A090353 -id:A090353 - cp's OEIS frontend

A090354 Self-convolution equals the binomial transform of A090353: A^2 = BINOMIAL(A090353).

1, 1, 3, 19, 190, 2574, 43922, 903986, 21784659, 601478195, 18715354049, 647834803569, 24688869993252, 1027073272425876, 46309250293477020, 2249435671825385244, 117101538463333719891, 6503918951175618656403

Offset: 0

Paul D. Hanna, Nov 26 2003

nonn

See comments in A090353.

Vaclav Kotesovec, Table of n, a(n) for n = 0..370

Cf. A090353, A090355.

{a(n)=local(A); if(n<1,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A^3,x, x/(1-x))/(1-x)+x*O(x^n); A=A-A^4+B);B=subst(A,x, x/(1-x))/(1-x)+x*O(x^n); polcoeff(B^(1/2),n,x))}

a(n) ~ (n-1)! / (18 * log(4/3)^(n+1)). - Vaclav Kotesovec, May 28 2025

A090351 G.f. satisfies A^3 = BINOMIAL(A^2).

Original entry on oeis.org

1, 1, 3, 15, 108, 1032, 12388, 179572, 3052986, 59555338, 1310677726, 32114051862, 866766965308, 25547102523604, 816335926158372, 28107705687291892, 1037367351120788551, 40852168787823027351, 1709792654612819858341

Offset: 0

Paul D. Hanna, Nov 26 2003

In general, if A^n = BINOMIAL(A^(n-1)), then for all integer m>0 there exists an integer sequence B such that B^d = BINOMIAL(A^m) where d=gcd(m+1,n). Also, coefficients of A(k*x)^n = k-th binomial transform of coefficients in A(k*x)^(n-1) for all k>0.

			A^3 = BINOMIAL(A090352), since A090352=A^2.

Vaclav Kotesovec, Table of n, a(n) for n = 0..390

Cf. A004123, A050351, A084784, A090352, A090353, A090356, A090358, A088222.

m:=40;
f:= func< n,x | Exp((&+[(&+[2^(j-1)*Factorial(j)* StirlingSecond(k,j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;
R:=PowerSeriesRing(Rationals(), m+1);  // A090351
Coefficients(R!( f(m,x) )); // G. C. Greubel, Jun 08 2023

nmax = 18; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^3 - A[x/(1 - x)]^2/(1 - x) + 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-1)*j!* StirlingS2[k,j], {j,k}]*x^k/k, {k,m+1}]], {x,0,m}], x]] (* G. C. Greubel, Jun 08 2023 *)

{a(n) = my(A); if(n<0,0,A = 1+x +x*O(x^n); for(k=1,n, B = subst(A^2,x,x/(1-x))/(1-x) +x*O(x^n); A = A - A^3 + B); polcoef(A,n,x))}
for(n=0,25,print1(a(n),", "))

m=50
def f(n, x): return exp(sum(sum(2^(j-1)*factorial(j)* stirling_number2(k,j)*x^k/k for j in range(1,k+1)) for k in range(1,n+2)))
def A090351_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m,x) ).list()
A090351_list(m-9) # G. C. Greubel, Jun 08 2023

G.f. satisfies: A(x)^3 = A(x/(1-x))^2 / (1-x).
a(n) ~ (n-1)! / (6 * (log(3/2))^(n+1)). - Vaclav Kotesovec, Nov 18 2014
O.g.f. A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = Sum_{k = 1..n} k!*Stirling2(n,k)*2^(k-1) = A050351(n) = 1/2*A004123(n+1) for n >= 1. - Peter Bala, May 26 2015
G.f. satisfies [x^n] 1/A(x)^(2*n-2) = [x^n] 1/A(x)^(3*n-3) = -(n-1)*A088222(n) for n >= 0. - Paul D. Hanna, Apr 28 2025
G.f.: Product_{k>=1} 1/(1 - k*x)^((1/6) * (2/3)^k). - Seiichi Manyama, May 26 2025

A090358 G.f. satisfies A^6 = BINOMIAL(A^5).

Original entry on oeis.org

1, 1, 6, 66, 1071, 23151, 627236, 20452976, 779947641, 34050858041, 1674497370602, 91575747294582, 5512402585832847, 362148111801511407, 25783279860096503952, 1977349647140061768364, 162508269041154881377519

Offset: 0

Paul D. Hanna, Nov 26 2003

In general, if A^n = BINOMIAL(A^(n-1)), then for all integer m>0 there exists an integer sequence B such that B^d = BINOMIAL(A^m) where d=gcd(m+1,n). Also, coefficients of A(k*x)^n = k-th binomial transform of coefficients in A(k*x)^(n-1) for all k>0.

In general, if g.f. satisfies A(x)^(k+1) = A(x/(1-x))^k / (1-x), k>0, then a(n) ~ (n-1)! / (k*(k+1) * (log((k+1)/k))^(n+1)). - Vaclav Kotesovec, Nov 19 2014

			A^6 = BINOMIAL(A090362), since A090362=A^5. Also,
BINOMIAL(A) = A090359^2 since 2=gcd(1+1,6),
BINOMIAL(A^2) = A090360^3 since 3=gcd(2+1,6) and
BINOMIAL(A^3) = A090361^2 since 2=gcd(3+1,6).

Vaclav Kotesovec, Table of n, a(n) for n = 0..280

Cf. A084784, A090351, A090353, A090356, A090359, A090360, A090361, A090362, A094418.

m:=40;
f:= func< n,x | Exp((&+[(&+[5^(j-1)*Factorial(j)*StirlingSecond(k,j) *x^k/k: j in [1..k]]): k in [1..n+2]])) >;
R:=PowerSeriesRing(Rationals(), m+1); // A090358
Coefficients(R!( f(m,x) )); // G. C. Greubel, Jun 08 2023

nmax = 16; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^6 - A[x/(1 - x)]^5/(1 - x) + 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[5^(j-1)*j!* StirlingS2[k,j], {j,k}]*x^k/k, {k,m+1}]], {x,0,m}], x]] (* G. C. Greubel, Jun 08 2023 *)

{a(n)=local(A); if(n<1,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A^5,x,x/(1-x))/(1-x)+x*O(x^n); A=A-A^6+B);polcoeff(A,n,x))}

m=50
def f(n, x): return exp(sum(sum( 5^(j-1)*factorial(j)* stirling_number2(k,j)*x^k/k for j in range(1,k+1)) for k in range(1,n+2)))
def A090358_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m,x) ).list()
A090358_list(m-5) # G. C. Greubel, Jun 08 2023

G.f. satisfies: A(x)^6 = A(x/(1-x))^5/(1-x).
a(n) ~ (n-1)! / (30 * (log(6/5))^(n+1)). - Vaclav Kotesovec, Nov 19 2014
O.g.f.: A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = Sum_{k = 1..n} k!*Stirling2(n,k)*5^(k-1) = 1/5*A094418(n) for n >= 1. - Peter Bala, May 26 2015
G.f.: Product_{k>=1} 1/(1 - k*x)^((1/30) * (5/6)^k). - Seiichi Manyama, May 26 2025

A090356 G.f. A(x) satisfies A(x)^5 = BINOMIAL(A(x)^4); that is, the binomial transform of the coefficients in A(x)^4 yields the coefficients in A(x)^5.

Original entry on oeis.org

1, 1, 5, 45, 595, 10475, 231255, 6148495, 191276600, 6815243040, 273601200136, 12217471594856, 600580173151560, 32224787998758280, 1873909224391774760, 117388347849375956328, 7880739469498103077588, 564440024187816634143380

Offset: 0

Paul D. Hanna, Nov 26 2003

In general, if A^n = BINOMIAL(A^(n-1)), then for all integer m>0 there exists an integer sequence B such that B^d = BINOMIAL(A^m) where d=gcd(m+1,n). Also, coefficients of A(k*x)^n = k-th binomial transform of coefficients in A(k*x)^(n-1) for all k>0.

			G.f.: A(x) = 1 + x + 5*x^2 + 45*x^3 + 595*x^4 + 10475*x^5 + 231255*x^6 + ...
The coefficients in A(x)^4 are given by A090357 and begin
A(x)^4: [1, 4, 26, 244, 3131, 52600, 1111940, ..., A090357(n), ...].
The binomial transform of A090357 yields the coefficients of A(x)^5:
A(x)^5: [1, 5, 35, 335, 4280, 70976, 1479800, ...]
as shown by
1 = 1*1,
5 = 1*1 + 1*4,
35 = 1*1 + 2*4 + 1*26,
335 = 1*1 + 3*4 + 3*26 + 1*244,
4280 = 1*1 + 4*4 + 6*26 + 4*244 + 1*3131, ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..310

Cf. A050353, A084784, A090353, A090357, A090358, A094417.

m:=40;
f:= func< n,x | Exp((&+[(&+[4^(j-1)*Factorial(j)* StirlingSecond(k,j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;
R:=PowerSeriesRing(Rationals(), m+1); // A090356
Coefficients(R!( f(m,x) )); // G. C. Greubel, Jun 09 2023

nmax = 17; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^5 - A[x/(1 - x)]^4/(1 - x) + 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[4^(j-1)*j!* StirlingS2[k,j], {j,k}]*x^k/k, {k,m+1}]], {x,0,m}], x]] (* G. C. Greubel, Jun 09 2023 *)

{a(n)=local(A); if(n<1,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A^4,x,x/(1-x))/(1-x)+x*O(x^n); A=A-A^5+B);polcoeff(A,n,x))}

m=40
def f(n, x): return exp(sum(sum(4^(j-1)*factorial(j)* stirling_number2(k,j)*x^k/k for j in range(1,k+1)) for k in range(1,n+2)))
def A090356_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m,x) ).list()
A090356_list(m) # G. C. Greubel, Jun 09 2023

G.f. satisfies: A(x)^5 = A(x/(1-x))^4/(1-x).
a(n) ~ (n-1)! / (20 * (log(5/4))^(n+1)). - Vaclav Kotesovec, Nov 19 2014
O.g.f.: A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = Sum_{k = 1..n} k!*Stirling2(n,k)*4^(k-1) = A050353(n) = 1/4*A094417(n) for n >= 1. - Peter Bala, May 26 2015
G.f.: Product_{k>=1} 1/(1 - k*x)^((1/20) * (4/5)^k). - Seiichi Manyama, May 26 2025

A090355 G.f. satisfies A^4 = BINOMIAL(A)^3.

Original entry on oeis.org

1, 3, 15, 109, 1086, 14178, 232906, 4647006, 109376595, 2967406345, 91130074437, 3123199831983, 118106517900868, 4883161763750820, 219076867059030300, 10597531747143624820, 549768536732090716371, 30443800514118532762329

Offset: 0

Paul D. Hanna, Nov 26 2003

See comments in A090353.

Cf. A090353, A090354; A019538, A032033, A050352, A201354.

Cf. A090352, A090357, A090362.

nmax = 17; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^4 - A[x/(1 - x)]^3/(1 - x)^3 + 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 *)

{a(n)=local(A); if(n<1,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A,x,x/(1-x))/(1-x)+x*O(x^n); A=A-A^4+B^3);polcoeff(A,n,x))}

G.f.: A(x)^4 = A(x/(1-x))^3/(1-x)^3.
From Peter Bala, May 26 2015: (Start)
O.g.f.: A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = Sum_{k = 1..n} k!*Stirling2(n,k)*3^k = A032033(n) = 3*A050352(n).
BINOMIAL(A(x)) = exp( Sum_{n >= 1} c(n)*x^n/n ) where c(n) = (-1)^n*Sum_{k = 1..n} k!*Stirling2(n,k)*4^k = A201354(n) = 4*A050352(n) for n >= 1. A(x) = B(x)^3 and BINOMIAL(A(x)) = B(x)^4 where B(x) = 1 + x + 4*x^2 + 28*x^3 + 286*x^4 + ... is the o.g.f. for A090353. See also A019538. (End)
G.f.: Product_{k>=1} 1/(1 - k*x)^((1/4) * (3/4)^k). - Seiichi Manyama, May 26 2025
a(n) ~ (n-1)! / (4 * log(4/3)^(n+1)). - Vaclav Kotesovec, May 28 2025

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

A381890 Expansion of Product_{k>=1} (1 + kx)^((1/12) (3/4)^k).

Original entry on oeis.org

1, 1, -3, 21, -225, 3207, -56821, 1202099, -29558466, 828401462, -26068940938, 910286433318, -34930741605414, 1461245816594058, -66187658069563710, 3227353484661602866, -168557942284281821933, 9388117645333487820387, -555463036269652132509113

Offset: 0

Seiichi Manyama, May 26 2025

sign

Cf. A384343, A384344, A384345.

Cf. A050352, A090353.

my(N=20, x='x+O('x^N)); Vec(exp(sum(k=1, N, (-1)^(k-1)*sum(j=0, k, 3^(j-1)*j!*stirling(k, j, 2))*x^k/k)))

G.f. A(x) satisfies A(x) = (1+x)^(1/4) * A(x/(1+x))^(3/4).
G.f.: exp(Sum_{k>=1} (-1)^(k-1) * A050352(k) * x^k/k).
G.f.: 1/B(-x), where B(x) is the g.f. of A090353.

A384325 Expansion of Product_{k>=1} 1/(1 - k*x)^((3/4)^k).

Original entry on oeis.org

1, 12, 114, 1084, 11319, 136920, 1981228, 34705656, 731268315, 18203860748, 524073230394, 17111173850652, 623571696107069, 25046605210733184, 1097919954149781264, 52109508350206511840, 2660615337817983390318, 145353541761618312219336

Offset: 0

Seiichi Manyama, May 26 2025

nonn

Cf. A084785, A384305, A384324, A384326.

Cf. A032033, A090353.

terms = 20; A[] = 1; Do[A[x] = -3*A[x] + 4*A[x/(1-x)]^(3/4) / (1-x)^3 + O[x]^j // Normal, {j, 1, terms}]; CoefficientList[A[x], x] (* Vaclav Kotesovec, May 27 2025 *)

my(N=20, x='x+O('x^N)); Vec(exp(4*sum(k=1, N, sum(j=0, k, 3^j*j!*stirling(k, j, 2))*x^k/k)))

G.f. A(x) satisfies A(x) = A(x/(1-x))^(3/4) / (1-x)^3.
G.f.: exp(4 * Sum_{k>=1} A032033(k) * x^k/k).
G.f.: B(x)^12, where B(x) is the g.f. of A090353.
a(n) ~ (n-1)! / log(4/3)^(n+1). - Vaclav Kotesovec, May 27 2025

cp's OEIS Frontend

A090354 Self-convolution equals the binomial transform of A090353: A^2 = BINOMIAL(A090353).

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A090351 G.f. satisfies A^3 = BINOMIAL(A^2).

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A090358 G.f. satisfies A^6 = BINOMIAL(A^5).

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A090356 G.f. A(x) satisfies A(x)^5 = BINOMIAL(A(x)^4); that is, the binomial transform of the coefficients in A(x)^4 yields the coefficients in A(x)^5.

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A090355 G.f. satisfies A^4 = BINOMIAL(A)^3.

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

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A381890 Expansion of Product_{k>=1} (1 + k*x)^((1/12) * (3/4)^k).

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A381890 Expansion of Product_{k>=1} (1 + kx)^((1/12) (3/4)^k).