A030528 -id:A030528 - cp's OEIS frontend

A049327 A convolution triangle of numbers generalizing Pascal's triangle A007318.

1, 15, 1, 120, 30, 1, 540, 465, 45, 1, 1296, 4680, 1035, 60, 1, 1296, 33192, 15795, 1830, 75, 1, 0, 171072, 176688, 37260, 2850, 90, 1, 0, 641520, 1521828, 563409, 72450, 4095, 105, 1, 0, 1710720, 10359360, 6686064, 1375605, 124740, 5565, 120, 1

Offset: 1

Wolfdieter Lang

			{1}; {15,1}; {120,30,1}; {540,465,45,1}; {1296,4680,1035,60,1}; ...

W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.

a(n, m) := s1(-5, n, m), a member of a sequence of triangles including s1(0, n, m)= A023531(n, m) (unit matrix) and s1(2, n, m)=A007318(n-1, m-1) (Pascal's triangle). s1(-1, n, m)= A030528.

Cf. A049351.

a(n, m) = 6*(6*m-n+1)*a(n-1, m)/n + m*a(n-1, m-1)/n, n >= m >= 1; a(n, m) := 0, nA033842(5, m)).

A179199 E.g.f. satisfies: A(x) = (1+x)/(1+2x)A(x+x^2) with A(0)=0.

Original entry on oeis.org

0, 1, -2, 9, -64, 620, -7536, 109032, -1809984, 33562944, -681799680, 14980204800, -354016189440, 9017296704000, -249422713344000, 7530733353024000, -246212297533440000, 8509848430274150400, -302719894872204902400

Offset: 0

Paul D. Hanna, Jul 09 2010

			E.g.f.: A(x) = x - 2*x^2/2! + 9*x^3/3! - 64*x^4/4! + 620*x^5/5! - 7536*x^6/6! + 109032*x^7/7! - 1809984*x^8/8! + 33562944*x^9/9! - 681799680*x^10/10! + 14980204800*x^11/11! - 354016189440*x^12/12! + ...
E.g.f. satisfies: A(x) = (1+x)/(1+2*x)*A(x+x^2) where:
. A(x+x^2) = x - 3*x^3/3! + 20*x^4/4! - 120*x^5/5! + 624*x^6/6! - 840*x^7/7! - 58752*x^8/8! + 1512000*x^9/9! - 25660800*x^10/10! + ...
E.g.f. A = A(x) satisfies:
. x = A + A*Dx(A)/2! + A*Dx(A*Dx(A))/3! + A*Dx(A*Dx(A*Dx(A)))/4! + ...
where Dx(F) = d/dx(x*F) and expansions begin:
. A*Dx(A) = 4*x^2/2! - 30*x^3/3! + 288*x^4/4! - 3500*x^5/5! +- ...
. A*Dx(A*Dx(A)) = 36*x^3/3! - 624*x^4/4! + 10680*x^5/5! -+ ...
. A*Dx(A*Dx(A*Dx(A))) = 576*x^4/4! - 18480*x^5/5! + 504000*x^6/6! -+ ...
. A*Dx(A*Dx(A*Dx(A*Dx(A)))) = 14400*x^5/5! - 751680*x^6/6! +- ...

G. C. Greubel, Table of n, a(n) for n = 0..300
Loïc Foissy, Cointeraction on noncrossing partitions and related polynomial invariants, arXiv:2501.18212 [math.CO], 2025. See p. 24.

Cf. A005119, A030528, A179198.

a[n_]:= a[n]= If[n<2, n, (-1/(n-1))*Sum[j!*Binomial[n, j]*Binomial[n-j+1, j+1]*a[n -j], {j, n-1}]];
Table[a[n], {n,0,40}] (* G. C. Greubel, Sep 03 2022 *)

/* E.g.f. satisfies: A(x) = (1+x)/(1+2*x)*A(x+x^2): */
{a(n)=local(A=x,B);for(m=2,n,B=(1+x)/(1+2*x+O(x^(n+3)))*subst(A,x,x+x^2+O(x^(n+3)));A=A-polcoeff(B,m+1)*x^m/(m-1));n!*polcoeff(A,n)}

/* Recurrence (slow): */
{a(n)=if(n<1, 0, if(n==1, 1, -n*(n-2)!*sum(i=1, n-1,binomial(n-i+1, i+1)*a(n-i)/(n-i)!)))}

/* x = A + A*Dx(A)/2! + A*Dx(A*Dx(A))/3! + A*Dx(A*Dx(A*Dx(A)))/4! +...: */
{a(n)=local(A=x+sum(m=2,n-1,a(m)*x^m/m!),G=1,R=0);R=sum(m=1,n,(G=A*deriv(x*G+x*O(x^n)))/m!);if(n==1,1,-n!*polcoeff(R,n))}

/* As column 0 of the matrix log of triangle A030528: */
{a(n)=local(A030528=matrix(n+1,n+1,r,c,if(r>=c,binomial(c,r-c))),LOG,ID=A030528^0);LOG=sum(m=1,n+1,-(ID-A030528)^m/m);n!*LOG[n+1,1]}

@CachedFunction
def a(n): # a = A179199
    if (n<2): return n
    else: return (-1/(n-1))*sum( factorial(j)*binomial(n,j)*binomial(n-j+1, j+1)*a(n-j) for j in (1..n-1) )
[a(n) for n in (0..40)] # G. C. Greubel, Sep 03 2022

E.g.f. A=A(x) satisfies: x = A + A*Dx(A)/2! + A*Dx(A*Dx(A))/3! + A*Dx(A*Dx(A*Dx(A)))/4! + ... where Dx(F) = d/dx(x*F).
...
a(n) = -n*(n-2)!*Sum_{i=1..n-1} C(n-i+1,i+1)*a(n-i)/(n-i)! for n>1 with a(1)=1.
...
a(n) = (-1)^(n-1)*n*A005119(n), where A005119 describes the infinitesimal generator of (x+x^2).
...
Equals column 0 of A179198, the matrix log of triangle A030528, where A030528(n,k) = C(k,n-k); the g.f. of column k in A030528 is (x+x^2)^(k+1)/x.
...
A179198(n,k) = (k+1)*a(n-k)/(n-1)! for n>0, k>=0, where A179198 = matrix log of triangle A030528.
...

A307500 Expansion of Product_{k>=1} 1/(1 - (x*(1 - x))^k).

Original entry on oeis.org

1, 1, 1, -1, -2, -4, 3, -1, 17, -16, 21, -57, 67, -130, 305, -536, 995, -1726, 2652, -4286, 7320, -13043, 24458, -45405, 81415, -141724, 239755, -400603, 676872, -1171076, 2072334, -3695550, 6519951, -11279015, 19188230, -32462795, 55334284, -95718737, 167673672, -294894076

Offset: 0

Ilya Gutkovskiy, Apr 11 2019

sign

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A000041, A030528, A218482, A238441, A286955, A307501.

nmax = 39; CoefficientList[Series[Product[1/(1 - (x (1 - x))^k), {k, 1, nmax}], {x, 0, nmax}], x]
nmax = 39; CoefficientList[Series[Exp[Sum[DivisorSigma[1, k] (x (1 - x))^k/k, {k, 1, nmax}]], {x, 0, nmax}], x]
Table[Sum[(-1)^(n - k) Binomial[k, n - k] PartitionsP[k], {k, 0, n}], {n, 0, 39}]

G.f.: exp(Sum_{k>=1} sigma(k)*(x*(1 - x))^k/k).

a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(k,n-k)*A000041(k).

A179200 E.g.f. equals the real part of the i-th iteration of (x + x^2), where i=sqrt(-1).

Original entry on oeis.org

0, 1, 0, -6, 60, -600, 5880, -38640, -624960, 45077760, -1773129600, 58531809600, -1657462435200, 33703750080000, 171919752076800, -76383384045696000, 6034124486347776000, -348318907415331840000, 15862493882862941184000

Offset: 0

Paul D. Hanna, Jul 02 2010

sign

Let H(x) equal the i-th iteration of (x + x^2), then
. the inverse of H(x) equals the conjugate of H(x);
. H(x+x^2) = H(x) + H(x)^2;
. H(x) = F(x) + i*G(x) where G(x) = e.g.f. of A179201 and F(x) = e.g.f. of this sequence, where H(F(x) - i*G(x)) = x;
. coefficients of H(x) form the first column of triangular matrix A030528 raised to the i-th power, where A030528(n,k) = C(k,n-k).

			E.g.f: F(x) = x - 6*x^3/3! + 60*x^4/4! - 600*x^5/5! + 5880*x^6/6! +...
The e.g.f. of A179201, G(x), begins:
G(x) = 2*x^2/2! - 6*x^3/3! + 12*x^4/4! + 200*x^5/5! - 6240*x^6/6! + 139440*x^7/7! - 2869440*x^8/8! +...
The i-th iteration of (x + x^2) = H(x) = F(x) + i*G(x), begins:
H(x) = x + i*x^2 - (1 + i)*x^3 + (5 + i)*x^4/2 - (15 - 5*i)*x^5/3 + (49 - 52*i)*x^6/6 - (23 - 83*i)*x^7/3 - (93 + 427*i)*x^8/6 + (15652 + 18537*i)*x^9/126 - (61567 + 24585*i)*x^10/126 + (369519 - 42094*i)*x^11/252 - (1743963 - 1222750*i)*x^12/504 + ...
where H(F(x) - i*G(x)) = x.

Cf. A179201, A030528.

{a(n)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial(c,r-c))),L=sum(k=1,#M,-(M^0-M)^k/k),N=sum(k=0,#L,(I*L)^k/k!));if(n<1,0,real(n!*N[n,1]))}

E.g.f.: F(x) satisfies:
. F(x) = (G(x+x^2)/G(x) - 1)/2
. G(x) = sqrt( F(x) + F(x)^2 - F(x+x^2) )
where G(x) is the e.g.f. of A179201.

A179201 E.g.f. equals the imaginary part of the i-th iteration of (x + x^2), where i=sqrt(-1).

Original entry on oeis.org

0, 0, 2, -6, 12, 200, -6240, 139440, -2869440, 53386560, -708048000, -6667689600, 1162101600000, -68789252563200, 3158414682259200, -118988867559744000, 3123174474201600000, 17680394964750336000, -10490102782572441600000

Offset: 0

Paul D. Hanna, Jul 02 2010

sign

Let H(x) equal the i-th iteration of (x + x^2), then
. the inverse of H(x) equals the conjugate of H(x);
. H(x+x^2) = H(x) + H(x)^2;
. H(x) = F(x) + i*G(x) where F(x) = e.g.f. of A179200 and G(x) = e.g.f. of this sequence, where H(F(x) - i*G(x)) = x;
. coefficients of H(x) form the first column of triangular matrix A030528 raised to the i-th power, where A030528(n,k) = C(k,n-k).

			E.g.f: G(x) = 2*x^2/2! - 6*x^3/3! + 12*x^4/4! + 200*x^5/5! +...
The e.g.f. of A179200, F(x), begins:
F(x) = x - 6*x^3/3! + 60*x^4/4! - 600*x^5/5! + 5880*x^6/6! - 38640*x^7/7! - 624960*x^8/8! +...
The i-th iteration of (x + x^2) = H(x) = F(x) + i*G(x), begins:
H(x) = x + i*x^2 - (1 + i)*x^3 + (5 + i)*x^4/2 - (15 - 5*i)*x^5/3 + (49 - 52*i)*x^6/6 - (23 - 83*i)*x^7/3 - (93 + 427*i)*x^8/6 + (15652 + 18537*i)*x^9/126 - (61567 + 24585*i)*x^10/126 + (369519 - 42094*i)*x^11/252 - (1743963 - 1222750*i)*x^12/504 + ...
where H(F(x) - i*G(x)) = x.

Cf. A179200, A030528.

{a(n)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial(c,r-c))),L=sum(k=1,#M,-(M^0-M)^k/k),N=sum(k=0,#L,(I*L)^k/k!));if(n<1,0,imag(n!*N[n,1]))}

E.g.f.: G(x) satisfies:
. G(x) = sqrt( F(x) + F(x)^2 - F(x+x^2) )
. F(x) = (G(x+x^2)/G(x) - 1)/2
where F(x) is the e.g.f. of A179200.

A339884 Triangle read by rows: T(n, m) gives the number of partitions of n with m parts and parts from {1, 2, 3}.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 0, 1, 2, 1, 1, 0, 1, 2, 2, 1, 1, 0, 0, 2, 2, 2, 1, 1, 0, 0, 1, 3, 2, 2, 1, 1, 0, 0, 1, 2, 3, 2, 2, 1, 1, 0, 0, 0, 2, 3, 3, 2, 2, 1, 1, 0, 0, 0, 1, 3, 3, 3, 2, 2, 1, 1, 0, 0, 0, 1, 2, 4, 3, 3, 2, 2, 1, 1

Offset: 1

Wolfdieter Lang, Jan 31 2021

Row sums give A001399(n), for n >= 1.

One could add the column [1,repeat 0] for m = 0 starting with n >= 0.

			The triangle T(n,m) begins:
  n\m  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ...
  1:   1
  2:   1 1
  3:   1 1 1
  4:   0 2 1 1
  5:   0 1 2 1 1
  6:   0 1 2 2 1 1
  7:   0 0 2 2 2 1 1
  8:   0 0 1 3 2 2 1 1
  9:   0 0 1 2 3 2 2 1 1
  10:  0 0 0 2 3 3 2 2 1  1
  11:  0 0 0 1 3 3 3 2 2  1  1
  12:  0 0 0 1 2 4 3 3 2  2  1  1
  13:  0 0 0 0 2 3 4 3 3  2  2  1  1
  14:  0 0 0 0 1 3 4 4 3  3  2  2  1  1
  15:  0 0 0 0 1 2 4 4 4  3  3  2  2  1  1
  16:  0 0 0 0 0 2 3 5 4  4  3  3  2  2  1  1
  17:  0 0 0 0 0 1 3 4 5  4  4  3  3  2  2  1  1
  18:  0 0 0 0 0 1 2 4 5  5  4  4  3  3  2  2  1  1
  19:  0 0 0 0 0 0 2 3 5  5  5  4  4  3  3  2  2  1  1
  20:  0 0 0 0 0 0 1 3 4  6  5  5  4  4  3  3  2  2  1  1
  ...
Row n = 6: the partitions of 6 with number of parts m = 1,2, ...., 6, and parts from {1,2,3} are (in Abramowitz-Stegun order): [] | [],[],[3,3] | [],[1,2,3],[2^3] | [1^3,3],[1^2,2^2] | [1^4,2] | 1^6, giving 0, 1, 2, 2, 1, 1.

Louis Comtet, Advanced Combinatorics, Reidel (1974)

Cf. A001399, A008284 (all parts), A145362 (parts 1, 2), A232539 (parts <=4), A291983.

Compositions: A007818, A030528 (parts 1, 2), A078803 (parts 1, 2, 3).

Sum_{k=0..n} (-1)^k * T(n,k) = A291983(n). - Alois P. Heinz, Feb 01 2021

G.f.: 1/((1-u*t)*(1-u*t^2)*(1-u*t^3)). [Comtet page 97 [2c]]. - R. J. Mathar, May 27 2025

A049323 Triangle of coefficients of certain polynomials (exponents in increasing order), equivalent to A033842.

Original entry on oeis.org

1, 1, 1, 1, 3, 3, 1, 6, 16, 16, 1, 10, 50, 125, 125, 1, 15, 120, 540, 1296, 1296, 1, 21, 245, 1715, 7203, 16807, 16807, 1, 28, 448, 4480, 28672, 114688, 262144, 262144, 1, 36, 756, 10206, 91854, 551124, 2125764, 4782969, 4782969, 1, 45, 1200, 21000, 252000

Offset: 0

Wolfdieter Lang

These polynomials p(n, x) appear in the W. Lang reference as c1(-(n+1);x), n >= 0 on p.12. The coefficients are given there in eq.(44) on p. 6. - Wolfdieter Lang, Nov 20 2015

			The triangle a(n, m) begins:
n\m 0  1   2    3     4      5      6      7 ...
0:  1
1:  1  1
2:  1  3   3
3:  1  6  16   16
4:  1 10  50  125  125
5:  1 15 120  540  1296  1296
6:  1 21 245 1715  7203  16807  16807
7:  1 28 448 4480 28672 114688 262144 262144
... reformatted. - Wolfdieter Lang, Nov 20 2015
E.g. the third row {1,3,3} corresponds to polynomial p(2,x)= 1 + 3*x + 3*x^2.

W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.

a(n, 0)= A000012 (powers of 1), a(n, 1)= A000217 (triangular numbers), a(n, n)= A000272(n+1), n >= 0 (diagonal), a(n, n-1)= A000272(n+1), n >= 1.
For n = 0..5 the row sequences a(n, m), m >= 0, are the first columns of the triangles A023531 (unit matrix), A030528, A049324, A049325, A049326, A049327, respectively.
Cf. A033842, A046757.

/* As triangle: */ [[Binomial(n+1, k+1)*(n+1)^(k-1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Nov 20 2015

seq(seq(binomial(n+1,m+1)*(n+1)^(m-1),m=0..n),n=0..10); # Robert Israel, Oct 19 2015

Table[Binomial[n + 1, k + 1] (n + 1)^(k - 1), {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 19 2015 *)

a(n, m) = A033842(n, n-m) = binomial(n+1, m+1)*(n+1)^{m-1}, n >= m >= 0, else 0.
p(k-1, -x)/(1-k*x)^k =(-1+1/(1-k*x)^k)/(x*k^2) is for k=1..5 G.f. for A000012, A001792, A036068, A036070, A036083, respectively.
From Werner Schulte, Oct 19 2015: (Start)
a(2*n,n) = A000108(n)*(2*n+1)^n;
a(3*n,2*n) = A001764(n)*(3*n+1)^(2*n);
a(p*n,(p-1)*n) = binomial(p*n,n)/((p-1)*n+1)*(p*n+1)^((p-1)*n) for p > 0;
Sum_{m=0..n} (m+1)*a(n,m) = (n+2)^n;
Sum_{m=0..n} (-1)^m*(m+1)*a(n,m) = (-n)^n where 0^0 = 1;
p(n,x) = Sum_{m=0..n} a(n,m)*x^m = ((1+(n+1)*x)^(n+1)-1)/((n+1)^2*x).
(End)

A125250 Square array, read by antidiagonals, where A(1,1) = A(2,2) = 1, A(1,2) = A(2,1) = 0, A(n,k) = 0 if n < 1 or k < 1, otherwise A(n,k) = A(n-2,k-2) + A(n-1,k-2) + A(n-2,k-1) + A(n-1,k-1).

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 2, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 1, 5, 1, 0, 0, 0, 0, 0, 5, 5, 0, 0, 0, 0, 0, 0, 3, 11, 3, 0, 0, 0, 0, 0, 0, 1, 13, 13, 1, 0, 0, 0, 0, 0, 0, 0, 9, 26, 9, 0, 0, 0, 0, 0, 0, 0, 0, 4, 32, 32, 4, 0, 0, 0, 0, 0, 0, 0, 0, 1, 26, 63, 26, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14, 80, 80, 14, 0, 0, 0, 0, 0

Offset: 1

Gerald McGarvey, Jan 15 2007

It appears that the main diagonal (1,1,2,5,11,...) is A051286 (Whitney number of level n of the lattice of the ideals of the fence of size 2 n) that the diagonals (0,1,2,5,13,...) adjacent to the main diagonal are A110320 (Number of blocks in all RNA secondary structures with n nodes) and that the n-th antidiagonal sum = A094686(n-1) (a Fibonacci convolution). The n-th row sum = A002605(n).

			Array starts as:
1 0 0 0  0  0  0 ...
0 1 1 0  0  0  0 ...
0 1 2 2  1  0  0 ...
0 0 2 5  5  3  1   0 ...
0 0 1 5 11 13  9   4   1   0...
0 0 0 3 13 26 32  26  14   5   1  0 ...
0 0 0 1  9 32 63  80  71  45  20  6  1 0 ...
0 0 0 0  4 26 80 153 201 191 135 71 27 7 1 0 ...
...

Cf. A051286, A110320, A002605, A026729, A030528, A057089, A109466.

T[n_, k_] := Sum[Binomial[i, n-i] Binomial[i, k-i], {i, Floor[(n+1)/2], k}];
Table[T[n-k, k], {n, 0, 13}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 12 2019 *)

A=matrix(22,22);A[1,1]=1;A[2,2]=1;A[2,1]=0;A[1,2]=0;A[3,2]=1;A[2,3]=1; for(n=3,22,for(k=3,22,A[n,k]=A[n-2,k-2]+A[n-1,k-2]+A[n-2,k-1]+A[n-1,k-1])); for(n=1,22,for(i=1,n,print1(A[n-i+1,i],", ")))

A(1,1) = A(2,2) = 1, A(1,2) = A(2,1) = 0, A(n,k) = 0 if n < 1 or k < 1, otherwise A(n,k) = A(n-2,k-2) + A(n-1,k-2) + A(n-2,k-1) + A(n-1,k-1).
From Peter Bala, Nov 07 2017: (Start)
T(n,k) = Sum_{i = floor((n+1)/2)..k} binomial(i,n-i)* binomial(i,k-i).
Square array = A026729 * transpose(A026729), where A026729 is viewed as a lower unit triangular array. Omitting the first row and column of square array = A030528 * transpose(A030528).
O.g.f. 1/(1 - t*(1 + t)*x - t*(1 + t)*x^2) = 1 + (t + t^2)*x + (t + 2*t^2 + 2*t^3 + t^4)*x^2 + .... Cf. A109466 with o.g.f. 1/(1 - t*x - t*x^2).
The n-th row polynomial R(n,t) satisfies R(n,t) = R(n,-1 - t).
R(n,t) = (-1)^n*sqrt(-t*(1 + t))^n*U(n, 1/2*sqrt(-t*(1 + t))), where U(n,x) denotes the n-th Chebyshev polynomial of the second kind.
The sequence of row polynomials R(n,t) is a divisibility sequence of polynomials, that is, if m divides n then R(m,t) divides R(n,t) in the polynomial ring Z[t].
R(n,1) = A002605; R(n,2) = A057089. (End)

A210460 Expansion of x(1+x)/(1-x-2x^2-2*x^3-x^4).

Original entry on oeis.org

1, 2, 4, 10, 23, 53, 123, 285, 660, 1529, 3542, 8205, 19007, 44030, 101996, 236275, 547334, 1267906, 2937120, 6803875, 15761261, 36511157, 84578549, 195927260, 453867933, 1051390708, 2435559643, 5642004185, 13069772820, 30276291184

Offset: 1

Perminova Maria, Jan 22 2013

Transform of Fibonacci numbers based on the triangle A030528.

Index entries for linear recurrences with constant coefficients, signature (1,2,2,1).

Cf. A000045, A030528, A123392.

[&+[Fibonacci(k)*Binomial(k,n-k): k in [Floor((n-1)/2)+1..n]]: n in [1..30]]; // Bruno Berselli, Jan 23 2013

CoefficientList[Series[(1 + x)/(1 - x - 2 x^2 - 2 x^3 - x^4), {x, 0, 30}], x] (* Bruno Berselli, Jan 23 2013 *)
LinearRecurrence[{1,2,2,1},{1,2,4,10},30] (* Harvey P. Dale, Mar 28 2015 *)

a(n) = sum(Fibonacci(k)*binomial(k,n-k), k=floor((n-1)/2)+1..n).
G.f.: x*(1+x)/(1-x-2*x^2-2*x^3-x^4).
a(n) = A123392(n-1)+A123392(n-2). [Bruno Berselli, Jan 23 2013]

A177040 Irregular triangle t(n,m) = binomial(m+1,n-m) read by rows floor((n+1)/2) <= m <= n.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 3, 4, 1, 6, 5, 1, 4, 10, 6, 1, 10, 15, 7, 1, 5, 20, 21, 8, 1, 15, 35, 28, 9, 1, 6, 35, 56, 36, 10, 1, 21, 70, 84, 45, 11, 1, 7, 56, 126, 120, 55, 12, 1, 28, 126, 210, 165, 66, 13, 1, 8, 84, 252, 330, 220, 78, 14, 1, 36, 210, 462, 495, 286, 91, 15, 1

Offset: 0

Roger L. Bagula, May 01 2010

Row sums are in A052952.

Contains the right half of each row of A030528. - R. J. Mathar, May 19 2013

			1;
1;
2, 1;
3, 1;
3, 4, 1;
6, 5, 1;
4, 10, 6, 1;
10, 15, 7, 1;
5, 20, 21, 8, 1;
15, 35, 28, 9, 1;
6, 35, 56, 36, 10, 1;
21, 70, 84, 45, 11, 1;
7, 56, 126, 120, 55, 12, 1;
28, 126, 210, 165, 66, 13, 1;
8, 84, 252, 330, 220, 78, 14, 1;
36, 210, 462, 495, 286, 91, 15, 1;

Cf. A180987 (read diagonally downwards), A098925, A026729, A085478, A165253

t[n_, m_] := Binomial[m + 1, n - m];
Table[Table[t[n, m], {m, Floor[(n + 1)/2], n}], {n, 0, 15}];
Flatten[%]

T(m,n)=binomial(n+1,m-n) \\ Charles R Greathouse IV, May 19 2013

cp's OEIS Frontend

A049327 A convolution triangle of numbers generalizing Pascal's triangle A007318.

Views

Author

Keywords

Examples

Links

Crossrefs

Formula

A179199 E.g.f. satisfies: A(x) = (1+x)/(1+2*x)*A(x+x^2) with A(0)=0.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

PARI

SageMath

Formula

A307500 Expansion of Product_{k>=1} 1/(1 - (x*(1 - x))^k).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A179200 E.g.f. equals the real part of the i-th iteration of (x + x^2), where i=sqrt(-1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula

A179201 E.g.f. equals the imaginary part of the i-th iteration of (x + x^2), where i=sqrt(-1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula

A339884 Triangle read by rows: T(n, m) gives the number of partitions of n with m parts and parts from {1, 2, 3}.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A049323 Triangle of coefficients of certain polynomials (exponents in increasing order), equivalent to A033842.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A125250 Square array, read by antidiagonals, where A(1,1) = A(2,2) = 1, A(1,2) = A(2,1) = 0, A(n,k) = 0 if n < 1 or k < 1, otherwise A(n,k) = A(n-2,k-2) + A(n-1,k-2) + A(n-2,k-1) + A(n-1,k-1).

Views

Author

Keywords

Comments

A179199 E.g.f. satisfies: A(x) = (1+x)/(1+2x)A(x+x^2) with A(0)=0.

A210460 Expansion of x(1+x)/(1-x-2x^2-2*x^3-x^4).