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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. 30
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

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

LINKS

Boris Putievskiy, Rows n = 1..140 of triangle, flattened

Rely Pellicer, David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4

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

FORMULA

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

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).

EXAMPLE

The start of the sequence as triangle 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;

PROG

(Python)

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

CROSSREFS

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).

Sequence in context: A133702 A080001 A178938 * A261326 A136743 A011387

Adjacent sequences:  A228193 A228194 A228195 * A228197 A228198 A228199

KEYWORD

nonn,tabl

AUTHOR

Boris Putievskiy, Aug 15 2013

EXTENSIONS

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

STATUS

approved

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Last modified February 24 07:50 EST 2018. Contains 299599 sequences. (Running on oeis4.)