A108447 -id:A108447 - cp's OEIS frontend

A118343 Triangle, read by rows, where diagonals are successive self-convolutions of A108447.

1, 1, 0, 1, 1, 0, 1, 2, 4, 0, 1, 3, 9, 20, 0, 1, 4, 15, 48, 113, 0, 1, 5, 22, 85, 282, 688, 0, 1, 6, 30, 132, 519, 1762, 4404, 0, 1, 7, 39, 190, 837, 3330, 11488, 29219, 0, 1, 8, 49, 260, 1250, 5516, 22135, 77270, 199140, 0, 1, 9, 60, 343, 1773, 8461, 37404, 151089, 532239, 1385904, 0

Offset: 0

Paul D. Hanna, Apr 26 2006

A108447 equals the central terms of pendular triangle A118340 and the diagonals of this triangle form the semi-diagonals of the triangle A118340. Row sums equal A054727, the number of forests of rooted trees with n nodes on a circle without crossing edges.

			Show: T(n,k) = T(n-1,k) - T(n-1,k-1) + T(n,k-1) + T(n+1,k-1)
at n=8,k=4: T(8,4) = T(7,4) - T(7,3) + T(8,3) + T(9,3)
or 837 = 519 - 132 + 190 + 260.
Triangle begins:
  1;
  1, 0;
  1, 1,  0;
  1, 2,  4,   0;
  1, 3,  9,  20,    0;
  1, 4, 15,  48,  113,    0;
  1, 5, 22,  85,  282,  688,     0;
  1, 6, 30, 132,  519, 1762,  4404,      0;
  1, 7, 39, 190,  837, 3330, 11488,  29219,      0;
  1, 8, 49, 260, 1250, 5516, 22135,  77270, 199140,       0;
  1, 9, 60, 343, 1773, 8461, 37404, 151089, 532239, 1385904, 0;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A054727 (row sums), A108447, A118340.

T:= proc(n, k) option remember;
      if k<0 or  k>n then 0;
    elif k=0 then 1;
    elif k=n then 0;
    else T(n-1, k) -T(n-1, k-1) +T(n, k-1) +T(n+1, k-1);
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Mar 17 2021

T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0, 1, If[k==n, 0, T[n-1, k] -T[n-1, k-1] +T[n, k-1] +T[n+1, k-1] ]]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Mar 17 2021 *)

{T(n,k)=polcoeff((serreverse(x*(1-x+sqrt((1-x)*(1-5*x)+x*O(x^k)))/2/(1-x))/x)^(n-k),k)}

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==0): return 1
    elif (k==n): return 0
    else: return T(n-1, k) -T(n-1, k-1) +T(n, k-1) +T(n+1, k-1)
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 17 2021

Since g.f. G=G(x) of A108447 satisfies: G = 1 - x*G + x*G^2 + x*G^3 then T(n,k) = T(n-1,k) - T(n-1,k-1) + T(n,k-1) + T(n+1,k-1). Also, a recurrence involving antidiagonals is: T(n,k) = T(n-1,k) + Sum_{j=1..k} [2*T(n-1+j,k-j) - T(n-2+j,k-j)] for n>k>=0.

Sum_{k=0..n} T(n,k) = [n=0] + A054727(n) = [n=0] + Sum_{j=1..n} binomial(n, j-1)*binomial(3*n-2*j-1, n-j)/(2*n-j). - G. C. Greubel, Mar 17 2021

A118342 Self-convolution cube of A108447.

Original entry on oeis.org

1, 3, 15, 85, 519, 3330, 22135, 151089, 1052805, 7458236, 53554548, 388913046, 2851368587, 21076979445, 156907929177, 1175381112901, 8853057357159, 67007699270040, 509390314298820, 3887605401119964, 29775366104932044

Offset: 0

Paul D. Hanna, Apr 25 2006

nonn

Cf. A108447, A118341, A118343.

{a(n)=polcoeff((serreverse(x*(1-x+sqrt((1-x)*(1-5*x)+x*O(x^n)))/2/(1-x))/x)^3,n)}

A118341 Self-convolution square of A108447.

Original entry on oeis.org

1, 2, 9, 48, 282, 1762, 11488, 77270, 532239, 3735488, 26617976, 192061278, 1400453568, 10303466638, 76391703591, 570195719792, 4281169118106, 32312317899352, 245016567557504, 1865677590201192, 14259825593908356

Offset: 0

Paul D. Hanna, Apr 25 2006

nonn

Cf. A108447, A118342, A118343.

{a(n)=polcoeff((serreverse(x*(1-x+sqrt((1-x)*(1-5*x)+x*O(x^n)))/2/(1-x))/x)^2,n)}

A364748 G.f. A(x) satisfies A(x) = 1 + xA(x)^5 / (1 - xA(x)).

Original entry on oeis.org

1, 1, 6, 47, 424, 4159, 43097, 464197, 5145475, 58313310, 672598269, 7869856070, 93183973405, 1114471042413, 13443614108307, 163372291277764, 1998239045199623, 24580340878055298, 303893356012560280, 3774099648814193998, 47061518776483143441

Offset: 0

Seiichi Manyama, Aug 05 2023

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..890

Cf. A000108, A001003, A108447, A364747, A378691, A378692.

Cf. A002295, A243667, A365192, A365193, A365194.

a(n) = if(n==0, 1, sum(k=0, n-1, binomial(n, k)*binomial(5*n-4*k, n-1-k))/n);

a(n, r=1, s=1, t=5, u=1) = r*sum(k=0, n, binomial(t*k+u*(n-k)+r, k)*binomial(n+(s-1)*k-1, n-k)/(t*k+u*(n-k)+r)); \\ Seiichi Manyama, Dec 05 2024

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n,k) * binomial(5*n-4*k,n-1-k) for n > 0.
From Seiichi Manyama, Dec 05 2024: (Start)
G.f. A(x) satisfies A(x) = 1/(1 - x*A(x)^4/(1 - x*A(x))).
If g.f. satisfies A(x) = ( 1 + x*A(x)^(t/r) / (1 - x*A(x)^(u/r))^s )^r, then a(n) = r * Sum_{k=0..n} binomial(t*k+u*(n-k)+r,k) * binomial(n+(s-1)*k-1,n-k)/(t*k+u*(n-k)+r). (End)

A118340 Pendular triangle, read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + T(n-1,k), for n>=k>0, with T(n,0) = 1 and T(n,n) = 0^n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 1, 0, 1, 4, 9, 5, 1, 0, 1, 5, 15, 20, 6, 1, 0, 1, 6, 22, 48, 28, 7, 1, 0, 1, 7, 30, 85, 113, 37, 8, 1, 0, 1, 8, 39, 132, 282, 169, 47, 9, 1, 0, 1, 9, 49, 190, 519, 688, 237, 58, 10, 1, 0, 1, 10, 60, 260, 837, 1762, 1074, 318, 70, 11, 1, 0

Offset: 0

Paul D. Hanna, Apr 25 2006

Definitions. A pendular triangle is a triangle in which row n is generated from the pendular sums of row n-1. Pendular sums of a row are partial sums taken in back-and-forth order, starting with the leftmost term, jumping to the rightmost term, back to the leftmost unused term, then forward to the rightmost unused term, etc.
In each pass, the partial sum is placed in the new row directly under the term most recently used in the sum. Continue in this way until all the terms of the prior row have been used and then complete the new row by appending a zero at the end. Pendular sums are so named because the process resembles a swinging pendulum that slows down on each pass until it eventually comes to rest in the center.
In the simplest case, pendular triangles obey the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + p*T(n-1,k), for n>=k>0, with T(n,0)=1 and T(n,n)=0^n, for some fixed number p.
In which case the g.f. G=G(x) of the central terms satisfies: G = 1 - p*x*G + p*x*G^2 + x*G^3.
More generally, a pendular triangle is defined by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + Sum_{j>=1} p(j)*T(n-1,k-1+j), for n>=k>0, with T(n,0)=1 and T(n,n)=0^n.
Remarkably, the g.f. G=G(x) of the central terms satisfies: G = 1 + x*G^3 + Sum_{j>=1} p(j)*x^j*[G^(2*j) - G^(2*j-1)].
Further, the g.f. of the m-th lower semi-diagonal equals G(x)^(m+1) for m>=0, where the m-th semi-diagonal consists of those terms located at m rows below the central terms.
For variants of pendular triangles, the main diagonal may be nonzero, but then the g.f.s of the semi-diagonals are more complex.

			Row 6 equals the pendular sums of row 5:
  [1,  4,  9,  5,  1,  0], where the sums proceed as follows:
  [1, __, __, __, __, __], T(6,0) = T(5,0) = 1;
  [1, __, __, __, __,  1], T(6,5) = T(6,0) + T(5,5) = 1 + 0 = 1;
  [1,  5, __, __, __,  1], T(6,1) = T(6,5) + T(5,1) = 1 + 4 = 5;
  [1,  5, __, __,  6,  1], T(6,4) = T(6,1) + T(5,4) = 5 + 1 = 6;
  [1,  5, 15, __,  6,  1], T(6,2) = T(6,4) + T(5,2) = 6 + 9 = 15;
  [1,  5, 15, 20,  6,  1], T(6,3) = T(6,2) + T(5,3) = 15 + 5 = 20;
  [1,  5, 15, 20,  6,  1, 0] finally, append a zero to obtain row 6.
Triangle begins:
  1;
  1,  0;
  1,  1,  0;
  1,  2,  1,   0;
  1,  3,  4,   1,    0;
  1,  4,  9,   5,    1,    0;
  1,  5, 15,  20,    6,    1,    0;
  1,  6, 22,  48,   28,    7,    1,    0;
  1,  7, 30,  85,  113,   37,    8,    1,   0;
  1,  8, 39, 132,  282,  169,   47,    9,   1,  0;
  1,  9, 49, 190,  519,  688,  237,   58,  10,  1,  0;
  1, 10, 60, 260,  837, 1762, 1074,  318,  70, 11,  1, 0;
  1, 11, 72, 343, 1250, 3330, 4404, 1568, 413, 83, 12, 1, 0; ...
Central terms are T(2*n,n) = A108447(n);
semi-diagonals form successive self-convolutions of the central terms:
T(2*n+1,n) = A118341(n) = [A108447^2](n),
T(2*n+2,n) = A118342(n) = [A108447^3](n).

G. C. Greubel, Rows n = 0..100 of the triangle, flattened(Rows n = 0..20 from Paul D. Hanna)

Cf. A108447 (central terms), A118341, A118343.
variants: A118344 (Catalan), A118362 (ternary), A118350, A118355.
Cf. A167763 (p=0), this sequence (p=1), A118345 (p=2), A118350 (p=3).

function T(n,k,p)
  if k lt 0 or n lt k then return 0;
  elif k eq 0 then return 1;
  elif k eq n then return 0;
  elif n gt 2*k then return T(n,n-k,p) + T(n-1,k,p);
  else return T(n,n-k-1,p) + p*T(n-1,k,p);
  end if;
  return T;
end function;
[T(n,k,1): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 17 2021

T[n_, k_, p_]:= T[n,k,p] = If[nG. C. Greubel, Feb 17 2021 *)

{T(n,k) = if(n2*k, T(n-1,k) + T(n,n-k), T(n-1,k) + T(n,n-1-k)))))}
for(n=0,12, for(k=0,n, print1(T(n,k),", "));print(""))

@CachedFunction
def T(n, k, p):
    if (k<0 or n2*k): return T(n,n-k,p) + T(n-1,k,p)
    else: return T(n, n-k-1, p) + p*T(n-1, k, p)
flatten([[T(n,k,1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 17 2021

T(2*n+m,n) = [A108447^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A108447; compare semi-diagonals to the diagonals of convolution triangle A118343.

A364747 G.f. A(x) satisfies A(x) = 1 + xA(x)^4 / (1 - xA(x)).

Original entry on oeis.org

1, 1, 5, 32, 234, 1854, 15490, 134380, 1198944, 10931761, 101412677, 954155059, 9083120975, 87326765375, 846709605539, 8269910074087, 81291388929027, 803592049667495, 7983612883739843, 79671910265120574, 798283229227457304, 8027625597750959053

Offset: 0

Seiichi Manyama, Aug 05 2023

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..972

Cf. A000108, A001003, A108447, A364748, A378691, A378692.

Cf. A002294, A243659, A364765, A364792.

a(n) = if(n==0, 1, sum(k=0, n-1, binomial(n, k)*binomial(4*n-3*k, n-1-k))/n);

a(n, r=1, s=1, t=4, u=1) = r*sum(k=0, n, binomial(t*k+u*(n-k)+r, k)*binomial(n+(s-1)*k-1, n-k)/(t*k+u*(n-k)+r)); \\ Seiichi Manyama, Dec 05 2024

a(n) = (1/n) * Sum_{k=0..n-1} binomial(n,k) * binomial(4*n-3*k,n-1-k) for n > 0.
From Seiichi Manyama, Dec 05 2024: (Start)
G.f. A(x) satisfies A(x) = 1/(1 - x*A(x)^3/(1 - x*A(x))).
If g.f. satisfies A(x) = ( 1 + x*A(x)^(t/r) / (1 - x*A(x)^(u/r))^s )^r, then a(n) = r * Sum_{k=0..n} binomial(t*k+u*(n-k)+r,k) * binomial(n+(s-1)*k-1,n-k)/(t*k+u*(n-k)+r). (End)

A364984 E.g.f. satisfies A(x) = 1 + xA(x)^3exp(x*A(x)).

Original entry on oeis.org

1, 1, 8, 117, 2596, 77705, 2936406, 134228059, 7204913528, 444331053873, 30963240318250, 2406301353714731, 206354828717754036, 19357367027097743449, 1971809610601104110942, 216754216326949771274715, 25575749384428387961718256, 3224227609551980271408565985

Offset: 0

Seiichi Manyama, Aug 15 2023

nonn

Cf. A364983, A364985, A364986.

Cf. A108447.

a(n) = n!*sum(k=0, n, k^(n-k)*binomial(n+2*k+1, k)/((n+2*k+1)*(n-k)!));

a(n) = n! * Sum_{k=0..n} k^(n-k) * binomial(n+2*k+1,k)/( (n+2*k+1)*(n-k)! ).

A378692 G.f. A(x) satisfies A(x) = 1 + xA(x)^7/(1 - xA(x)).

Original entry on oeis.org

1, 1, 8, 86, 1075, 14667, 211799, 3182454, 49243854, 779379652, 12558073022, 205312307834, 3397359326116, 56790504859929, 957574385205771, 16267419813629731, 278162968238908681, 4783813617177604232, 82691541747420586716, 1435895455224032519430, 25035634270828781060188

Offset: 0

Seiichi Manyama, Dec 04 2024

nonn

Cf. A000108, A001003, A108447, A364747, A364748, A378691.
Cf. A378686, A378690, A378693.
Cf. A378685, A378688, A378694.

a(n, r=1, s=1, t=7, u=1) = r*sum(k=0, n, binomial(t*k+u*(n-k)+r, k)*binomial(n+(s-1)*k-1, n-k)/(t*k+u*(n-k)+r));

G.f. A(x) satisfies A(x) = 1/(1 - x*A(x)^6/(1 - x*A(x))).

If g.f. satisfies A(x) = ( 1 + x*A(x)^(t/r) / (1 - x*A(x)^(u/r))^s )^r, then a(n) = r * Sum_{k=0..n} binomial(t*k+u*(n-k)+r,k) * binomial(n+(s-1)*k-1,n-k)/(t*k+u*(n-k)+r).

A336706 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(0,k) = 1 and T(n,k) = (1/n) * Sum_{j=1..n} binomial(n,j) * binomial(n+(k-1)*j,j-1) for n > 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 5, 4, 1, 1, 4, 11, 14, 9, 1, 1, 5, 20, 45, 42, 21, 1, 1, 6, 32, 113, 197, 132, 51, 1, 1, 7, 47, 234, 688, 903, 429, 127, 1, 1, 8, 65, 424, 1854, 4404, 4279, 1430, 323, 1, 1, 9, 86, 699, 4159, 15490, 29219, 20793, 4862, 835

Offset: 0

Seiichi Manyama, Aug 01 2020

			Square array begins:
   1,   1,   1,    1,     1,     1,      1, ...
   1,   1,   1,    1,     1,     1,      1, ...
   1,   2,   3,    4,     5,     6,      7, ...
   2,   5,  11,   20,    32,    47,     65, ...
   4,  14,  45,  113,   234,   424,    699, ...
   9,  42, 197,  688,  1854,  4159,   8192, ...
  21, 132, 903, 4404, 15490, 43097, 101538, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0-3 give: A001006(n-1), A000108, A001003, A108447.
Main diagonal gives A335871.
Cf. A336575, A336707, A336708, A336709.

T[0, k_] := 1; T[n_, k_] := Sum[Binomial[n, j] * Binomial[n + (k - 1)*j, j - 1], {j, 1, n}] / n; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Aug 01 2020 *)

{T(n, k) = if(n==0, 1, sum(j=1, n, binomial(n, j)*binomial(n+(k-1)*j, j-1))/n)}

{T(n, k) = local(A=1+x*O(x^n)); for(i=0, n, A=1+x*A^k/(1-x*A)); polcoef(A, n)}

G.f. A_k(x) of column k satisfies A_k(x) = 1 + x * A_k(x)^k / (1 - x * A_k(x)).

A378686 G.f. A(x) satisfies A(x) = ( 1 + xA(x)^(7/3)/(1 - xA(x)) )^3.

Original entry on oeis.org

1, 3, 27, 313, 4122, 58584, 875897, 13577139, 216224616, 3516601243, 58160887857, 975211608399, 16539799297342, 283243124783136, 4890858070498203, 85060240453556192, 1488653675438168001, 26197808077514204832, 463311206395709908936, 8229849868810254813378

Offset: 0

Seiichi Manyama, Dec 04 2024

nonn

Cf. A108447, A371581.
Cf. A349017, A369012.
Cf. A378690, A378692, A378693.
Cf. A378685.

a(n, r=3, s=1, t=7, u=3) = r*sum(k=0, n, binomial(t*k+u*(n-k)+r, k)*binomial(n+(s-1)*k-1, n-k)/(t*k+u*(n-k)+r));

G.f. A(x) satisfies A(x) = 1/( 1 - x*A(x)^2/(1 - x*A(x)) )^3.
G.f.: A(x) = B(x)^3 where B(x) is the g.f. of A378685.
If g.f. satisfies A(x) = ( 1 + x*A(x)^(t/r) / (1 - x*A(x)^(u/r))^s )^r, then a(n) = r * Sum_{k=0..n} binomial(t*k+u*(n-k)+r,k) * binomial(n+(s-1)*k-1,n-k)/(t*k+u*(n-k)+r).

cp's OEIS Frontend

A118343 Triangle, read by rows, where diagonals are successive self-convolutions of A108447.

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A118342 Self-convolution cube of A108447.

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A118341 Self-convolution square of A108447.

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A364748 G.f. A(x) satisfies A(x) = 1 + x*A(x)^5 / (1 - x*A(x)).

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A118340 Pendular triangle, read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) + T(n-1,k), for n>=k>0, with T(n,0) = 1 and T(n,n) = 0^n.

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A364747 G.f. A(x) satisfies A(x) = 1 + x*A(x)^4 / (1 - x*A(x)).

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A364984 E.g.f. satisfies A(x) = 1 + x*A(x)^3*exp(x*A(x)).

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A378692 G.f. A(x) satisfies A(x) = 1 + x*A(x)^7/(1 - x*A(x)).

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A336706 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(0,k) = 1 and T(n,k) = (1/n) * Sum_{j=1..n} binomial(n,j) * binomial(n+(k-1)*j,j-1) for n > 0.

A364748 G.f. A(x) satisfies A(x) = 1 + xA(x)^5 / (1 - xA(x)).

A364747 G.f. A(x) satisfies A(x) = 1 + xA(x)^4 / (1 - xA(x)).

A364984 E.g.f. satisfies A(x) = 1 + xA(x)^3exp(x*A(x)).

A378692 G.f. A(x) satisfies A(x) = 1 + xA(x)^7/(1 - xA(x)).

A378686 G.f. A(x) satisfies A(x) = ( 1 + xA(x)^(7/3)/(1 - xA(x)) )^3.