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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A055248 Triangle of partial row sums of triangle A007318(n,m) (Pascal's triangle). Triangle A008949 read backwards. Riordan (1/(1-2x), x/(1-x)).

Original entry on oeis.org

1, 2, 1, 4, 3, 1, 8, 7, 4, 1, 16, 15, 11, 5, 1, 32, 31, 26, 16, 6, 1, 64, 63, 57, 42, 22, 7, 1, 128, 127, 120, 99, 64, 29, 8, 1, 256, 255, 247, 219, 163, 93, 37, 9, 1, 512, 511, 502, 466, 382, 256, 130, 46, 10, 1, 1024, 1023, 1013, 968, 848, 638, 386, 176, 56, 11, 1
Offset: 0

Views

Author

Wolfdieter Lang, May 26 2000

Keywords

Comments

In the language of the Shapiro et al. reference (also given in A053121) such a lower triangular (ordinary) convolution array, considered as matrix, belongs to the Riordan-group. The g.f. for the row polynomials p(n,x) (increasing powers of x) is 1/((1-2*z)*(1-x*z/(1-z))).
Binomial transform of the all 1's triangle: as a Riordan array, it factors to give (1/(1-x),x/(1-x))(1/(1-x),x). Viewed as a number square read by antidiagonals, it has T(n,k) = Sum_{j=0..n} binomial(n+k,n-j) and is then the binomial transform of the Whitney square A004070. - Paul Barry, Feb 03 2005
Riordan array (1/(1-2x), x/(1-x)). Antidiagonal sums are A027934(n+1), n >= 0. - Paul Barry, Jan 30 2005; edited by Wolfdieter Lang, Jan 09 2015
Eigensequence of the triangle = A005493: (1, 3, 10, 37, 151, 674, ...); row sums of triangles A011971 and A159573. - Gary W. Adamson, Apr 16 2009
Read as a square array, this is the generalized Riordan array ( 1/(1 - 2*x), 1/(1 - x) ) as defined in the Bala link (p. 5), which factorizes as ( 1/(1 - x), x/(1 - x) )*( 1/(1 - x), x )*( 1, 1 + x ) = P*U*transpose(P), where P denotes Pascal's triangle, A007318, and U is the lower unit triangular array with 1's on or below the main diagonal. - Peter Bala, Jan 13 2016

Examples

			The triangle a(n,m) begins:
n\m    0    1    2   3   4   5   6   7  8  9 10 ...
0:     1
1:     2    1
2:     4    3    1
3:     8    7    4   1
4:    16   15   11   5   1
5:    32   31   26  16   6   1
6:    64   63   57  42  22   7   1
7:   128  127  120  99  64  29   8   1
8:   256  255  247 219 163  93  37   9  1
9:   512  511  502 466 382 256 130  46 10  1
10: 1024 1023 1013 968 848 638 386 176 56 11  1
... Reformatted. - _Wolfdieter Lang_, Jan 09 2015
Fourth row polynomial (n=3): p(3,x)= 8 + 7*x + 4*x^2 + x^3.
The matrix inverse starts
   1;
  -2,   1;
   2,  -3,   1;
  -2,   5,  -4,    1;
   2,  -7,   9,   -5,    1;
  -2,   9, -16,   14,   -6,    1;
   2, -11,  25,-  30,   20,   -7,    1;
  -2,  13, -36,   55,  -50,   27,   -8,    1;
   2, -15,  49,  -91,  105,  -77,   35,   -9,  1;
  -2,  17, -64,  140, -196,  182, -112,   44, -10,   1;
   2, -19,  81, -204,  336, -378,  294, -156,  54, -11, 1;
   ...
which may be related to A029653. - _R. J. Mathar_, Mar 29 2013
From _Peter Bala_, Dec 23 2014: (Start)
With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
/1      \ /1        \ /1       \       /1       \
|2 1     ||0 1       ||0 1      |      |2  1     |
|4 3 1   ||0 2 1     ||0 0 1    |... = |4  5 1   |
|8 7 4 1 ||0 4 3 1   ||0 0 2 1  |      |8 19 9 1 |
|...     ||0 8 7 4 1 ||0 0 4 3 1|      |...      |
|...     ||...       ||...      |      |         |
= A143494. (End)
Matrix factorization of square array as P*U*transpose(P):
/1      \ /1        \ /1 1 1 1 ...\    /1  1  1  1 ...\
|1 1     ||1 1       ||0 1 2 3 ... |   |2  3  4  5 ... |
|1 2 1   ||1 1 1     ||0 0 1 3 ... | = |4  7 11 16 ... |
|1 3 3 1 ||1 1 1 1   ||0 0 0 1 ... |   |8 15 26 42 ... |
|...     ||...       ||...         |   |...            |
- _Peter Bala_, Jan 13 2016
		

Crossrefs

Column sequences: A000079 (powers of 2, m=0), A000225 (m=1), A000295 (m=2), A002662 (m=3), A002663 (m=4), A002664 (m=5), A035038 (m=6), A035039 (m=7), A035040 (m=8), A035041 (m=9), A035042 (m=10).
Row sums: A001792(n) = A055249(n, 0).
Alternating row sums: A011782.
Cf. A011971, A159573. - Gary W. Adamson, Apr 16 2009

Programs

  • Haskell
    a055248 n k = a055248_tabl !! n !! k
    a055248_row n = a055248_tabl !! n
    a055248_tabl = map reverse a008949_tabl
    -- Reinhard Zumkeller, Jun 20 2015
  • Maple
    T := (n,k) -> 2^n - (1/2)*binomial(n, k-1)*hypergeom([1, n + 1], [n-k + 2], 1/2).
    seq(seq(simplify(T(n,k)), k=0..n),n=0..10); # Peter Luschny, Oct 10 2019
  • Mathematica
    a[n_, m_] := Sum[ Binomial[n, m + j], {j, 0, n}]; Table[a[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 05 2013, after Paul Barry *)
    T[n_, k_] := Binomial[n, k] * Hypergeometric2F1[1, k - n, k + 1, -1];
    Flatten[Table[T[n, k], {n, 0, 7}, {k, 0, n}]]  (* Peter Luschny, Oct 06 2023 *)

Formula

a(n, m) = A008949(n, n-m), if n > m >= 0.
a(n, m) = Sum_{k=m..n} A007318(n, k) (partial row sums in columns m).
Column m recursion: a(n, m) = Sum_{j=m..n-1} a(j, m) + A007318(n, m) if n >= m >= 0, a(n, m) := 0 if n
G.f. for column m: (1/(1-2*x))*(x/(1-x))^m, m >= 0.
a(n, m) = Sum_{j=0..n} binomial(n, m+j). - Paul Barry, Feb 03 2005
Inverse binomial transform (by columns) of A112626. - Ross La Haye, Dec 31 2006
T(2n,n) = A032443(n). - Philippe Deléham, Sep 16 2009
From Peter Bala, Dec 23 2014: (Start)
Exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(8 + 7*x + 4*x^2/2! + x^3/3!) = 8 + 15*x + 26*x^2/2! + 42*x^3/3! + 64*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
Let M denote the present triangle. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A143494 (but with a different offset). See the Example section. Cf. A106516. (End)
a(n,m) = Sum_{p=m..n} 2^(n-p)*binomial(p-1,m-1), n >= m >= 0, else 0. - Wolfdieter Lang, Jan 09 2015
T(n, k) = 2^n - (1/2)*binomial(n, k-1)*hypergeom([1, n+1], [n-k+2], 1/2). - Peter Luschny, Oct 10 2019
T(n, k) = binomial(n, k)*hypergeom([1, k - n], [k + 1], -1). - Peter Luschny, Oct 06 2023
n-th row polynomial R(n, x) = (2^n - x*(1 + x)^n)/(1 - x). These polynomials can be used to find series acceleration formulas for the constants log(2) and Pi. - Peter Bala, Mar 03 2025