A228576 A triangle formed like generalized Pascal's triangle. The rule is T(n,k) = 2*T(n-1,k-1) + T(n-1,k), the left border is n and the right border is n^2 instead of 1.

0, 1, 1, 2, 3, 4, 3, 7, 10, 9, 4, 13, 24, 29, 16, 5, 21, 50, 77, 74, 25, 6, 31, 92, 177, 228, 173, 36, 7, 43, 154, 361, 582, 629, 382, 49, 8, 57, 240, 669, 1304, 1793, 1640, 813, 64, 9, 73, 354, 1149, 2642, 4401, 5226, 4093, 1690, 81, 10, 91, 500, 1857, 4940, 9685, 14028, 14545, 9876, 3461, 100

Offset: 1

Author

Boris Putievskiy, Aug 26 2013

Keywords

• nonn
• tabl

Examples

			The start of the sequence as triangle array read by rows:
  0;
  1,  1;
  2,  3,  4;
  3,  7, 10,  9;
  4, 13, 24, 29, 16;
  5, 21, 50, 77, 74, 25;
...
		

Links

• Boris Putievskiy, Rows n = 1..140 of triangle, flattened
• Rely Pellicer, David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4
• Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
• Rattanapol Wasutharat and Kantaphon Kuhapatanakul, The Generalized Pascal-Like Triangle and Applications Int. J. Contemp. Math. Sciences, Vol. 7, 2012, no. 41, pp. 1989 - 1992
• Index entries for triangles and arrays related to Pascal's triangle

Crossrefs

Cf. We denote generalized Pascal's like triangle with coefficients a, b and with L(n) on the left border and R(n) on the right border by (a,b,L(n),R(n)). The list of sequences for (1,1,L(n),R(n)) see A228196;

A038207 (1,2,2^n,1), A105728 (1, 2, 1, n+1), A112468 (1,-1,1,1), A112626 (1,2,3^n,1), A119258 (2,1,1,1), A119673 (3,1,1,1), A119725 (3,2,1,1), A119726 (4,2,1,1), A119727 (5,2,1,1), A209705 (2,1,n+1,0);

A002061 (column 2), A000244 (sums of rows r of triangle array - (r-2)(r+1)/2).

Programs

GAP

T:= function(n,k)
    if k=0 then return n;
    elif k=n then return n^2;
    else return 2*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 13 2019

Magma

function T(n,k)
  if k eq 0 then return n;
  elif k eq n then return n^2;
  else return 2*T(n-1,k-1) + T(n-1,k);
  end if;
  return T;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 13 2019

Maple

T := proc(n, k) option remember;
if k = 0 then RETURN(n) fi;
if k = n then RETURN(n^2) fi;
2*T(n-1, k-1) + T(n-1, k) end:
seq(seq(T(n,k),k=0..n),n=0..9);  # Peter Luschny, Aug 26 2013

Mathematica

T[n_, 0]:= n; T[n_, n_]:= n^2; T[n_, k_]:= T[n, k] = 2*T[n-1, k-1]+T[n-1, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 25 2014 *)

PARI

T(n,k) = if(k==0, n, if(k==n, n^2, 2*T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 13 2019

Sage

@CachedFunction
def T(n, k):
    if (k==0): return n
    elif (k==n): return n^2
    else: return 2*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 13 2019

Formula

T(n, k) = 2*T(n-1, k-1) + T(n-1, k) for n,k >=0, with T(n,0) = n, T(n,n) = n^2.

Closed-form formula for generalized Pascal's triangle. Let a,b be any numbers. The rule is T(n, k) = a*T(n-1, k-1) + b*T(n-1, k) for n,k >0. Let L(m) and R(m) be the left border and the right border generalized Pascal's triangle, respectively.

As table read by antidiagonals T(n,k) = Sum_{m1=1..n} a^(n-m1) * b^k*R(m1)*C(n+k-m1-1,n-m1) + Sum_{m2=1..k} a^n*b^(k-m2)*L(m2)*C(n+k-m2-1,k-m2); n,k >=0.

As linear sequence a(n) = Sum_{m1=1..i} a^(i-m1)*b^j*R(m1)*C(i+j-m1-1,i-m1) + Sum_{m2=1..j} a^i*b^(j-m2)*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 a=b=1, then the closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196.

If a=0, then as table read by antidiagonals T(n,k)=b*R(n), as linear sequence a(n)=b*R(i), where i=n-t*(t+1)/2-1, t=floor((-1+sqrt(8*n-7))/2); n>0. The sequence a(n) is the reluctant sequence of sequence b*R(n) - a(n) is triangle array read by rows: row number k coincides with first k elements of the sequence b*R(n). Similarly for b=0, we get T(n,k)=a*L(k).

For this sequence L(m)=m and R(m)=m^2, a=2, b=1. As table read by antidiagonals T(n,k) = Sum_{m1=1..n} 2^(n-m1)*m1^2*C(n+k-m1-1,n-m1) + Sum_{m2=1..k} 2^n*m2*C(n+k-m2-1,k-m2); n,k >=0.

As linear sequence a(n) = Sum_{m1=1..i} 2^(i-m1)*m1^2*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} 2^i*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.