A227550 -id:A227550 - cp's OEIS frontend

A227791 Central terms of the triangle in A227550.

1, 2, 8, 36, 176, 940, 5568, 37128, 280992, 2410812, 23250080, 249164344, 2934303264, 37617633976, 521009920256, 7748175156240, 123095897716800, 2080205257723740, 37253560076385120, 704703668205036120, 14039778681732928800, 293831851498842784680

Offset: 0

Reinhard Zumkeller, Aug 05 2013

nonn

Cf. A074911, A225621, A227550.

Haskell
```
a227791 n = a227550 (2 * n) n
```

b:= proc(x, y) option remember; `if`(x=0, y!,
      b(x-1, y)+b(sort([x, y-1])[]))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..26);  # Alois P. Heinz, Jul 14 2021

T[n_, 0] := n!; T[n_, n_] := n!;
T[n_, k_] /; 0Jean-François Alcover, Nov 03 2022 *)

a(n) = A074911(2*n,n).

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.

Original entry on oeis.org

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

A074911 Triangle generated by Pascal's rule, except begin and end the n-th row with n!.

Original entry on oeis.org

1, 2, 2, 6, 4, 6, 24, 10, 10, 24, 120, 34, 20, 34, 120, 720, 154, 54, 54, 154, 720, 5040, 874, 208, 108, 208, 874, 5040, 40320, 5914, 1082, 316, 316, 1082, 5914, 40320, 362880, 46234, 6996, 1398, 632, 1398, 6996, 46234, 362880

Offset: 1

Joseph L. Pe, Oct 01 2002

			Triangle begins:
1;
2, 2;
6, 4, 6;
24, 10, 10, 24;
120, 34, 20, 34, 120;
720, 154, 54, 54, 154, 720;
5040, 874, 208, 108, 208, 874, 5040;

Reinhard Zumkeller, Rows n = 1..100 of table, flattened

Cf. A227550, A225621 (central terms).

a074911 n k = a074911_tabl !! (n-1) !! (k-1)
a074911_row n = a074911_tabl !! (n-1)
a074911_tabl = map fst $ iterate
   (\(vs, w:ws) -> (zipWith (+) ([w] ++ vs) (vs ++ [w]), ws))
   ([1], tail a001563_list)
-- Reinhard Zumkeller, Aug 05 2013

T[n_, 1] := n!;
T[n_, n_] := n!;
T[n_, k_] /; 1Jean-François Alcover, Nov 03 2022 *)

t(n, k) = {if (k<1 || k>n, return (0)); if (k==1 || k==n, return (n!)); return (t(n-1, k-1) + t(n-1, k));}
tabl(nn) = {for (n=1, nn, for (k=1, n, print1(t(n, k), ", ");); print(););} \\ Michel Marcus, May 19 2013

More terms from Michel Marcus, May 19 2013

A140710 Number of maximal initial consecutive columns ending at the same level, summed over all deco polyominoes of height n.

Original entry on oeis.org

1, 3, 10, 38, 172, 944, 6208, 47696, 417952, 4101824, 44491648, 528068096, 6804155392, 94559581184, 1409615239168, 22434345998336, 379633330204672, 6805952938041344, 128854632579186688, 2568966172926181376

Offset: 1

Emeric Deutsch, Jun 03 2008

nonn

A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column.

			a(3)=10 because the 6 deco polyominoes of height 3 have columns ending at levels 3, 22, 12, 111, 22, 122, respectively and 1+2+1+3+2+1=10.

G. C. Greubel, Table of n, a(n) for n = 1..440
E. Barcucci, A. Del Lungo and R. Pinzani, "Deco" polyominoes, permutations and random generation, Theoretical Computer Science, 159, 1996, 29-42.

Cf. A000142, A011782, A140709.

Row sums of A227550/2.

[2^(n-1)*(&+[j*Factorial(j)/2^j: j in [1..n-1]]): n in [1..30]]; // G. C. Greubel, May 02 2021

a:=proc(n) options operator, arrow: 2^(n-1)*(1+sum(j^2*factorial(j-1)/2^j, j= 1..n-1)) end proc: seq(a(n),n=1..20);

Table[2^(n-1)*(1 + Sum[j*j!/2^j, {j,n-1}]), {n,30}] (* G. C. Greubel, May 02 2021 *)

[2^(n-1)*sum(j*factorial(j)/2^j for j in (1..n-1)) for n in (1..30)] # G. C. Greubel, May 02 2021

a(n) = 2^(n-1) * (1 + Sum_{j=1..n-1} j*j!/2^j ).
a(n) = (n-1)!*(n-1) + 2*a(n-1) with a(1) = 1.
a(n) = Sum_{k=1..n} k*A140709(n,k).
(1 + x + 2*x^2 + 4*x^3 + 8*x^4 + ...)*(1 + 2*x + 6*x^2 + 24*x^3 + 120*x^4 + ...) = (1 + 3*x + 10*x^2 + 38*x^3 + 172*x^4 + ...) which is (Sum_{n>=0} A011782(n)*x^n) * (Sum_{n>=0} A000142(n+1)*x^n) = Sum_{n>=0} a(n+1)*x^n. - Gary W. Adamson, Feb 24 2012
a(n) = Sum_{j=0..n} (j+1)!*A011782(n-j) = (n+1)! + Sum_{j=0..n-1} 2^(n-k-1)*(j+1)!. - G. C. Greubel, May 03 2021
D-finite with recurrence a(n) +(-n-3)*a(n-1) +3*n*a(n-2) +2*(-n+2)*a(n-3)=0. - R. J. Mathar, Jul 26 2022

A347584 Triangle formed by Pascal's rule, except that the n-th row begins and ends with the n-th Lucas number.

Original entry on oeis.org

2, 1, 1, 3, 2, 3, 4, 5, 5, 4, 7, 9, 10, 9, 7, 11, 16, 19, 19, 16, 11, 18, 27, 35, 38, 35, 27, 18, 29, 45, 62, 73, 73, 62, 45, 29, 47, 74, 107, 135, 146, 135, 107, 74, 47, 76, 121, 181, 242, 281, 281, 242, 181, 121, 76, 123, 197, 302, 423, 523, 562, 523, 423, 302, 197, 123

Offset: 0

Noah Carey and Greg Dresden, Sep 07 2021

nonn

Similar in spirit to the Fibonacci-Pascal triangle A074829, which uses Fibonacci numbers instead of Lucas numbers at the ends of each row.

If we consider the top of the triangle to be the 0th row, then the sum of terms in n-th row is 2*(2^(n+1) - Lucas(n+1)). This sum also equals 2*A027973(n-1) for n>0.

			The first two Lucas numbers (for n=0 and n=1) are 2 and 1, so the first two rows (again, for n=0 and n=1) of the triangle are 2 and 1, 1 respectively.
Triangle begins:
               2;
             1,  1;
           3,  2,  3;
         4,  5,  5,  4;
       7,  9, 10,  9,  7;
    11, 16, 19, 19, 16, 11;
  18, 27, 35, 38, 35, 27, 18;

Index entries for triangles and arrays related to Pascal's triangle

Cf. A000032, A027973.
Cf. A227550, A228196 (general formula).
Fibonacci borders: A074829, A108617, A316938, A316939.

T[n_, 0] := LucasL[n]; T[n_, n_] := LucasL[n];
T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten

a(n) = 2*A074829(n+1) - A108617(n).

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A227791 Central terms of the triangle in A227550.

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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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A074911 Triangle generated by Pascal's rule, except begin and end the n-th row with n!.

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A140710 Number of maximal initial consecutive columns ending at the same level, summed over all deco polyominoes of height n.

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A347584 Triangle formed by Pascal's rule, except that the n-th row begins and ends with the n-th Lucas number.

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