A317023 Square array A(n,k), n >= 0, k >= 0, read by ascending antidiagonals, where the sequence of row n is the expansion of (1-x^(n+1))/((1-x)^(n+1)).

1, 1, 0, 1, 2, 0, 1, 3, 2, 0, 1, 4, 6, 2, 0, 1, 5, 10, 9, 2, 0, 1, 6, 15, 20, 12, 2, 0, 1, 7, 21, 35, 34, 15, 2, 0, 1, 8, 28, 56, 70, 52, 18, 2, 0, 1, 9, 36, 84, 126, 125, 74, 21, 2, 0, 1, 10, 45, 120, 210, 252, 205, 100, 24, 2, 0, 1, 11, 55, 165, 330, 462, 461, 315, 130, 27, 2, 0, 1, 12, 66

Offset

0,5

Comments

Conjecture: alternating row sums of the triangle give A106510 for n >= 0.

Links

Table of n, a(n) for n=0..80.

Formula

A(n,k) = binomial(n+k,k) - binomial(k-1,k-1-n) for n >= 0 and k >= 0 with binomial(i,j) = 0 if i < j or j < 0.

G.f.: Sum_{k>=0,n>=0} A(n,k)*x^k*y^n = ((1-x)^2)/((1-x-y)*(1-x-x*y)).

Seen as a triangle T(n,k) = A(n-k,k) = binomial(n,k)-binomial(k-1,2*k-1-n) for 0 <= k <= n with binomial(i,j) = 0 if i < j or j < 0.

Mirror image of the triangle equals A173265 except column 0.

Example

The square array A(n,k) begins:
  n\k |  0  1  2   3    4    5    6     7     8     9     10
  ====+=====================================================
   0  |  1  0  0   0    0    0    0     0     0     0      0
   1  |  1  2  2   2    2    2    2     2     2     2      2
   2  |  1  3  6   9   12   15   18    21    24    27     30
   3  |  1  4 10  20   34   52   74   100   130   164    202
   4  |  1  5 15  35   70  125  205   315   460   645    875
   5  |  1  6 21  56  126  252  461   786  1266  1946   2877
   6  |  1  7 28  84  210  462  924  1715  2996  4977   7924
   7  |  1  8 36 120  330  792 1716  3432  6434 11432  19412
   8  |  1  9 45 165  495 1287 3003  6435 12870 24309  43749
   9  |  1 10 55 220  715 2002 5005 11440 24310 48620  92377
  10  |  1 11 66 286 1001 3003 8008 19448 43758 92378 184756
  etc.
The triangle T(n,k) begins:
  n\k |  0  1  2   3   4   5   6   7   8   9 10 11 12
  ====+==============================================
   0  |  1
   1  |  1  0
   2  |  1  2  0
   3  |  1  3  2   0
   4  |  1  4  6   2   0
   5  |  1  5 10   9   2   0
   6  |  1  6 15  20  12   2   0
   7  |  1  7 21  35  34  15   2   0
   8  |  1  8 28  56  70  52  18   2   0
   9  |  1  9 36  84 126 125  74  21   2   0
  10  |  1 10 45 120 210 252 205 100  24   2  0
  11  |  1 11 55 165 330 462 461 315 130  27  2  0
  12  |  1 12 66 220 495 792 924 786 460 164 30  2  0
  etc.

Mathematica

Table[SeriesCoefficient[(1 - x^(# + 1))/((1 - x)^(# + 1)), {x, 0, k}] &[n - k], {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 20 2018 *)

Prog

(GAP) nmax:=15;; A:=List([0..nmax], n->List([0..nmax], k->Binomial(n+k, k)-Binomial(k-1, k-1-n)));;   b:=List([2..nmax], n->OrderedPartitions(n, 2));;
a:=Flat(List([1..Length(b)], i->List([1..Length(b[i])], j->A[b[i][j][2]][b[i][j][1]]))); # Muniru A Asiru, Jul 20 2018
(PARI) T(n, k) = binomial(n+k, k) - binomial(k-1, k-1-n); \\ Michel Marcus, Aug 07 2018

Crossrefs

Row sums of the triangle give A099036 for n >= 0.

Cf. A000984 (main diagonal), A000012 (column 0), A087156 (column 1).

Cf. A099036, A106510.

In the square array; row 0..12 are: A000007, A040000, A008486, A005893, A008487, A008488, A008489, A008490, A008491, A008492, A008493, A008494, A008495.

A173265 is based on the same square array, but is read by descending antidiagonals with special treatment of column 0.

Sequence in context: A334892 A216235 A306914 * A319284 A338526 A182703

Adjacent sequences: A317020 A317021 A317022 * A317024 A317025 A317026

Keyword

nonn,easy,tabl

Author

Werner Schulte, Jul 19 2018

Status

approved