A090351 -id:A090351 - cp's OEIS frontend

A090352 G.f. satisfies A^3 = BINOMIAL(A)^2, where A = A090351^2.

1, 2, 7, 36, 255, 2370, 27713, 393352, 6582068, 126888632, 2767912036, 67362737168, 1808596304964, 53083358012760, 1690443996202428, 58039582729688320, 2136931230333535178, 83981145793974066484

Offset: 0

Paul D. Hanna, Nov 26 2003

See comments in A090351.

G. C. Greubel, Table of n, a(n) for n = 0..390

Cf. A004123, A019538, A050351, A084785, A090351, A201339.

Cf. A090355, A090357, A090362.

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);  // A090352
Coefficients(R!( f(m, x) )); // G. C. Greubel, Jul 07 2023

nmax = 17; 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)^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 *)

{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^3+B^2); polcoeff(A,n,x))}

m=50
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)))
def A090352_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m, x) ).list()
A090352_list(m-9) # G. C. Greubel, Jul 07 2023

G.f. satisfies: A(x)^3 = A(x/(1-x))^2/(1-x)^2.
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)*2^k = A004123(n+1) = 2*A050351(n) for n >= 1. Cf. A084785.
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)*(-3)^k = A201339(n) = 3*A050351(n) for n >= 1.
A(x) = B(x)^2 and BINOMIAL(A(x)) = B(x)^3 where B(x) = 1 + x + 3*x^2 + 15*x^3 + 108*x^4 + ... is the o.g.f. for A090351. See also A019538. (End)
G.f.: Product_{k>=1} 1/(1 - k*x)^((1/3) * (2/3)^k). - Seiichi Manyama, May 26 2025
a(n) ~ (n-1)! / (3 * log(3/2)^(n+1)). - Vaclav Kotesovec, May 28 2025

A111342 BINOMIAL transform of A090351.

Original entry on oeis.org

1, 2, 6, 28, 191, 1758, 20552, 292336, 4902711, 94682414, 2068050286, 50376290220, 1353408261849, 39741665020434, 1266001712189172, 43477659901096872, 1601086339395521724, 62931760277350226040, 2629518833911705359016, 116382349367962203872528, 5439115134893599838307636

Offset: 0

N. J. A. Sloane, Nov 08 2005

nonn

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

Cf. A090351, A111343.

G.f. A(x/(1-x))/(1-x), where A = g.f. for A090351.

a(n) ~ (n-1)! / (4 * log(3/2)^(n+1)). - Vaclav Kotesovec, Oct 30 2017

A111343 G.f. A(x/(1-x)), where A = g.f. for A090351.

Original entry on oeis.org

1, 1, 4, 22, 163, 1567, 18794, 271784, 4610375, 89779703, 1973367872, 48308239934, 1303031971629, 38388256758585, 1226260047168738, 42211658188907700, 1557608679494424852, 61330673937954704316, 2566587073634355132976, 113752830534050498513512, 5322732785525637634435108

Offset: 0

N. J. A. Sloane, Nov 08 2005

nonn

a(n >= 1) is binomial transform of A090351(n >= 1). - Michael Somos, Nov 11 2005

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

A090353 G.f. satisfies A^4 = BINOMIAL(A^3).

Original entry on oeis.org

1, 1, 4, 28, 286, 3886, 66260, 1361972, 32784353, 904412593, 28124223808, 973106096392, 37073604836768, 1541948625066176, 69513081435903392, 3376138396206853792, 175739519606046355540, 9760024269508314079444

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^4 = BINOMIAL(A090355), since A090355=A^3. Also, BINOMIAL(A) = A090354^2.

Vaclav Kotesovec, Table of n, a(n) for n = 0..320

Cf. A032033, A050352, A084784, A090351, A090354, A090355, A090356, A090358.

m:=40;
f:= func< n,x | Exp((&+[(&+[3^(j-1)*Factorial(j)* StirlingSecond(k,j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;
R:=PowerSeriesRing(Rationals(), m+1); // A090353
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]^4 - A[x/(1 - x)]^3/(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[3^(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) = my(A); if(n<0,0,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); polcoef(A,n,x))}
for(n=0,20,print1(a(n),", "))

m=50
def f(n, x): return exp(sum(sum(3^(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 A090353_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m,x) ).list()
A090353_list(m-9) # G. C. Greubel, Jun 09 2023

G.f. satisfies: A(x)^4 = A(x/(1-x))^3/(1-x).
a(n) ~ (n-1)! / (12 * (log(4/3))^(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)*3^(k-1) = A050352(n) = 1/3*A032033(n) for n >= 1. - Peter Bala, May 26 2015
G.f. satisfies [x^n] 1/A(x)^(3*n-3) = [x^n] 1/A(x)^(4*n-4) for n >= 0. - Paul D. Hanna, Apr 28 2025
G.f.: Product_{k>=1} 1/(1 - k*x)^((1/12) * (3/4)^k). - Seiichi Manyama, May 26 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

A384324 Expansion of Product_{k>=1} 1/(1 - k*x)^((2/3)^k).

Original entry on oeis.org

1, 6, 33, 200, 1428, 12408, 132604, 1730160, 27043866, 495026316, 10388326986, 245555445888, 6446710871724, 185904786328920, 5836500883321164, 198054400887909264, 7220679972923312487, 281402128806812402490, 11671796413017231008663

Offset: 0

Seiichi Manyama, May 26 2025

nonn

Cf. A084785, A384305, A384325, A384326.

Cf. A004123, A090351, A090352.

terms = 20; A[] = 1; Do[A[x] = -2*A[x] + 3*A[x/(1-x)]^(2/3) / (1-x)^2 + 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(3*sum(k=1, N, sum(j=0, k, 2^j*j!*stirling(k, j, 2))*x^k/k)))

G.f. A(x) satisfies A(x) = A(x/(1-x))^(2/3) / (1-x)^2.
G.f.: exp(3 * Sum_{k>=1} A004123(k+1) * x^k/k).
G.f.: B(x)^6, where B(x) is the g.f. of A090351.
a(n) ~ (n-1)! / log(3/2)^(n+1). - Vaclav Kotesovec, May 27 2025

A384344 Expansion of Product_{k>=1} (1 + kx)^((1/6) (2/3)^k).

Original entry on oeis.org

1, 1, -2, 10, -77, 787, -9972, 150552, -2637729, 52615903, -1177590290, 29228602546, -796945212035, 23681656958269, -761803800466856, 26376749702235900, -978091742247376932, 38674335439691203644, -1624351949069462807480, 72221688529265896447384

Offset: 0

Seiichi Manyama, May 26 2025

sign

Cf. A381890, A384343, A384345.

Cf. A050351, A090351.

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

G.f. A(x) satisfies A(x) = (1+x)^(1/3) * A(x/(1+x))^(2/3).
G.f.: exp(Sum_{k>=1} (-1)^(k-1) * A050351(k) * x^k/k).
G.f.: 1/B(-x), where B(x) is the g.f. of A090351.

A384413 Expansion of Product_{k>=1} 1/(1 - k^3 * x)^((1/6) * (2/3)^k).

Original entry on oeis.org

1, 37, 33987, 169103895, 2499834885228, 81779253109721484, 5002571587280667349252, 513188808423273125116834036, 81795428604490137664191461936826, 19140816569244304756404266108586220066, 6295058477497449841660364475294196843864030, 2810342651288539045376339873565157506716615522598

Offset: 0

Seiichi Manyama, May 28 2025

nonn

Cf. A004123, A090351.

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

G.f.: exp((1/2) * Sum_{k>=1} b(3*k) * x^k/k), where b(n) = Sum_{k=0..n} 2^k * k! * Stirling2(n,k).

cp's OEIS Frontend

A090352 G.f. satisfies A^3 = BINOMIAL(A)^2, where A = A090351^2.

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A111342 BINOMIAL transform of A090351.

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A111343 G.f. A(x/(1-x)), where A = g.f. for A090351.

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

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A090353 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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Maple

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A384324 Expansion of Product_{k>=1} 1/(1 - k*x)^((2/3)^k).

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A384344 Expansion of Product_{k>=1} (1 + k*x)^((1/6) * (2/3)^k).

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A384344 Expansion of Product_{k>=1} (1 + kx)^((1/6) (2/3)^k).