A084783 -id:A084783 - cp's OEIS frontend

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

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

A084786 Row sums of the triangle (A084783) and the differences of the main diagonal (A084785) and the first column (A084784).

Original entry on oeis.org

1, 3, 10, 41, 211, 1354, 10620, 99327, 1081744, 13443065, 187538132, 2899087774, 49149083790, 906169148064, 18044322039456, 385825735367745, 8814867042465387, 214270073007359704, 5520898403200292418, 150290771692227728963, 4309813692713979537503

Offset: 0

Paul D. Hanna, Jun 13 2003

nonn

In the triangle (A084783), the diagonal (A084785) is the self-convolution of the first column (A084784) and the row sums (this sequence) gives the differences of the diagonal and the first column.

Vaclav Kotesovec, Table of n, a(n) for n = 0..400

Cf. A084783, A084784, A084785.

A084784= With[{m=60}, CoefficientList[Series[Exp[Sum[Sum[ j!*StirlingS2[k, j], {j, k}]*x^k /k , {k, m + 1}]], {x,0,m}], x]];
T[n_, k_]:= T[n, k]= If[k==0, A084784[[n+1]], T[n, k-1] + T[n-1, k-1]]; (* A084783 *)
A084786[n_]:= A084786[n]= Sum[T[n, k], {k,0,n}];
Table[A084786[n], {n,0,40}] (* 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(rs)); \\ David Wasserman, Jan 06 2005

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)
@CachedFunction
def T(n,k): # T = A084783
    if k==0: return b[n]
    else: return T(n, k-1) + T(n-1, k-1)
def A084786(n): return sum(T(n, k) for k in range(n+1))
[A084786(n) for n in range(m-9)] # G. C. Greubel, Jun 08 2023

a(n) ~ n! / (2 * (log(2))^(n+2)). - Vaclav Kotesovec, Nov 19 2014

More terms from David Wasserman, Jan 06 2005

A258377 O.g.f. satisfies A^2(z) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(z))) ).

Original entry on oeis.org

1, 3, 13, 79, 649, 6955, 93813, 1539991, 29884881, 669628819, 17005862301, 482399018527, 15108642099673, 517599894435643, 19247498583665029, 771922934908235751, 33206411983713679009, 1525025984109289947171, 74466779211331635306029, 3852255519421356879419631

Offset: 0

Peter Bala, May 28 2015

For a fixed integer N, Hanna has considered the problem of finding an o.g.f. of the form A(z) = 1 + N*z + a(2)*z^2 + a(3)*z^3 + ..., with integer coefficients a(2), a(3), ... dependent on the parameter N, which is a solution to the functional equation A^(N+1) = ( BINOMIAL(A) )^N. Here BINOMIAL(F(z))= 1/(1 - z)*F(z/(1 - z)) denotes the binomial transform of the o.g.f. F(z).
The function A(z) is related to the triangle of ordered Stirling numbers A019538 via logarithmic differentiation. It can be shown that z*A'(z)/A(z) = Sum_{k >= 1} R(k,N)*z^k, where R(k,x) denotes the k-th row polynomial of A019538; equivalently, A(z) = exp( Sum_{k >= 1} R(k,N)*z^k/k ).
Cases include A084784 (N = 1), A090352 (N = 2), A090355 (N = 3), A090357 (N = 4), A090362 (N = 5) and a signed version of A084785 (N = -2).
It turns out that the o.g.f. B(z) := A(z)^(1/N) also has integer coefficients. It satisfies the functional equation B^(N+1) = BINOMIAL(B^N). For cases see A084784 (N = 1), A090351 (N = 2), A090353 (N = 3), A090356 (N = 4), A090358 (N = 5) and A084784 (N = -2).
There are similar results to the above associated with triangle A145901, which can be viewed as a type B analog of A019538.
For a fixed integer N, consider the problem of finding an o.g.f. with integer coefficients (depending on the parameter N) of the form A(z) = 1 + N*z + a(2)*z^2 + a(3)*z^3 + ..., which is a solution to the functional equation A^(N+1)(z) = 1/(1 - z) * ( BINOMIAL(BINOMIAL(A(z))) )^N; equivalently A^(N+1)(z) = 1/(1 - z)* 1/(1 - 2*z)^N*A^N(z/(1 - 2*z)). This is the type B analog of Hanna's type A functional equation above.
It can be shown that z*A'(z)/A(z) = Sum_{k >= 1} P(k,N)*z^k, where P(k,x) denotes the k-th row polynomial of A145901. However, unlike the type A situation, the type B function A(z)^(1/N) does not have all integer coefficients.
The present sequence is the case N = 1. For further examples of solutions to the type B functional equation see A258378 (N = 2), A258379 (N = 3), A258380 (N = 4) and A258381 (N = 5).
From Peter Bala, Dec 06 2017: (Start)
a(n) appears to be always odd. Calculation suggests that for k = 1,2,3,..., the sequence a(n) (mod 2^k) is purely periodic with period 2^(k-1). For example, a(n) (mod 4) = (1,3,1,3,...) seems to be purely periodic with period 2 and a(n) (mod 8) = (1,3,5,7,1,3,5,7,...) seems to be purely periodic with period 4 (both checked up to n = 1000).
(End)

Alois P. Heinz, Table of n, a(n) for n = 0..383
N. J. A. Sloane, Transforms.

Cf. A019538, A080253, A084784, A084785, A090351, A090352, A090353, A090355, A090356, A090357, A090358, A090362, A145901, A258378 (N = 2), A258379 (N = 3), A258380 (N = 4), A258381 (N = 5).

Cf. A084783.

#A258377
with(combinat):
#recursively define row polynomials R(n,x) of A145901
R := proc (n, x) option remember; if n = 0 then 1 else 1 + x*add(binomial(n, i)*2^(n-i)*R(i,x), i = 0..n-1) end if; end proc:
#define a family of sequences depending on an integer parameter k
a := proc (n, k) option remember; if n = 0 then 1 else 1/n*add(R(i+1,k)*a(n-1-i,k), i = 0..n-1) end if; end proc:
# display the case k = 1
seq(a(n,1), n = 0..19);

R[n_, x_] := R[n, x] = If[n==0, 1, 1+x*Sum[Binomial[n, i]*2^(n-i)*R[i, x], {i, 0, n-1}]];
a[n_, k_] := a[n, k] = If[n==0, 1, 1/n*Sum[R[i+1, k]*a[n-1-i, k], {i, 0, n-1}]];
a[n_] := a[n, 1];
a /@ Range[0, 19] (* Jean-François Alcover, Oct 02 2019 *)

a(0) = 1 and for n >= 1, a(n) = 1/n*Sum_{i = 0..n-1} R(i+1,1)*a(n-1-i), where R(n,x) denotes the n-th row polynomial of A145901.
O.g.f.: A(z) = 1 + 3*z + 13*z^2 + 79*z^3 + 649*z^4 + ... satisfies A^2(z) = 1/(1 - z)*1/(1 - 2*z)*A(z/(1 - 2*z)).
O.g.f.: A(z) = exp( Sum_{k >= 1} R(k,1)*z^k/k ).
1 + z*A'(z)/A(z) = 1 + 3*z + 17*z^2 + 147*z^3 + 1697*z^4 + ... is the o.g.f. for A080253.
a(n) = Sum_{j=0..n} binomial(n,j) * A084783(n,n-j). - Alois P. Heinz, Jun 09 2023
a(n) ~ (n-1)! * 2^(n - 1/2) / log(2)^(n+1). - Vaclav Kotesovec, May 28 2025

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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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A084786 Row sums of the triangle (A084783) and the differences of the main diagonal (A084785) and the first column (A084784).

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A258377 O.g.f. satisfies A^2(z) = 1/(1 - z)*( BINOMIAL(BINOMIAL(A(z))) ).

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