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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. 19
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

1,4

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]

Rattanapol Wasutharat, 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

FORMULA

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

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.

EXAMPLE

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;

...

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-Fran├žois Alcover, Feb 25 2014 *)

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).

Sequence in context: A058267 A048259 A198461 * A211507 A283832 A102086

Adjacent sequences:  A228573 A228574 A228575 * A228577 A228578 A228579

KEYWORD

nonn,tabl

AUTHOR

Boris Putievskiy, Aug 26 2013

STATUS

approved

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Last modified February 17 16:03 EST 2018. Contains 299296 sequences. (Running on oeis4.)