A027826 -id:A027826 - cp's OEIS frontend

A152193 Triangle read by rows, A034839 * (A027826 * 0^(n-k)).

1, 1, 1, 1, 1, 3, 1, 6, 2, 1, 10, 10, 1, 15, 30, 4, 1, 21, 70, 28, 1, 28, 140, 112, 9, 1, 36, 252, 336, 81, 1, 45, 420, 840, 405, 21, 1, 55, 660, 1848, 1485, 231, 1, 66, 990, 3696, 4455, 1386, 50

Offset: 0

Gary W. Adamson, Nov 28 2008

Row sums = A027826, the eigensequence of triangle A034839.

			First few rows of the triangle =
1;
1;
1, 1;
1, 3;
1, 6, 2;
1, 10, 10;
1, 15, 30, 4;
1, 21, 70, 28;
1, 28, 140, 112, 9;
1, 36, 252, 336, 81;
1, 45, 420, 840, 405, 21;
1, 55, 660, 1848, 1485, 231;
1, 66, 990, 3696, 4455, 1386, 50;
...

Cf. A034839, A027826.

A034839 * (A027826 * 0^(n-k)) = A034839 * an infinite lower triangular matrix with A027826: (1, 1, 2, 4, 9, 21, 50, 120,...) and the rest zeros.

A152226 Binomial transform of A027826.

Original entry on oeis.org

1, 2, 5, 14, 42, 132, 428, 1416, 4745, 16034, 54505, 186166, 638538, 2198692, 7598624, 26350944, 91670946, 319825796, 1118686730, 3921801820, 13775938132, 48473233032, 170815501576, 602710956464, 2128981351316, 7527515059432, 26637591204996, 94331697697688

Offset: 0

Gary W. Adamson, Nov 29 2008

nonn

			a(3) = 14 = (1, 1, 2, 4) dot (1, 3, 3, 1) = (1 + 3 + 6 + 4).

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A027826.

b:= proc(n) option remember; add(`if`(k=0, 1,
      `if`(k::odd, 0, b(k/2)))*binomial(n, k), k=0..n)
    end:
a:= n-> add(b(k)*binomial(n, k), k=0..n):
seq(a(n), n=0..40);  # Alois P. Heinz, Jul 08 2015

b[n_] := b[n] = Sum[If[k == 0, 1, If[OddQ[k], 0, b[k/2]]]*Binomial[n, k], {k, 0, n}];
a[n_] := Sum[b[k]*Binomial[n, k], {k, 0, n}];
Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Jun 04 2018, from Maple *)

A007318 * A027826 as a vector. A027826 = [1, 1, 2, 4, 9, 21, 50, 120, 290,...].

A-Number corrected by R. J. Mathar, Apr 04 2012

A051163 Sequence is defined by property that (a0,a1,a2,a3,...) = binomial transform of (a0,a0,a1,a1,a2,a2,a3,a3,...).

Original entry on oeis.org

1, 2, 5, 12, 30, 76, 194, 496, 1269, 3250, 8337, 21428, 55184, 142376, 367916, 952000, 2466014, 6393372, 16586678, 43054344, 111801908, 290412296, 754543052, 1960808160, 5096293794, 13247503540, 34440553562, 89549255592, 232868582328, 605646682144

Offset: 0

Jonas Wallgren

Equals the self-convolution of A027826. Also equals antidiagonal sums of symmetric square array A100936. - Paul D. Hanna, Nov 22 2004

Equals eigensequence of triangle A152198. - Gary W. Adamson, Nov 28 2008

Alois P. Heinz, Table of n, a(n) for n = 0..1000
N. J. A. Sloane, Transforms

Cf. A051164, A051165, A051166, A027826, A100936, A100937, A152198.

a:= proc(n) option remember; add(`if`(k<2, 1,
      a(iquo(k, 2)))*binomial(n, k), k=0..n)
    end:
seq(a(n), n=0..40);  # Alois P. Heinz, Jul 08 2015

a[n_] := a[n] = 1 + Sum[Binomial[k, j]*Binomial[n-k, j]*a[j], {k, 1, n}, {j, 0, n-k}]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 11 2015 *)

a(n)=1+sum(k=1,n,sum(j=0,n-k,binomial(k,j)*binomial(n-k,j)*a(j)))

a(n)=local(A,m); if(n<0,0,m=1; A=1+O(x); while(m<=n,m*=2; A=subst(A,x,(x/(1-x))^2)/(1-x)); polcoeff(A^2,n))
for(n=0,40,print1(a(n),", ")) \\ Paul D. Hanna, Nov 22 2004

a(n) = 1 + Sum_{k=1..n} Sum_{j=0..n-k} C(k, j)*C(n-k, j)*a(j). - Paul D. Hanna, Nov 22 2004
G.f. A(x) satisfies: A(x) = A(x^2/(1-x)^2)/(1-x)^2 and A(x^2) = A(x/(1+x))/(1+x)^2. - Paul D. Hanna, Nov 22 2004
a(0) = 1; a(n) = Sum_{k=0..floor(n/2)} binomial(n+1,2*k+1) * a(k). - Ilya Gutkovskiy, Apr 07 2022

More terms from Vladeta Jovovic, Jul 26 2002

A260492 Pascal's triangle aerated with columns of zeros.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 2, 0, 1, 1, 0, 3, 0, 3, 0, 1, 1, 0, 4, 0, 6, 0, 4, 0, 1, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1, 1, 0, 6, 0, 15, 0, 20, 0, 15, 0, 6, 0, 1, 1, 0, 7, 0, 21, 0, 35, 0, 35, 0, 21, 0, 7, 0, 1, 1, 0, 8, 0, 28, 0, 56, 0, 70, 0, 56, 0, 28, 0, 8, 0, 1, 1, 0, 9, 0, 36, 0, 84, 0, 126, 0, 126, 0, 84, 0, 36, 0, 9, 0, 1, 1, 0, 10, 0, 45, 0, 120, 0, 210, 0, 252, 0, 210, 0, 120, 0, 45, 0, 10, 0, 1

Offset: 0

Peter Bala, Aug 15 2015

To obtain this array we take Pascal's triangle A007318 and insert a column of zeros at columns 1, 3, 5, .... Thus the n-th row of this array gives the coefficients in the expansion of (1 + x^2)^n.
The transpose of this array is a generalized Riordan array (1, 1 + x^2) as defined by Wilson in 2005 (see the Bala link for details). Note that these are not the same as the generalized Riordan arrays introduced by Wang and Wang in 2008.
Call this array C and let p(x) = 1 + x^2. Then C^2 gives the coefficients in the expansion of the polynomials ( p(p(x)) )^n, C^3 gives the coefficients in the expansion of the polynomials ( p(p(p(x))) )^n and so on.
See A204293 for Pascal's triangle aerated by both row and columns.
The triangle is read by rows of lengths equal to the odd numbers 2*n + 1 = A005408(n), n >= 0; although it is then (from the 3rd line of comments & formula on) considered as an infinite square matrix C, the upper right being filled with zeros. - M. F. Hasler, Aug 19 2015

			Table begins
n\k 0  1  2  3  4  5  6  7  8  9 10 11 12
- - - - - - - - - - - - - - - - - - - - -
0   1
1   1  0  1
2   1  0  2  0  1
3   1  0  3  0  3  0  1
4   1  0  4  0  6  0  4  0  1
5   1  0  5  0 10  0 10  0  5  0  1
6   1  0  6  0 15  0 20  0 15  0  6  0  1
...

P. Bala, Notes on generalized Riordan arrays
Evgeniy Burlachenko, Riordan arrays and generalized Euler polynomials, arXiv:1709.02229 [math.NT], 2017.
Evgeniy Burlachenko, Riordan arrays, Chebyshev polynomials, Fibonacci bases, arXiv:1903.10435 [math.NT], 2019.
Alan D. Sokal, How to generalize (and not to generalize) the Chu-Vandermonde identity, arXiv:1804.08919 [math.CO], 2018.
W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.

Cf. A003095, A027826, A007318, A034870, A072191, A161869, A204293.

#A260492
#define the aerated Pascal matrix (with indexing starting at 1)
C := Matrix(30, 30, (i,j) -> (1 - mod(j-1, 2))*binomial(i-1, floor((j-1)/2))):
for n from 1 to 12 do seq(C(n, k), k = 1 .. 2*n-1) end do;
# alternative program using row polynomials
rowpoly:= proc(n) option remember; expand((1 + x^2)^n) end:
T := (n, k)-> coeff(rowpoly(n), x, k):
seq(seq(T(n, k), k = 0..2*n), n = 0..11);

T[n_, k_] := If[OddQ[k], 0, Binomial[n, k/2]];
Table[T[n, k], {n, 0, 10}, {k, 0, 2n}] // Flatten (* Jean-François Alcover, Mar 23 2018 *)

C=matrix(20,20,m,n,if(n%2,binomial(m-1,n\2))); a=concat(vector(5,i,vector(i*2-1,j,C[i,j]))) \\ (C is read by rows, not antidiagonals.) - M. F. Hasler, Aug 19 2015

T(n,k) = (1 - k mod 2) binomial(n,floor(k/2)).
O.g.f.: 1/(1 - (1 + x^2)*t) = 1 + (1 + x^2)*t + (1 + 2*x^2 + x^4)*t^2 + ....
Let C denote this array.
Row sums of C = 2^n; Row sums of C^2 = 5^n; Row sums of C^3 = 26^n; Row of sums C^4 = 677^n. In general the row sums of C^m = A003095(m)^n.
First column of C^2 = 2^n; first column of C^3 = 5^n, first column of C^4 = 26^n and so on.
Let P denote Pascal's triangle A007318. Then C * transpose(C) = P * transpose(P) = the square symmetric Pascal matrix.
For n >= 0, (P^n)*C is the array P^(n+1) aerated by columns.
First column of P*C = 2^n; first column of (P*C)^2 = 6^n; first column of (P*C)^3 = 38^n, and so on, where [2, 6, 38, ...] is A072191.
Let R equal Pascal's triangle aerated with rows of zeros. Then C*R = P^2.
R*P*C is P^3 aerated by both rows and columns
Conjecturally, the limit of R^n * C^n as n -> oo has as its first column an aerated version of A027826, with zeros elsewhere in the array.
Aeration by rows/columns amounts to multiplication to the left/right by the identity matrix aerated by rows/column: R = J*P, C = P*J'. This makes obvious that P^n*C = P^(n+1)*J', R*P^n = J*P^(n+1), R*P^n*C = J*P^(n+2)*J' (aerated by both rows and columns). - M. F. Hasler, Aug 19 2015

A351970 a(0) = 1; a(n) = Sum_{k=0..floor(n/3)} binomial(n,3k) a(k).

Original entry on oeis.org

1, 1, 1, 2, 5, 11, 22, 43, 85, 171, 351, 738, 1590, 3498, 7827, 17757, 40725, 94113, 218439, 507852, 1180395, 2739312, 6342084, 14642100, 33703182, 77344686, 176984406, 403901722, 919533655, 2089000936, 4737194867, 10726221713, 24257223605, 54805216567

Offset: 0

Ilya Gutkovskiy, Feb 26 2022

nonn

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

Cf. A027826, A351971.

a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, 3 k] a[k], {k, 0, Floor[n/3]}]; Table[a[n], {n, 0, 33}]
nmax = 33; A[] = 1; Do[A[x] = A[x^3/(1 - x)^3]/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

G.f. A(x) satisfies: A(x) = A(x^3/(1 - x)^3) / (1 - x).

E.g.f.: exp(x) * Sum_{n>=0} a(n) * x^(3*n) / (3*n)!.

A351971 a(0) = 1; a(n) = Sum_{k=0..floor(n/4)} binomial(n,4k) a(k).

Original entry on oeis.org

1, 1, 1, 1, 2, 6, 16, 36, 72, 136, 256, 496, 992, 2016, 4096, 8256, 16513, 32913, 65689, 131785, 266482, 544230, 1124344, 2352188, 4984968, 10699656, 23244976, 51072256, 113370112, 253930816, 573005056, 1300390016, 2962852353, 6766967329, 15472840497

Offset: 0

Ilya Gutkovskiy, Feb 26 2022

nonn

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

Cf. A027826, A351970, A352066.

a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, 4 k] a[k], {k, 0, Floor[n/4]}]; Table[a[n], {n, 0, 34}]
nmax = 34; A[] = 1; Do[A[x] = A[x^4/(1 - x)^4]/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

G.f. A(x) satisfies: A(x) = A(x^4/(1 - x)^4) / (1 - x).

E.g.f.: exp(x) * Sum_{n>=0} a(n) * x^(4*n) / (4*n)!.

A352044 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-1,2k) a(k).

Original entry on oeis.org

1, 1, 1, 2, 4, 8, 16, 33, 71, 159, 367, 864, 2058, 4934, 11870, 28608, 69020, 166652, 402748, 974464, 2361416, 5733656, 13954488, 34054016, 83351472, 204659440, 504162416, 1246065920, 3089752480, 7685498976, 19174255584, 47971104769, 120326714335

Offset: 0

Ilya Gutkovskiy, Mar 01 2022

nonn

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

Cf. A027826, A352045, A352066.

a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, 2 k] a[k], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 32}]
nmax = 32; A[] = 0; Do[A[x] = 1 + x A[x^2/(1 - x)^2]/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

G.f. A(x) satisfies: A(x) = 1 + x * A(x^2/(1 - x)^2) / (1 - x).

E.g.f.: Integral exp(x) * Sum_{n>=0} a(n) * x^(2*n) / (2*n)! dx.

A161869 Convergent of an infinite product of Pascal's triangles aerated by rows.

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 33, 71, 160, 376, 912, 2256, 5633, 14093, 35170, 87344, 215680, 529568, 1293633, 3146515, 7627208, 18441476, 44510160, 107310480, 258566402, 622900466, 1500717220, 3616471960, 8717948688, 21023129472, 50713990918, 122374025914, 295366777856

Offset: 0

Gary W. Adamson, Jun 20 2009

The sequence may be the binomial transform of A024493 interleaved with zeros. A024493 = (1, 1, 1, 2, 5, 11, 22,...); so the conjecture succeeds through a(12) = A007318 * [1, 0, 1, 0, 1, 0, 2, 0, 5, 0, 11,...].

Calculating more terms of the sequence shows the above conjecture is incorrect. - Peter Bala, Jul 07 2015

			Pascal(1)     Pascal(2)    Pascal(3)
1              1            1
1 1            0 0          0 0
1 2 1          1 1 0        0 0 0
1 3 3 1        0 0 0 0      1 1 0 0
1 4 6 4 1      1 2 1 0 0    0 0 0 0 0
...
First columns of
Pascal(1):
1, 1, 1, 1, 1, 1, 1, 1,...
Pascal(1)*Pascal(2):
1, 1, 2, 4, 8, 16, 32, 64,...
Pascal(1)*Pascal(2)*Pascal(3):
1, 1, 2, 4, 8, 16, 33, 71,...
Pascal(1)*Pascal(2)*Pascal(3)*Pascal(4):
1, 1, 2, 4, 8, 16, 33, 71,...
...
converging to A161869.

Cf. A024493, A027826 (from first column of Pascal(2)^n as n -> inf).

#A161869
#define aerated Pascal matrices (note indexing starts at 1)
Pascal := proc (n) local i, j, r;
Matrix(33, 33, (i, j) -> (product(r-(mod(i-1, n)), r = 1 .. n-1))*binomial(floor((i-1)/n), j-1) )/factorial(n-1) end proc:
#it suffices to take the product of the first four aerated
#Pascal arrays to get 33 correct terms of the sequence
seq((Pascal(1).Pascal(2).Pascal(3).Pascal(4))(n, 1), n = 1 .. 33);
# Peter Bala, Jul 07 2015

From Peter Bala, Jul 07 2015: (Start)
Construct an infinite set of Pascal's triangles aerated by rows, denoted Pascal(1), Pascal(2), ..., where Pascal(1) = A007318, Pascal(2) is an aerated version of Pascal(1) with alternate rows 1, 3, 5, ... set equal to (0, 0, 0, ...), Pascal(3) is a further aeration of Pascal(1) with now two adjacent rows set equal to (0, 0, 0, ...), and so on.
The infinite product Pascal(1)*Pascal(2)*Pascal(3)*... is well-defined. This sequence is the first column of the infinite product - all the other entries in the product are zero. (End)

Terms a(12) through a(32) added by Peter Bala, Jul 07 2015

A162533 a(n) = Sum_{k=0..n} binomial(n,2k)*A002426(k).

Original entry on oeis.org

1, 1, 2, 4, 10, 26, 68, 176, 454, 1174, 3052, 7976, 20932, 55108, 145448, 384704, 1019462, 2706214, 7194956, 19155896, 51065260, 136284236, 364097912, 973654240, 2605983772, 6980545276, 18712478072, 50196568144, 134739960904, 361892443592, 972537193168

Offset: 0

Paul Barry, Jul 05 2009

Hankel transform is (-1)^binomial(n,2)*(-2)^A128054(n) (see A128055).

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A027826.

b[n_] := If[n < 0, 0, 3^n Hypergeometric2F1[1/2, -n, 1, 4/3]]; Table[Sum[Binomial[n, 2*k]*b[k], {k, 0, n}], {n, 0, 50}] (* or *) CoefficientList[Series[(1-x)/sqrt(1-4*x+4*x^2-4*x^4), {x, 0, 50}], x] (* G. C. Greubel, Feb 27 2017 *)

x='x+O('x^50); Vec((1-x)/sqrt(1-4*x+4*x^2-4*x^4)) \\ G. C. Greubel, Feb 27 2017

G.f.: (1-x)/((1-x)^2-x^2-2x^4/((1-x)^2-x^2-x^4/((1-x)^2-x^2-x^4/(1-... (continued fraction).
G.f.: (1-x)/sqrt(1-4*x+4*x^2-4*x^4) = (1-x)/sqrt((1-2*x)^2-4*x^4) = (1-x)/sqrt((1-x-2*x^2)*(1-x+2*x^2)). - Paul Barry, Oct 13 2009
Conjecture: n*a(n) + (4-5*n)*a(n-1) + 2*(4*n-7)*a(n-2) + 4*(3-n)*a(n-3) + 4*(2-n)*a(n-4) + 4*(n-4)*a(n-5) = 0. - R. J. Mathar, Nov 16 2011
a(n) ~ 3^(1/4) * (1 + sqrt(3))^(n + 1/2) / (2^(3/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Jun 08 2019

A219226 Number of rooted unlabeled ordered (plane) trees with 2n leaves such that i) every internal node has an even number of children and ii) every path from the root to a leaf is the same length.

Original entry on oeis.org

0, 1, 2, 3, 6, 13, 29, 65, 147, 337, 785, 1857, 4452, 10789, 26365, 64833, 160167, 397025, 986593, 2456193, 6123726, 15286021, 38198573, 95555937, 239294222, 599914489, 1505750425, 3783967201, 9521244242, 23988787485, 60520345765, 152889244033, 386752047956

Offset: 0

Geoffrey Critzer, Nov 15 2012

Alois P. Heinz, Table of n, a(n) for n = 0..2401
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 91

Cf. A014535, A027826.

a:= proc(n) option remember; add(`if`(k=0, 1,
     `if`(k::odd, a((k+1)/2)*binomial(n-1, k), 0)), k=0..n-1)
    end:
seq(a(n), n=0..35);  # Alois P. Heinz, Feb 26 2022

nn=60;f[x_]:=Sum[a[n]x^n,{n,0,nn}];sol=SolveAlways[0 == Series[f[x]-x-f[x^2/(1-x^2)],{x,0,nn}],x];a[0]=0;Table[a[n],{n,0,nn,2}]/.sol

O.g.f. satisfies A(x) = x + A(x^2/(1-x^2)).

cp's OEIS Frontend

A152193 Triangle read by rows, A034839 * (A027826 * 0^(n-k)).

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A152226 Binomial transform of A027826.

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A051163 Sequence is defined by property that (a0,a1,a2,a3,...) = binomial transform of (a0,a0,a1,a1,a2,a2,a3,a3,...).

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A260492 Pascal's triangle aerated with columns of zeros.

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A351970 a(0) = 1; a(n) = Sum_{k=0..floor(n/3)} binomial(n,3*k) * a(k).

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A351971 a(0) = 1; a(n) = Sum_{k=0..floor(n/4)} binomial(n,4*k) * a(k).

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A352044 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-1,2*k) * a(k).

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A161869 Convergent of an infinite product of Pascal's triangles aerated by rows.

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A351970 a(0) = 1; a(n) = Sum_{k=0..floor(n/3)} binomial(n,3k) a(k).

A351971 a(0) = 1; a(n) = Sum_{k=0..floor(n/4)} binomial(n,4k) a(k).

A352044 a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-1,2k) a(k).