cp's OEIS Frontend

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.

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A051601 Rows of triangle formed using Pascal's rule except we begin and end the n-th row with n.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 4, 4, 3, 4, 7, 8, 7, 4, 5, 11, 15, 15, 11, 5, 6, 16, 26, 30, 26, 16, 6, 7, 22, 42, 56, 56, 42, 22, 7, 8, 29, 64, 98, 112, 98, 64, 29, 8, 9, 37, 93, 162, 210, 210, 162, 93, 37, 9, 10, 46, 130, 255, 372, 420, 372, 255, 130, 46, 10
Offset: 0

Views

Author

Keywords

Comments

The number of spotlight tilings of an m X n rectangle missing the southeast corner. E.g., there are 2 spotlight tilings of a 2 X 2 square missing its southeast corner. - Bridget Tenner, Nov 10 2007
T(n,k) = A134636(n,k) - A051597(n,k). - Reinhard Zumkeller, Nov 23 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

Examples

			From _Roger L. Bagula_, Feb 17 2009: (Start)
Triangle begins:
   0;
   1,  1;
   2,  2,   2;
   3,  4,   4,   3;
   4,  7,   8,   7,    4;
   5, 11,  15,  15,   11,    5;
   6, 16,  26,  30,   26,   16,   6;
   7, 22,  42,  56,   56,   42,   22,    7;
   8, 29,  64,  98,  112,   98,   64,   29,   8;
   9, 37,  93, 162,  210,  210,  162,   93,   37,   9;
  10, 46, 130, 255,  372,  420,  372,  255,  130,  46,  10;
  11, 56, 176, 385,  627,  792,  792,  627,  385, 176,  56, 11;
  12, 67, 232, 561, 1012, 1419, 1584, 1419, 1012, 561, 232, 67, 12. ... (End)
		

Crossrefs

Row sums give A000918(n+1).
Columns from 2 to 9, respectively: A000124; A000125, A055795, A027660, A055796, A055797, A055798, A055799 (except 1 for the last seven). [Bruno Berselli, Aug 02 2013]
Cf. A001477, A162551 (central terms).

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k->  Binomial(n, k+1) + Binomial(n, n-k+1) ))); # G. C. Greubel, Nov 12 2019
  • Haskell
    a051601 n k = a051601_tabl !! n !! k
    a051601_row n = a051601_tabl !! n
    a051601_tabl = iterate
                   (\row -> zipWith (+) ([1] ++ row) (row ++ [1])) [0]
    -- Reinhard Zumkeller, Nov 23 2012
    
  • Magma
    /* As triangle: */ [[Binomial(n,m+1)+Binomial(n,n-m+1): m in [0..n]]: n in [0..12]]; // Bruno Berselli, Aug 02 2013
    
  • Maple
    seq(seq(binomial(n,k+1) + binomial(n, n-k+1), k=0..n), n=0..12); # G. C. Greubel, Nov 12 2019
  • Mathematica
    T[n_, k_]:= T[n, k] = Binomial[n, k+1] + Binomial[n, n-k+1];
    Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* Roger L. Bagula, Feb 17 2009; modified by G. C. Greubel, Nov 12 2019 *)
  • PARI
    T(n,k) = binomial(n, k+1) + binomial(n, n-k+1);
    for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Nov 12 2019
    
  • Sage
    [[binomial(n, k+1) + binomial(n, n-k+1) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019
    

Formula

T(m,n) = binomial(m+n,m) - 2*binomial(m+n-2,m-1), up to offset and transformation of array to triangular indices. - Bridget Tenner, Nov 10 2007
T(n,k) = binomial(n, k+1) + binomial(n, n-k+1). - Roger L. Bagula, Feb 17 2009
T(0,n) = T(n,0) = n, T(n,k) = T(n-1,k) + T(n-1,k-1), 0 < k < n.

A108561 Triangle read by rows: T(n,0)=1, T(n,n)=(-1)^n, T(n+1,k)=T(n,k-1)+T(n,k) for 0 < k < n.

Original entry on oeis.org

1, 1, -1, 1, 0, 1, 1, 1, 1, -1, 1, 2, 2, 0, 1, 1, 3, 4, 2, 1, -1, 1, 4, 7, 6, 3, 0, 1, 1, 5, 11, 13, 9, 3, 1, -1, 1, 6, 16, 24, 22, 12, 4, 0, 1, 1, 7, 22, 40, 46, 34, 16, 4, 1, -1, 1, 8, 29, 62, 86, 80, 50, 20, 5, 0, 1, 1, 9, 37, 91, 148, 166, 130, 70, 25, 5, 1, -1, 1, 10, 46, 128, 239, 314, 296, 200, 95, 30, 6, 0, 1, 1, 11, 56, 174, 367
Offset: 0

Views

Author

Reinhard Zumkeller, Jun 10 2005

Keywords

Comments

Sum_{k=0..n} T(n,k) = A078008(n);
Sum_{k=0..n} abs(T(n,k)) = A052953(n-1) for n > 0;
T(n,1) = n - 2 for n > 1;
T(n,2) = A000124(n-3) for n > 2;
T(n,3) = A003600(n-4) for n > 4;
T(n,n-6) = A001753(n-6) for n > 6;
T(n,n-5) = A001752(n-5) for n > 5;
T(n,n-4) = A002623(n-4) for n > 4;
T(n,n-3) = A002620(n-1) for n > 3;
T(n,n-2) = A008619(n-2) for n > 2;
T(n,n-1) = n mod 2 for n > 0;
T(2*n,n) = A072547(n+1).
Sum_{k=0..n} T(n,k)*x^k = A232015(n), A078008(n), A000012(n), A040000(n), A001045(n+2), A140725(n+1) for x = 2, 1, 0, -1, -2, -3 respectively. - Philippe Deléham, Nov 17 2013, Nov 19 2013
(1,a^n) Pascal triangle with a = -1. - Philippe Deléham, Dec 27 2013
T(n,k) = A112465(n,n-k). - Reinhard Zumkeller, Jan 03 2014

Examples

			From _Philippe Deléham_, Nov 17 2013: (Start)
Triangle begins:
  1;
  1, -1;
  1,  0,  1;
  1,  1,  1, -1;
  1,  2,  2,  0,  1;
  1,  3,  4,  2,  1, -1;
  1,  4,  7,  6,  3,  0,  1; (End)
		

Crossrefs

Cf. A007318 (a=1), A008949(a=2), A164844(a=10).
Similar to the triangles A035317, A059259, A080242, A112555.
Cf. A072547 (central terms).

Programs

  • GAP
    Flat(List([0..13],n->List([0..n],k->Sum([0..k],i->Binomial(n,i)*(-2)^(k-i))))); # Muniru A Asiru, Feb 19 2018
  • Haskell
    a108561 n k = a108561_tabl !! n !! k
    a108561_row n = a108561_tabl !! n
    a108561_tabl = map reverse a112465_tabl
    -- Reinhard Zumkeller, Jan 03 2014
    
  • Maple
    A108561 := (n, k) -> add(binomial(n, i)*(-2)^(k-i), i = 0..k):
    seq(seq(A108561(n,k), k = 0..n), n = 0..12); # Peter Bala, Feb 18 2018
  • Mathematica
    Clear[t]; t[n_, 0] = 1; t[n_, n_] := t[n, n] = (-1)^Mod[n, 2]; t[n_, k_] := t[n, k] = t[n-1, k] + t[n-1, k-1]; Table[t[n, k], {n, 0, 13}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)
  • Sage
    def A108561_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            if k==0: return 0
            return -prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
        return [(-1)^k*prec(n, k) for k in (1..n-1)]+[(-1)^(n+1)]
    for n in (1..12): print(A108561_row(n)) # Peter Luschny, Mar 16 2016
    

Formula

G.f.: (1-y*x)/(1-x-(y+y^2)*x). - Philippe Deléham, Nov 17 2013
T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 17 2013
From Peter Bala, Feb 18 2018: (Start)
T(n,k) = Sum_{i = 0..k} binomial(n,i)*(-2)^(k-i), 0 <= k <= n.
The n-th row polynomial is the n-th degree Taylor polynomial of the rational function (1 + x)^n/(1 + 2*x) about 0. For example, for n = 4, (1 + x)^4/(1 + 2*x) = 1 + 2*x + 2*x^2 + x^4 + O(x^5). (End)

Extensions

Definition corrected by Philippe Deléham, Dec 26 2013

A093560 (3,1) Pascal triangle.

Original entry on oeis.org

1, 3, 1, 3, 4, 1, 3, 7, 5, 1, 3, 10, 12, 6, 1, 3, 13, 22, 18, 7, 1, 3, 16, 35, 40, 25, 8, 1, 3, 19, 51, 75, 65, 33, 9, 1, 3, 22, 70, 126, 140, 98, 42, 10, 1, 3, 25, 92, 196, 266, 238, 140, 52, 11, 1, 3, 28, 117, 288, 462, 504, 378, 192, 63, 12, 1, 3, 31, 145, 405, 750, 966, 882, 570, 255, 75, 13, 1
Offset: 0

Views

Author

Wolfdieter Lang, Apr 22 2004

Keywords

Comments

The array F(3;n,m) gives in the columns m >= 1 the figurate numbers based on A016777, including the pentagonal numbers A000326 (see the W. Lang link).
This is the third member, d=3, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal (d=1), A029653 (d=2).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(1+2*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(1+2*z)/(1-(1+x)*z).
The SW-NE diagonals give the Lucas numbers A000032: L(n) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with L(0)=2. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Triangle T(n,k), read by rows, given by [3,-2,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 17 2009
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From Wolfdieter Lang, Jan 09 2015: (Start)
The signed lower triangular matrix (-1)^(n-1)*a(n,m) is the inverse of the Riordan matrix A106516; that is Riordan ((1-2*x)/(1+x),x/(1+x)).
See the Peter Bala comment from Dec 23 2014 in A106516 for general Riordan triangles of the type (g(x), x/(1-x)): exp(x)*r(n,x) = d(n,x) with the e.g.f. r(n,x) of row n and the e.g.f. of diagonal n.
Similarly, for general Riordan triangles of the type (g(x), x/(1+x)): exp(x)*r(n,-x) = d(n,x). (End)
The n-th row polynomial is (3 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018
Binomial(n-2,k)+2*Binomial(n-3,k) is also the number of permutations avoiding both 123 and 132 with k double descents, i.e., positions with w[i]>w[i+1]>w[i+2]. - Lara Pudwell, Dec 19 2018

Examples

			Triangle begins
  1,
  3,  1,
  3,  4,  1,
  3,  7,  5,   1,
  3, 10, 12,   6,   1,
  3, 13, 22,  18,   7,   1,
  3, 16, 35,  40,  25,   8,   1,
  3, 19, 51,  75,  65,  33,   9,  1,
  3, 22, 70, 126, 140,  98,  42, 10,  1,
  3, 25, 92, 196, 266, 238, 140, 52, 11, 1,
		

References

  • Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
  • Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Crossrefs

Cf. Column sequences for m=1..9: A016777, A000326 (pentagonal), A002411, A001296, A051836, A051923, A050494, A053367, A053310;
A007318 (Pascal's triangle), A029653 ((2,1) Pascal triangle), A093561 ((4,1) Pascal triangle), A228196, A228576.

Programs

  • GAP
    Concatenation([1],Flat(List([1..11],n->List([0..n],k->Binomial(n,k)+2*Binomial(n-1,k))))); # Muniru A Asiru, Dec 20 2018
    
  • Haskell
    a093560 n k = a093560_tabl !! n !! k
    a093560_row n = a093560_tabl !! n
    a093560_tabl = [1] : iterate
                   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [3, 1]
    -- Reinhard Zumkeller, Aug 31 2014
    
  • Python
    from math import comb, isqrt
    def A093560(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<1))//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

a(n, m)=F(3;n-m, m) for 0<= m <= n, otherwise 0, with F(3;0, 0)=1, F(3;n, 0)=3 if n>=1 and F(3;n, m):=(3*n+m)*binomial(n+m-1, m-1)/m if m>=1.
G.f. column m (without leading zeros): (1+2*x)/(1-x)^(m+1), m>=0.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=3 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
T(n, k) = C(n, k) + 2*C(n-1, k). - Philippe Deléham, Aug 28 2005
Equals M * A007318, where M = an infinite triangular matrix with all 1's in the main diagonal and all 2's in the subdiagonal. - Gary W. Adamson, Dec 01 2007
Sum_{k=0..n} T(n,k) = A151821(n+1). - Philippe Deléham, Sep 17 2009
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(3 + 7*x + 5*x^2/2! + x^3/3!) = 3 + 10*x + 22*x^2/2! + 40*x^3/3! + 65*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
G.f.: (-1-2*x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015

Extensions

Incorrect connection with A046055 deleted by N. J. A. Sloane, Jul 08 2009

A071919 Number of monotone nondecreasing functions [n]->[m] for n >= 0, m >= 0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 1, 5, 10, 10, 5, 1, 0, 1, 6, 15, 20, 15, 6, 1, 0, 1, 7, 21, 35, 35, 21, 7, 1, 0, 1, 8, 28, 56, 70, 56, 28, 8, 1, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 0, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 0, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 0
Offset: 0

Views

Author

Michele Dondi (bik.mido(AT)tiscalinet.it), Jun 14 2002

Keywords

Comments

Sometimes called a Riordan array.
Number of different partial sums of 1 + [2,3] + [3,4] + [4,5] + ... - Jon Perry, Jan 01 2004
Triangle T(n,k), 0 <= k <= n, read by rows, given by [1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [0, 1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 05 2005
T(n,k)=abs(A110555(n,k)), A110555(n,k)=T(n,k)*(-1)^k. - Reinhard Zumkeller, Jul 27 2005
(1,0)-Pascal triangle. - Philippe Deléham, Nov 21 2006
A129186*A007318 as infinite lower triangular matrices. - Philippe Deléham, Mar 07 2009
Let n>=0 index the rows and m>=0 index the columns of this rectangular array. R(n,m) is "m multichoose n", the number of multisets of length n on m symbols. R(n,m) = Sum_{i=0..n} R(i,m-1). The summation conditions on the number of members in a size n multiset that are not the element m (an arbitrary element in the set of m symbols). R(n,m) = Sum_{i=1..m} R(n-1,i). The summation conditions on the largest element in a size n multiset on {1,2,...,m}. - Geoffrey Critzer, Jun 03 2009
Sum_{k=0..n} T(n,k)*B(k) = B(n), n>=0, with the Bell numbers B(n):=A000110(n) (eigensequence). See, e.g., the W. Lang link, Corollary 4. - Wolfdieter Lang, Jun 23 2010
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

Examples

			   1,    1,    1,    1,    1,    1,    1,    1,    1, ...
   0,    1,    2,    3,    4,    5,    6,    7,    8, ...
   0,    1,    3,    6,   10,   15,   21,   28,   36, ...
   0,    1,    4,   10,   20,   35,   56,   84,  120, ...
   0,    1,    5,   15,   35,   70,  126,  210,  330, ...
   0,    1,    6,   21,   56,  126,  252,  462,  792, ...
   0,    1,    7,   28,   84,  210,  462,  924, 1716, ...
   0,    1,    8,   36,  120,  330,  792, 1716, 3432, ...
   0,    1,    9,   45,  165,  495, 1287, 3003, 6435, ...
		

Crossrefs

Main diagonal gives A088218.

Programs

  • Maple
    A:= (n, m)-> binomial(n+m-1, n):
    seq(seq(A(n, d-n), n=0..d), d=0..14);  # Alois P. Heinz, Jan 13 2017
  • Mathematica
    Table[Table[Binomial[m - 1 + n, n], {m, 0, 10}], {n, 0, 10}] // Grid (* Geoffrey Critzer, Jun 03 2009 *)
    a[n_, m_] := Binomial[m - 1 + n, n]; Table[Table[a[n, m - n], {n, 0, m}], {m, 0, 10}] // Flatten (* G. C. Greubel, Nov 22 2017 *)
  • PARI
    { n=20; v=vector(n); for (i=1,n,v[i]=vector(2^(i-1))); v[1][1]=1; for (i=2,n, k=length(v[i-1]); for (j=1,k, v[i][j]=v[i-1][j]+i; v[i][j+k]=v[i-1][j]+i+1)); c=vector(n); for (i=1,n, for (j=1,2^(i-1), if (v[i][j]<=n, c[v[i][j]]++))); c } \\ Jon Perry
    
  • PARI
    {a(n) = my(m); if( n<1, n==0, m = (sqrtint(8*n+1) - 1)\2; binomial(m-1, n - m*(m+1)/2))}; /* Michael Somos, Aug 20 2006 */

Formula

Limit_{k->infinity} A071919^k = (A000110,0,0,0,0,...) with the Bell numbers in the first column. For a proof see, e.g., the W. Lang link, proposition 12.
A(n,k) = binomial(n+k-1,n). - Reinhard Zumkeller, Jul 27 2005
G.f.: 1 + x + x^3(1+x) + x^6(1+x)^2 + x^10(1+x)^3 + ... . - Michael Somos, Aug 20 2006
G.f. of the triangular interpretation: (-1+x*y)/(-1+x*y+x). - R. J. Mathar, Aug 11 2015

A095660 Pascal (1,3) triangle.

Original entry on oeis.org

3, 1, 3, 1, 4, 3, 1, 5, 7, 3, 1, 6, 12, 10, 3, 1, 7, 18, 22, 13, 3, 1, 8, 25, 40, 35, 16, 3, 1, 9, 33, 65, 75, 51, 19, 3, 1, 10, 42, 98, 140, 126, 70, 22, 3, 1, 11, 52, 140, 238, 266, 196, 92, 25, 3, 1, 12, 63, 192, 378, 504, 462, 288, 117, 28, 3, 1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3
Offset: 0

Views

Author

Wolfdieter Lang, May 21 2004

Keywords

Comments

This is the third member, q=3, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with T(0,0)=2, not 1).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} T(n,m)*x^m is G(z,x) = g(z)/(1-x*z*f(z)). Here: g(x) = (3-2*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (3-2*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} T(n-1-k,k) = A000285(n-2), n>=2, with n=1 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Central terms: T(2*n,n) = A028329(n) = A100320(n) for n > 0, A028329 are the central terms of triangle A028326. - Reinhard Zumkeller, Apr 08 2012
Let P be Pascal's triangle, A007318 and R the Riordan array, A097805. Then Pascal triangle (1,q) = ((q-1) * R) + P. Example: Pascal triangle (1,3) = (2 * R) + P. - Gary W. Adamson, Sep 12 2015

Examples

			Triangle starts:
  3;
  1,  3;
  1,  4,  3;
  1,  5,  7,   3;
  1,  6, 12,  10,   3;
  1,  7, 18,  22,  13,   3;
  1,  8, 25,  40,  35,  16,   3;
  1,  9, 33,  65,  75,  51,  19,   3;
  1, 10, 42,  98, 140, 126,  70,  22,   3;
  1, 11, 52, 140, 238, 266, 196,  92,  25,   3;
  1, 12, 63, 192, 378, 504, 462, 288, 117,  28,  3;
  1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3;
		

Crossrefs

Row sums: A000079(n+1), n>=1, 3 if n=0. Alternating row sums are [3, -2, followed by 0's].
Column sequences (without leading zeros) give for m=1..9 with n>=0: A000027(n+3), A055998(n+1), A006503(n+1), A095661, A000574, A095662, A095663, A095664, A095665.
Cf. A097805.

Programs

  • Haskell
    a095660 n k = a095660_tabl !! n !! k
    a095660_row n = a095660_tabl !! n
    a095660_tabl = [3] : iterate
       (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,3]
    -- Reinhard Zumkeller, Apr 08 2012
    
  • Magma
    A095660:= func< n,k | n eq 0 select 3 else (1+2*k/n)*Binomial(n,k) >;
    [A095660(n,k): k in [0..n], n in [1..12]]; // G. C. Greubel, May 02 2021
    
  • Maple
    T(n,k):=piecewise(n=0,3,0Mircea Merca, Apr 08 2012
  • Mathematica
    {3}~Join~Table[(1 + 2 k/n) Binomial[n, k], {n, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 14 2015 *)
  • Sage
    def A095660(n,k): return 3 if n==0 else (1+2*k/n)*binomial(n,k)
    flatten([[A095660(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 02 2021

Formula

Recursion: T(n, m)=0 if m>n, T(0, 0)= 3; T(n, 0)=1 if n>=1; T(n, m) = T(n-1, m) + T(n-1, m-1).
G.f. column m (without leading zeros): (3-2*x)/(1-x)^(m+1), m>=0.
T(n,k) = (1+2*k/n) * binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012
Closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013

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.

Original entry on oeis.org

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

Views

Author

Boris Putievskiy, Aug 26 2013

Keywords

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;
...
		

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.

A093561 (4,1) Pascal triangle.

Original entry on oeis.org

1, 4, 1, 4, 5, 1, 4, 9, 6, 1, 4, 13, 15, 7, 1, 4, 17, 28, 22, 8, 1, 4, 21, 45, 50, 30, 9, 1, 4, 25, 66, 95, 80, 39, 10, 1, 4, 29, 91, 161, 175, 119, 49, 11, 1, 4, 33, 120, 252, 336, 294, 168, 60, 12, 1, 4, 37, 153, 372, 588, 630, 462, 228, 72, 13, 1, 4, 41, 190, 525, 960, 1218
Offset: 0

Views

Author

Wolfdieter Lang, Apr 22 2004

Keywords

Comments

The array F(4;n,m) gives in the columns m >= 1 the figurate numbers based on A016813, including the hexagonal numbers A000384 (see the W. Lang link).
This is the fourth member, d=4, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653 and A093560, for d=1..3.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} a(n,m)*x^m is G(z,x) = (1+3*z)/(1-(1+x)*z).
The SW-NE diagonals give A000285(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018

Examples

			Triangle begins
  [1];
  [4, 1];
  [4, 5, 1];
  [4, 9, 6, 1];
  ...
		

References

  • Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
  • Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Crossrefs

Cf. Row sums: A020714(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 3 for n=2 and 0 otherwise.
Columns m=1..9: A016813, A000384 (hexagonal), A002412, A002417, A034263, A051947, A050483, A052181, A055843.

Programs

  • Haskell
    a093561 n k = a093561_tabl !! n !! k
    a093561_row n = a093561_tabl !! n
    a093561_tabl = [1] : iterate
                   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [4, 1]
    -- Reinhard Zumkeller, Aug 31 2014
    
  • Python
    from math import comb, isqrt
    def A093561(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+3*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

a(n, m) = F(4;n-m, m) for 0<= m <= n, otherwise 0, with F(4;0, 0)=1, F(4;n, 0)=4 if n>=1 and F(4;n, m) = (4*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=4 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. row m (without leading zeros): (1+3*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 3*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(4 + 9*x + 6*x^2/2! + x^3/3!) = 4 + 13*x + 28*x^2/2! + 50*x^3/3! + 80*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A093563 (6,1)-Pascal triangle.

Original entry on oeis.org

1, 6, 1, 6, 7, 1, 6, 13, 8, 1, 6, 19, 21, 9, 1, 6, 25, 40, 30, 10, 1, 6, 31, 65, 70, 40, 11, 1, 6, 37, 96, 135, 110, 51, 12, 1, 6, 43, 133, 231, 245, 161, 63, 13, 1, 6, 49, 176, 364, 476, 406, 224, 76, 14, 1, 6, 55, 225, 540, 840, 882, 630, 300, 90, 15, 1, 6, 61, 280, 765, 1380
Offset: 0

Views

Author

Wolfdieter Lang, Apr 22 2004

Keywords

Comments

The array F(6;n,m) gives in the columns m >= 1 the figurate numbers based on A016921, including the octagonal numbers A000567, (see the W. Lang link).
This is the sixth member, d=6, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-2, for d=1..5.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=(1+5*z)/(1-(1+x)*z).
The SW-NE diagonals give A022096(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

Examples

			Triangle begins
  1;
  6,  1;
  6,  7,  1;
  6, 13,  8,  1;
  6, 19, 21,  9,  1;
  6, 25, 40, 30, 10,  1;
  ...
		

References

  • Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
  • Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Crossrefs

Row sums: A005009(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 5 for n=2 and 0 else.
The column sequences give for m=1..9: A016921, A000567 (octagonal), A002414, A002419, A051843, A027810, A034265, A054487, A055848.

Programs

  • Haskell
    a093563 n k = a093563_tabl !! n !! k
    a093563_row n = a093563_tabl !! n
    a093563_tabl = [1] : iterate
                   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [6, 1]
    -- Reinhard Zumkeller, Aug 31 2014
    
  • Mathematica
    lim = 11; s = Series[(1 + 5*x)/(1 - x)^(m + 1), {x, 0, lim}]; t = Table[ CoefficientList[s, x], {m, 0, lim}]; Flatten[ Table[t[[j - k + 1, k]], {j, lim + 1}, {k, j, 1, -1}]] (* Jean-François Alcover, Sep 16 2011, after g.f. *)
  • Python
    from math import comb, isqrt
    def A093563(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+5*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

a(n, m)=F(6;n-m, m) for 0<= m <= n, otherwise 0, with F(6;0, 0)=1, F(6;n, 0)=6 if n>=1 and F(6;n, m):= (6*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=6 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+5*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 5*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(6 + 13*x + 8*x^2/2! + x^3/3!) = 6 + 19*x + 40*x^2/2! + 70*x^3/3! + 110*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A051597 Rows of triangle formed using Pascal's rule except begin and end n-th row with n+1.

Original entry on oeis.org

1, 2, 2, 3, 4, 3, 4, 7, 7, 4, 5, 11, 14, 11, 5, 6, 16, 25, 25, 16, 6, 7, 22, 41, 50, 41, 22, 7, 8, 29, 63, 91, 91, 63, 29, 8, 9, 37, 92, 154, 182, 154, 92, 37, 9, 10, 46, 129, 246, 336, 336, 246, 129, 46, 10, 11, 56, 175, 375, 582, 672, 582, 375, 175, 56, 11
Offset: 0

Views

Author

Keywords

Comments

Row sums give A033484(n).
The number of spotlight tilings of an (m+1) X (n+1) rectangle, read by antidiagonals. - Bridget Tenner, Nov 09 2007
T(n,k) = A134636(n,k) - A051601(n,k). - Reinhard Zumkeller, Nov 23 2012
T(n,k) = A209561(n+2,k+1), 0 <= k <= n. - Reinhard Zumkeller, Dec 26 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

Examples

			Triangle begins as:
  1;
  2,  2;
  3,  4,  3;
  4,  7,  7,  4;
  5, 11, 14, 11, 5;
		

Crossrefs

Stripped variant of A072405, A122218.

Programs

  • GAP
    T:= function(n,k)
        if k<0 or k>n then return 0;
        elif k=0 or k=n then return n+1;
        else return 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 18 2019
  • Haskell
    a051597 n k = a051597_tabl !! n !! k
    a051597_row n = a051597_tabl !! n
    a051597_tabl = iterate (\row -> zipWith (+) ([1] ++ row) (row ++ [1])) [1]
    -- Reinhard Zumkeller, Nov 23 2012
    
  • Magma
    function T(n,k)
        if k lt 0 or k gt n then return 0;
      elif k eq 0 or k eq n then return n+1;
      else return 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 18 2019
    
  • Maple
    T:= proc(n, k) option remember;
          `if`(k<0 or k>n, 0,
          `if`(k=0 or k=n, n+1,
             T(n-1, k-1) + T(n-1, k) ))
        end:
    seq(seq(T(n, k), k=0..n), n=0..14);  # Alois P. Heinz, May 27 2013
  • Mathematica
    NestList[Append[ Prepend[Map[Apply[Plus, #] &, Partition[#, 2, 1]], #[[1]] + 1], #[[1]] + 1] &, {1}, 10] // Grid  (* Geoffrey Critzer, May 26 2013 *)
    T[n_, k_] := T[n, k] = If[k<0 || k>n, 0, If[k==0 || k==n, n+1, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n, 0, 14}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 09 2016, after Alois P. Heinz *)
  • PARI
    T(n,k) = if(k<0 || k>n, 0, if(k==0 || k==n, n+1, T(n-1, k-1) + T(n-1, k) ));
    for(n=0, 12, for(k=0, n, print1(T(n,k), ", "))) \\ G. C. Greubel, Nov 18 2019
    
  • Sage
    @CachedFunction
    def T(n, k):
        if (k<0 or k>n): return 0
        elif (k==0 or k==n): return n+1
        else: return 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 18 2019
    

Formula

T(2n,n) = A051924(n+1). - Philippe Deléham, Nov 26 2006
T(m,n) = binomial(m+n,m) - binomial(m+n-2,m-1) (correct up to offset and transformation of square indices to triangular indices). - Bridget Tenner, Nov 09 2007
T(0,n) = T(n,0) = n+1, T(n,k) = T(n-1,k) + T(n-1,k-1), 0 < k < n.
From Peter Bala, Feb 28 2013: (Start)
T(n,k) = binomial(n,k-1) + binomial(n,k) + binomial(n,k+1) for 0 <= k <= n.
O.g.f.: (1 - xt^2)/((1 - t)(1 - xt)(1 - (1+x)t)) = 1 + (2 + 2x)t + (3 + 4x + 3x^2)t^2 + ....
Row polynomials: ((1+x+x^2)*(1+x)^n - 1 - x^(n+2))/x. (End)

A093562 (5,1) Pascal triangle.

Original entry on oeis.org

1, 5, 1, 5, 6, 1, 5, 11, 7, 1, 5, 16, 18, 8, 1, 5, 21, 34, 26, 9, 1, 5, 26, 55, 60, 35, 10, 1, 5, 31, 81, 115, 95, 45, 11, 1, 5, 36, 112, 196, 210, 140, 56, 12, 1, 5, 41, 148, 308, 406, 350, 196, 68, 13, 1, 5, 46, 189, 456, 714, 756, 546, 264, 81, 14, 1, 5, 51, 235, 645, 1170
Offset: 0

Views

Author

Wolfdieter Lang, Apr 22 2004

Keywords

Comments

This is the fifth member, d=5, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-1, for d=1..4.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=(1+4*z)/(1-(1+x)*z).
The SW-NE diagonals give A022095(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 4. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
The array F(5;n,m) gives in the columns m >= 1 the figurate numbers based on A016861, including the heptagonal numbers A000566 (see the W. Lang link).
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018

Examples

			Triangle begins
  [1];
  [5,  1];
  [5,  6,  1];
  [5, 11,  7,  1];
  ...
		

References

  • Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
  • Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Crossrefs

Cf. Row sums: A007283(n-1), n>=1, 1 for n=0. A082505(n+1), alternating row sums are 1 for n=0, 4 for n=2 and 0 else.
Column sequences give for m=1..9: A016861, A000566 (heptagonal), A002413, A002418, A027800, A051946, A050484, A052255, A055844.

Programs

  • Haskell
    a093562 n k = a093562_tabl !! n !! k
    a093562_row n = a093562_tabl !! n
    a093562_tabl = [1] : iterate
                   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [5, 1]
    -- Reinhard Zumkeller, Aug 31 2014
    
  • Python
    from math import comb, isqrt
    def A093562(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<2))//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

a(n, m) = F(5;n-m, m) for 0<= m <= n, otherwise 0, with F(5;0, 0)=1, F(5;n, 0)=5 if n>=1 and F(5;n, m):=(5*n+m)*binomial(n+m-1, m-1)/m if m>=1.
G.f. column m (without leading zeros): (1+4*x)/(1-x)^(m+1), m>=0.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=5 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
T(n, k) = C(n, k) + 4*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(5 + 11*x + 7*x^2/2! + x^3/3!) = 5 + 16*x + 34*x^2/2! + 60*x^3/3! + 95*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
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