A084786 -id:A084786 - cp's OEIS frontend

A084783 Triangle, read by rows, such that the diagonal (A084785) is the self-convolution of the first column (A084784) and the row sums (A084786) gives the differences of the diagonal and the first column.

1, 1, 2, 2, 3, 5, 6, 8, 11, 16, 25, 31, 39, 50, 66, 137, 162, 193, 232, 282, 348, 944, 1081, 1243, 1436, 1668, 1950, 2298, 7884, 8828, 9909, 11152, 12588, 14256, 16206, 18504, 77514, 85398, 94226, 104135, 115287, 127875, 142131, 158337, 176841

Offset: 0

Paul D. Hanna, Jun 13 2003

			Triangle begins:
      1;
      1,     2;
      2,     3,     5;
      6,     8,    11,     16;
     25,    31,    39,     50,     66;
    137,   162,   193,    232,    282,    348;
    944,  1081,  1243,   1436,   1668,   1950,   2298;
   7884,  8828,  9909,  11152,  12588,  14256,  16206,  18504;
  77514, 85398, 94226, 104135, 115287, 127875, 142131, 158337, 176841;
  ...

Alois P. Heinz, Rows n = 0..150, flattened (first 45 rows from Paul D. Hanna)

Cf. A084784, A084785, A084786.

m:=50;
f:= func< n,x | Exp((&+[(&+[Factorial(j)*StirlingSecond(k,j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;
R:=PowerSeriesRing(Rationals(), m+1);
b:=Coefficients(R!( f(m,x) )); // b = A084784
function T(n,k) // T = A084783
  if k eq 0 then return b[n+1];
  else return T(n,k-1) + T(n-1,k-1);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jun 08 2023

T:= proc(n, k) option remember; `if`(k=0, 1+add(T(j, 0)*
     (binomial(n, j)-T(n-j, 0)), j=1..n-1), T(n, k-1)+T(n-1, k-1))
    end:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Jun 09 2023

b[n_]:= b[n]= If[n<1, Boole[n==0], Module[{A= 1/x -1/x^2}, Do[A=2A - Normal@Series[(x A^2)/. x-> x-1, {x, Infinity, k+1}], {k,2,n}]; (-1)^n Coefficient[A, x, -n-1]]]; (* b = A084784 *)
T[n_, k_]:= T[n, k]= If[k==0, b[n], T[n, k-1] +T[n-1, k-1]];
Table[T[n,k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jun 07 2023 *)

{A084784(n) = local(A); if( n<0, 0, A=1; for(k=1, n, A = truncate(A + O(x^k)) + x * O(x^k); A += A - 1 / subst(A^-2, x, x /(1 + x)) / (1 + x); ); polcoeff(A, n))}; /* After Michael Somos */
{T(n,k)=if(k==0,if(n==0,1,A084784(n)),T(n, k-1)+T(n-1, k-1))}
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

def f(n, x): return exp(sum(sum( factorial(j)*stirling_number2(k,j) *x^k/k for j in range(1,k+1)) for k in range(1,n+2)))
m=50
def A084784_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m,x) ).list()
b=A084784_list(m)
def T(n,k): # T = A084783
    if k==0: return b[n]
    else: return T(n, k-1) + T(n-1, k-1)
flatten([[T(n, k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Jun 08 2023

T(0,0) = 1, T(n,0) = A084784(n), T(n,n) = A084785(n), T(n,k) = T(n,k-1) + T(n-1,k-1) for n>0, k>0.

A084784 Binomial transform = self-convolution: first column of the triangle (A084783).

Original entry on oeis.org

1, 1, 2, 6, 25, 137, 944, 7884, 77514, 877002, 11218428, 160010244, 2516742498, 43260962754, 806650405800, 16213824084864, 349441656710217, 8037981040874313, 196539809431339642, 5090276002949080318, 139202688233361310841, 4008133046329085884137

Offset: 0

Paul D. Hanna, Jun 13 2003

nonn

In the triangle (A084783), the diagonal (A084785) is the self-convolution of this sequence and the row sums (A084786) gives the differences of the diagonal and this sequence.
Ramanujan considers the continued fraction phi(x) = 1 / (x + 1 - 1^2 / (x + 3 - 2^2 / (x + 5 - 3^2 / (x + 7 - 4^2 / ...)))) and states that phi(x+1) approaches x phi(x)^2 as x gets large. The asymptotic expansion is phi(x) = 1/x - 1/x^2 + 2/x^3 - 6/x^4 + 24/x^5 - ... + (-1)^n * n! / x^(n+1) + ... but if we replace this with f(x) = a(0)/x - a(1)/x^2 + a(2)/x^3 - a(3)/x^4 + ... then formally f(x+1) = x f(x)^2 which is similar to my Feb 16 2006 formula. - Michael Somos, Jun 20 2015
This is also the Euler transform of A060223. - Gus Wiseman, Oct 16 2016

			G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 25*x^4 + 137*x^5 + 944*x^6 + ...
where
A(x) = (1-x)^(-1/4)*(1-2*x)^(-1/8)*(1-3*x)^(-1/16)*(1-4*x)^(-1/32)*...
Also,
log(A(x)) = x + 3*x^2/2 + 13*x^3/3 + 75*x^4/4 + 541*x^5/5 + 4683*x^6/6 + ... + A000670(n)*x^n/n + ...
thus, the logarithmic derivative equals the series:
A'(x)/A(x) = 1/(1-x) + 2!*x/((1-x)*(1-2*x)) + 3!*x^2/((1-x)*(1-2*x)*(1-3*x)) + 4!*x^3/((1-x)*(1-2*x)*(1-3*x)*(1-4*x)) + ...

S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, Bombay 1957 Vol. 1, see page 223.

Paul D. Hanna, Table of n, a(n) for n = 0..200

Cf. A000670, A060223, A084783, A084785, A084786, A195983, A088791.

m:=50;
f:= func< n,x | Exp((&+[(&+[Factorial(j)*StirlingSecond(k,j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;
R:=PowerSeriesRing(Rationals(), m+1); // A084784
Coefficients(R!( f(m,x) )); // G. C. Greubel, Jun 08 2023

a:= proc(n) option remember;
      1+add(a(j)*(binomial(n,j)-a(n-j)), j=1..n-1)
    end:
seq(a(n), n=0..25);  # Alois P. Heinz, Jun 09 2023

a[ n_]:= If[n<1, Boole[n==0], Module[{A= 1/x - 1/x^2}, Do [A= 2 A - Normal @ Series[ (x A^2) /. x -> x-1, {x, Infinity, k+1}], {k,2,n}]; (-1)^n Coefficient[A, x, -n-1]]]; (* Michael Somos, Jun 20 2015 *)
nn=20;CoefficientList[Series[Exp[Sum[Times[1/k,i!,StirlingS2[k,i],x^k],{k,nn},{i,k}]],{x,0,nn}],x] (* Gus Wiseman, Oct 18 2016 *)

{a(n) = my(A); if( n<0, 0, A=1; for(k=1, n, A = truncate(A + O(x^k)) + x * O(x^k); A += A - 1 / subst(A^-2, x, x / (1 + x)) / (1 + x);); polcoeff(A, n))}; /* Michael Somos, Feb 18 2006 */

/* Using o.g.f. exp( Sum_{n>=1} A000670(n)*x^n/n ): */
{a(n) = polcoef(exp(intformal(sum(m=1, n+1, m!*x^(m-1)/prod(k=1, m, 1-k*x+x*O(x^n))))), n)}
for(n=0,30,print1(a(n),", "))

# after Alois P. Heinz
from functools import cache
from math import comb as binomial
@cache
def a(n: int) -> int:
    return 1 + sum((binomial(n, j) - a(n - j)) * a(j) for j in range(1, n))
print([a(n) for n in range(22)])  # Peter Luschny, Jun 09 2023

m=40
def f(n, x): return exp(sum(sum(factorial(j)*stirling_number2(k,j) *x^k/k for j in range(1,k+1)) for k in range(1,n+2)))
def A084784_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m,x) ).list()
A084784_list(m) # G. C. Greubel, Jun 08 2023

G.f. satisfies A(n*x)^2 = n-th binomial transform of A(n*x).
G.f. A(x) satisfies 1 + x = A(x/(1 + x))^2 / A(x). - Michael Somos, Feb 16 2006
G.f.: A(x) = Product_{n>=1} 1/(1 - n*x)^(1/2^(n+1)). - Paul D. Hanna, Jun 16 2010
G.f.: A(x) = exp( Sum_{n>=1} A000670(n)*x^n/n ) where Sum_{n>=0} A000670(n)*x^n = Sum_{n>=0} n!*x^n/Product_{k=0..n} (1-k*x). - Paul D. Hanna, Sep 26 2011
a(n) ~ (n-1)! / (2 * (log(2))^(n+1)). - Vaclav Kotesovec, Nov 18 2014
G.f. satisfies [x^n] 1/A(x)^(n-1) = [x^n] 1/A(x)^(2*n-2) = -(n-1)*A088791(n) for n >= 0. - Paul D. Hanna, Apr 28 2025

A084785 Diagonal of the triangle (A084783) and the self-convolution of the first column (A084784).

Original entry on oeis.org

1, 2, 5, 16, 66, 348, 2298, 18504, 176841, 1958746, 24661493, 347548376, 5415830272, 92410046544, 1712819553864, 34258146124320, 735267392077962, 16852848083339700, 410809882438699346, 10611174406149372736, 289493459925589039804, 8317946739043065421640

Offset: 0

Paul D. Hanna, Jun 13 2003

In the triangle (A084783), the diagonal (this sequence) is the self-convolution of the first column (A084784) and the row sums (A084786) gives the differences of the diagonal and the first column.

			G.f.: A(x) = (1-x)^(-1/2)*(1-2*x)^(-1/4)*(1-3*x)^(-1/8)*(1-4*x)^(-1/16)*... - _Paul D. Hanna_, Jun 16 2010

Vaclav Kotesovec, Table of n, a(n) for n = 0..350
Chao-Ping Chen, Sharp inequalities and asymptotic series related to Somos' quadratic recurrence constant, Journal of Number Theory, 2016, Volume 172, March 2017, Pages 145-159.
Olivier Golinelli, Remote control system of a binary tree of switches - II. balancing for a perfect binary tree, arXiv:2405.16968 [cs.DM], 2024. See p. 17.

Cf. A000629, A000670, A019538, A084783, A084784, A084786, A090352.

m:=40;
f:= func< n,x | Exp((&+[(&+[(-2)^j*Factorial(j)*StirlingSecond(k,j)*(-x)^k/k: j in [1..k]]): k in [1..n+2]])) >;
R:=PowerSeriesRing(Rationals(), m+1);  // A084785
Coefficients(R!( f(m,x) )); // G. C. Greubel, Jun 08 2023

nmax = 19; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[(1+x)^2 * A[x] - A[x/(1+x)]^2 + O[x]^(n+1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 02 2019 *)
With[{m=40}, CoefficientList[Series[Exp[Sum[Sum[(-2)^j*j!*StirlingS2[k, j], {j,k}]*(-x)^k /k, {k,m+1}]], {x,0,m}], x]] (* G. C. Greubel, Jun 08 2023 *)

A = matrix(25, 25); A[1, 1] = 1; rs = 1; print(1); for (n=2, 25, sc = sum(i=2, n-1, A[i, 1]*A[n+1-i, 1]); A[n, 1] = rs - sc; rs = A[n, 1]; for (k=2, n, A[n, k] = A[n, k-1] + A[n-1, k-1]; rs += A[n, k]); print(A[n, n])); \\ David Wasserman, Jan 06 2005

{a(n)=local(A); if(n<0, 0, A=1; for(k=1,n, A=truncate(A+O(x^k))+x*O(x^k); A+=A-(subst(1/A,x,x/(1+x))*(1+x))^-2;); polcoeff(A,n))} /* Michael Somos, Feb 18 2006 */

def f(n, x): return exp(sum(sum( (-2)^j*factorial(j)* stirling_number2(k,j)*(-x)^k/k for j in range(1,k+1)) for k in range(1,n+2)))
m=50
def A084785_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m,x) ).list()
A084785_list(m-9) # G. C. Greubel, Jun 08 2023

G.f. A(x) satisfies (1+x)^2 = A(x/(1+x))^2/A(x). - Michael Somos, Feb 16 2006
G.f.: A(x) = Product_{n>=1} 1/(1 - n*x)^(1/2^n). - Paul D. Hanna, Jun 16 2010
a(n) ~ (n-1)! / (log(2))^(n+1). - Vaclav Kotesovec, Nov 19 2014
From Peter Bala, May 26 2001: (Start)
O.g.f.: A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = (-1)^n*Sum_{k = 1..n} k!*Stirling2(n,k)*(-2)^k = A000629(n) = 2*A000670(n) for n >= 1. Cf. A090352.
sqrt(A(x)) = 1/(1 + x)*A(x/(1 + x)) = 1 + x + 2*x^2 + 6*x^3 + 25*x^4 + 137*x^5 + ... is the o.g.f. for A084784. See also A019538. (End)

More terms from David Wasserman, Jan 06 2005

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A084783 Triangle, read by rows, such that the diagonal (A084785) is the self-convolution of the first column (A084784) and the row sums (A084786) gives the differences of the diagonal and the first column.

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A084784 Binomial transform = self-convolution: first column of the triangle (A084783).

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A084785 Diagonal of the triangle (A084783) and the self-convolution of the first column (A084784).

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