) for i, j >= 0 and edges (i, j)-to-(i+1, j+1) for i=0, j >= 1 and odd and for j=0, i >= 1 and odd.

See A228053 for a sequence with many terms in common with this one. - T. D. Noe, Aug 07 2013

			Triangle begins as:
  1;
  1,  1;
  1,  2,  1;
  1,  4,  4,   1;
  1,  5,  8,   5,   1;
  1,  7, 13,  13,   7,   1;
  1,  8, 20,  26,  20,   8,   1;
  1, 10, 28,  46,  46,  28,  10,   1;
  1, 11, 38,  74,  92,  74,  38,  11,  1;
  1, 13, 49, 112, 166, 166, 112,  49, 13,  1;
  1, 14, 62, 161, 278, 332, 278, 161, 62, 14,  1;

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

Cf. A026638, A026639, A026640, A026641, A026642, A026643, A026644.
Cf. A026966, A026967, A026968, A026969, A026970, A228053.
Sums include: A000007 (alternating sign row), A026644 (row), A026645, A026646, A026647 (diagonal).

a026637 n k = a026637_tabl !! n !! k
a026637_row n = a026637_tabl !! n
a026637_tabl = [1] : [1,1] : map (fst . snd)
   (iterate f (0, ([1,2,1], [0,1,1,0]))) where
   f (i, (xs, ws)) = (1 - i,
     if i == 1 then (ys, ws) else (zipWith (+) ys ws, ws'))
        where ys = zipWith (+) ([0] ++ xs) (xs ++ [0])
              ws' = [0,1,0,0] ++ drop 2 ws
-- Reinhard Zumkeller, Aug 08 2013

function T(n,k) // T = A026637
   if k eq 0 or k eq n then return 1;
   elif k eq 1 or k eq n-1 then return Floor((3*n-1)/2);
   else return T(n-1, k) + T(n-1, k-1);
   end if;
end function;
[T(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jun 28 2024

A026637 := proc(n,k)
      option remember;
      if k=0 or k=n then
        1
    elif k=1 or k=n-1 then
        floor((3*n-1)/2) ;
    elif k <0 or k > n then
        0;
    else
        procname(n-1,k-1)+procname(n-1,k) ;
    end if;
end proc: # R. J. Mathar, Apr 26 2015

T[n_, k_] := T[n, k] = Which[k == 0 || k == n, 1, k == 1 || k == n-1, Floor[(3n-1)/2], k < 0 || k > n, 0, True, T[n-1, k-1] + T[n-1, k]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 30 2018 *)

def T(n,k): # T = A026637
    if k==0 or k==n: return 1
    elif k==1 or k==n-1: return ((3*n-1)//2)
    else: return T(n-1, k) + T(n-1, k-1)
flatten([[T(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Jun 28 2024

From G. C. Greubel, Jun 28 2024: (Start)
T(n, n-k) = T(n, k).
T(2*n-1, n-1) = A026641(n), n >= 1.
Sum_{k=0..n} T(n, k) = A026644(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n). (End)

A227075 A triangle formed like Pascal's triangle, but with 3^n on the borders instead of 1.

Original entry on oeis.org

1, 3, 3, 9, 6, 9, 27, 15, 15, 27, 81, 42, 30, 42, 81, 243, 123, 72, 72, 123, 243, 729, 366, 195, 144, 195, 366, 729, 2187, 1095, 561, 339, 339, 561, 1095, 2187, 6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561, 19683, 9843, 4938, 2556, 1578, 1578, 2556, 4938

Offset: 0

T. D. Noe, Aug 01 2013

All rows except the zeroth are divisible by 3. Is there a closed-form formula for these numbers, like for binomial coefficients?

Let b=3 and T(n,k) = A(n-k,k) be the associated reading of the symmetric array A by antidiagonals, then A(n,k) = sum_{r=1..n} b^r*A178300(n-r,k) + sum_{c=1..k} b^c*A178300(k-c,n). Similarly with b=4 and b=5 for A227074 and A227076. - R. J. Mathar, Aug 10 2013

			Triangle:
1,
3, 3,
9, 6, 9,
27, 15, 15, 27,
81, 42, 30, 42, 81,
243, 123, 72, 72, 123, 243,
729, 366, 195, 144, 195, 366, 729,
2187, 1095, 561, 339, 339, 561, 1095, 2187,
6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A166060 (row sums: 4*3^n - 3*2^n), A227074 (4^n edges), A227076 (5^n edges).

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 3^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

A227074 A triangle formed like Pascal's triangle, but with 4^n on the borders instead of 1.

Original entry on oeis.org

1, 4, 4, 16, 8, 16, 64, 24, 24, 64, 256, 88, 48, 88, 256, 1024, 344, 136, 136, 344, 1024, 4096, 1368, 480, 272, 480, 1368, 4096, 16384, 5464, 1848, 752, 752, 1848, 5464, 16384, 65536, 21848, 7312, 2600, 1504, 2600, 7312, 21848, 65536, 262144, 87384, 29160

Offset: 0

T. D. Noe, Aug 06 2013

All rows except the zeroth are divisible by 4. Is there a closed-form formula for these numbers, like for binomial coefficients?

			Triangle begins:
  1,
  4, 4,
  16, 8, 16,
  64, 24, 24, 64,
  256, 88, 48, 88, 256,
  1024, 344, 136, 136, 344, 1024,
  4096, 1368, 480, 272, 480, 1368, 4096,
  16384, 5464, 1848, 752, 752, 1848, 5464, 16384,
  65536, 21848, 7312, 2600, 1504, 2600, 7312, 21848, 65536

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A165665 (row sums: 3*4^n - 2*2^n), A227075 (3^n edges), A227076 (5^n edges).

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 4^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

A227076 A triangle formed like Pascal's triangle, but with 5^n on the borders instead of 1.

Original entry on oeis.org

1, 5, 5, 25, 10, 25, 125, 35, 35, 125, 625, 160, 70, 160, 625, 3125, 785, 230, 230, 785, 3125, 15625, 3910, 1015, 460, 1015, 3910, 15625, 78125, 19535, 4925, 1475, 1475, 4925, 19535, 78125, 390625, 97660, 24460, 6400, 2950, 6400, 24460, 97660, 390625

Offset: 0

T. D. Noe, Aug 06 2013

All rows except the zeroth are divisible by 5. Is there a closed-form formula for these numbers, as there is for binomial coefficients?

			Triangle begins as:
       1;
       5,     5;
      25,    10,    25;
     125,    35,    35,  125;
     625,   160,    70,  160,  625;
    3125,   785,   230,  230,  785, 3125;
   15625,  3910,  1015,  460, 1015, 3910, 15625;
   78125, 19535,  4925, 1475, 1475, 4925, 19535, 78125;
  390625, 97660, 24460, 6400, 2950, 6400, 24460, 97660, 390625;

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A083585 (row sums: (8*5^n - 5*2^n)/3), A227074 (4^n edges), A227075 (3^n edges).
Cf. A000351.

function T(n,k) // T = A227076
  if k eq 0 or k eq n then return 5^n;
  else return T(n-1,k) + T(n-1,k-1);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 10 2025

A227076 := proc(n,k)
    if k = 0 or k = n then
        5^n ;
    elif k < 0 or k > n then
        0;
    else
        procname(n-1,k)+procname(n-1,k-1) ;
    end if;
end proc: # R. J. Mathar, Aug 09 2013

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 5^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

from sage.all import *
@CachedFunction
def T(n,k): # T = A227076
    if k==0 or k==n: return pow(5,n)
    else: return T(n-1,k) + T(n-1,k-1)
print(flatten([[T(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 10 2025

From R. J. Mathar, Aug 09 2013: (Start)
T(n,0) = 5^n.
T(n,1) = 5*A047850(n-1).
T(n,2) = 5*(5^n/80 + 3*n/4 + 51/16).
T(n,3) = 5*(5^n/320 + 45*n/16 + 3*n^2/8 + 819/64). (End)
Sum_{k=0..n} (-1)^k*T(n, k) = 20*(1+(-1)^n)*A009969(floor((n-1)/2)) - (3/5)*[n = 0]. - G. C. Greubel, Jan 10 2025

cp's OEIS Frontend

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

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A026637 Triangular array T read by rows: T(n,0) = T(n,n) = 1 for n >= 0, T(n,1) = T(n,n-1) = floor((3*n-1)/2) for n >= 1, otherwise T(n,k) = T(n-1,k-1) + T(n-1,k) for 2 <= k <= n-2, n >= 4.

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A227075 A triangle formed like Pascal's triangle, but with 3^n on the borders instead of 1.

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A227074 A triangle formed like Pascal's triangle, but with 4^n on the borders instead of 1.

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A227076 A triangle formed like Pascal's triangle, but with 5^n on the borders instead of 1.

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