A232690 -id:A232690 - cp's OEIS frontend

A333044 Exponential self-convolution of A232690.

1, 2, 6, 26, 146, 1014, 8374, 80282, 875874, 10719734, 145422182, 2166117018, 35143677106, 616926078326, 11650276119702, 235510563023642, 5074238094097538, 116081921186069622, 2810093148179120710, 71767508789469271322, 1928439105298059705042

Offset: 0

Vaclav Kotesovec, Mar 06 2020

nonn

			E.g.f.: A(x) = 1 + 2*x + 6*x^2/2! + 26*x^3/3! + 146*x^4/4! + 1014*x^5/5! + ...

Cf. A232690.

CoefficientList[Simplify[Assuming[Element[x, Reals], Series[LambertW[-1, (4*x - 3)*E^(-3)]/(4*x - 3), {x, 0, 20}]]], x] * Range[0, 20]!

{a(n)=my(A = 1+2*x); for(i=1, n, A = exp(2/sqrt(A)*intformal(A^(3/2) + x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))

E.g.f. satisfies: A(x) = exp(2/sqrt(A(x)) * Integral A(x)^(3/2) dx).
E.g.f. LambertW(-1, (4*x-3)*exp(-3))/(4*x-3).
a(n) = Sum_{k=0..n} binomial(n,k) * A232690(k) * A232690(n-k).

A232687 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals Sum_{k=0..n} C(n,k)^3 = A000172(n) (Franel numbers), for n>=0.

Original entry on oeis.org

1, 1, 3, 7, 20, 66, 244, 980, 4182, 18674, 86353, 410541, 1996214, 9888844, 49760925, 253767097, 1309154825, 6822023553, 35865392690, 190038440422, 1014015337209, 5444707218851, 29401289997403, 159584901816255, 870267544114291, 4766246752344215, 26206635040151511

Offset: 0

Paul D. Hanna, Dec 05 2013

nonn

Compare to: Sum_{k=0..n} [x^k] 1/(1-x)^n = Sum_{k=0..n} C(n,k)^2 = (2*n)!/n!^2.

a(n+1)/a(n) tends to 6.0295... - Vaclav Kotesovec, Jan 22 2014

			G.f.: A(x) = 1 + x + 3*x^2 + 7*x^3 + 20*x^4 + 66*x^5 + 244*x^6 + 980*x^7 +...
ILLUSTRATION OF INITIAL TERMS.
If we form an array of coefficients of x^k in A(x)^n, n>=0, like so:
A^0: [1],0,  0,   0,    0,    0,     0,      0,      0, ...;
A^1: [1, 1], 3,   7,   20,   66,   244,    980,   4182, ...;
A^2: [1, 2,  7], 20,   63,  214,   789,   3124,  13112, ...;
A^3: [1, 3, 12,  40], 138,  492,  1848,   7326,  30531, ...;
A^4: [1, 4, 18,  68,  255], 960,  3716,  14920,  62295, ...;
A^5: [1, 5, 25, 105,  425, 1691], 6785,  27805, 117165, ...;
A^6: [1, 6, 33, 152,  660, 2772, 11560], 48588, 207774, ...;
A^7: [1, 7, 42, 210,  973, 4305, 18676,  80746],351792, ...;
A^8: [1, 8, 52, 280, 1378, 6408, 28916, 128808, 573311], ...; ...
then the sum of the coefficients of x^k, k=0..n, in A(x)^n (shown above in brackets) equals Sum_{k=0..n} C(n,k)^3 = A000172(n):
A000172(0) = 1 = 1;
A000172(1) = 1 + 1 = 2;
A000172(2) = 1 + 2 +  7 = 10;
A000172(3) = 1 + 3 + 12 +  40 = 56;
A000172(4) = 1 + 4 + 18 +  68 + 255 = 346;
A000172(5) = 1 + 5 + 25 + 105 + 425 + 1691 = 2252;
A000172(6) = 1 + 6 + 33 + 152 + 660 + 2772 + 11560 = 15184; ...

Paul D. Hanna and Vaclav Kotesovec, Table of n, a(n) for n = 0..460 (first 200 terms from Paul D. Hanna)

Cf. A232606, A232683, A232690, A000172.

Franel[n_] := Sum[Binomial[n, k]^3, {k, 0, n}];
a[0] = 1; a[n_] := Module[{B, G}, B = Sum[Franel[k]*x^k, {k, 0, n+1}] + x^3*O[x]^n; G = 1+x*O[x]^n; For[i=1, i <= n, i++, G = 1+Integrate[(B-1)* (G/x)-B*G^2, x]]; SeriesCoefficient[x/InverseSeries[x*G, x], {x, 0, n}]];
Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Jan 15 2018, translated from 2nd PARI program *)

/* By Definition (slow): */
{Franel(n)=sum(k=0,n,binomial(n,k)^3)}
{a(n)=if(n==0, 1, (Franel(n) - sum(k=0, n, polcoeff(sum(j=0, min(k, n-1), a(j)*x^j)^n + x*O(x^k), k)))/n)}
for(n=0, 20, print1(a(n), ", "))

/* Faster, using series reversion: */
{Franel(n)=sum(k=0,n,binomial(n,k)^3)}
{a(n)=local(B=sum(k=0, n+1, Franel(k)*x^k)+x^3*O(x^n), G=1+x*O(x^n));
for(i=1, n, G = 1 + intformal( (B-1)*G/x - B*G^2)); polcoeff(x/serreverse(x*G), n)}
for(n=0, 30, print1(a(n), ", "))

Given g.f. A(x), Sum_{k=0..n} [x^k] A(x)^n = Sum_{k=0..n} C(n,k)^3 = A000172(n).

Given g.f. A(x), let G(x) = A(x*G(x)) then (G(x) + x*G'(x)) / (G(x) - x*G(x)^2) = Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)/(1-2*x)^(3*n+1) = Sum_{n>=0} A000172(n)*x^n.

A232692 E.g.f. satisfies: A(x) = exp( 1/A(x)^3 * Integral A(x)^8 dx ).

Original entry on oeis.org

1, 1, 3, 24, 213, 3096, 46071, 967608, 20251809, 555747048, 15004870731, 508165972056, 16810393586733, 677183788645704, 26523956467895103, 1238567261126084856, 56056407696184372281, 2976966230117448265128, 152872356339113679491859, 9098430770913969095416728

Offset: 0

Paul D. Hanna, Dec 06 2013

nonn

Compare e.g.f. to: B(x) = exp( 1/B(x)^3 * Integral B(x)^3 dx ) where B(y) = Bessel polynomial y_n(-3) (cf. A065923).
Note that G(x) = exp(1/G(x)^3 * Integral G(x)^7 dx) has negative coefficients.
CONJECTURE:
Given G(x,n,k) = G such that G = exp( 1/G^n * Integral G^k dx ) then G(x,n,k) consists solely of positive coefficients when k >= A047399(n) where A047399 lists numbers that are congruent to {0,3,6} mod 8.

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 24*x^3/3! + 213*x^4/4! + 3096*x^5/5! +...
Related expansions:
log(A(x)) = x + 2*x^2/2! + 17*x^3/3! + 120*x^4/4! + 1905*x^5/5! + 23640*x^6/6! +...
Integral A(x)^8 dx = x + 8*x^2/2! + 80*x^3/3! + 1032*x^4/4! + 16320*x^5/5! +...
1/A(x)^3 = 1 - 3*x + 3*x^2/2! - 24*x^3/3! + 117*x^4/4! - 2088*x^5/5! +...

Paul D. Hanna, Table of n, a(n) for n = 0..300

Cf. A232690, A232691.

seq(n! * coeff(series((3*LambertW(-1, (25*x-8)/3*exp(-8/3))/(25*x-8))^(1/5), x, n+1), x, n), n=0..20) # Vaclav Kotesovec, Jan 05 2014

m = 20; A[] = 1; Do[A[x] = Exp[1/A[x]^3 Integrate[A[x]^8 + O[x]^m, x]] + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] Range[0, m-1]! (* Jean-François Alcover, Nov 03 2019 *)

{a(n)=local(A=1+x);for(i=1,n,A=exp(1/A^3*intformal(A^8+x*O(x^n))));n!*polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

E.g.f.: (3*LambertW(-1, (25*x-8)/3*exp(-8/3))/(25*x-8))^(1/5). - Vaclav Kotesovec, Jan 05 2014

A232691 E.g.f. satisfies: A(x) = exp( 1/A(x)^2 * Integral A(x)^6 dx ).

Original entry on oeis.org

1, 1, 3, 19, 161, 1857, 25843, 433891, 8378913, 185022049, 4565674115, 125075024211, 3755498096257, 122872235056993, 4345683577199283, 165338206044981091, 6730088764152273857, 291935651271257092161, 13440846879808207921027, 654704450541594973156627

Offset: 0

Paul D. Hanna, Dec 06 2013

nonn

Note that G(x) = exp(1/G(x)^2 * Integral G(x)^5 dx) has negative coefficients.

Compare e.g.f. to: B(x) = exp( 1/B(x)^2 * Integral B(x)^2 dx ) where B(y) = Bessel polynomial y_n(-2) (cf. A002119).

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 19*x^3/3! + 161*x^4/4! + 1857*x^5/5! +...
Related expansions:
log(A(x)) = x + 2*x^2/2! + 12*x^3/3! + 88*x^4/4! + 976*x^5/5! + 12576*x^6/6! +...
Integral A(x)^6 dx = x + 6*x^2/2! + 48*x^3/3! + 504*x^4/4! + 6576*x^5/5! +...
1/A(x)^2 = 1 - 2*x - 8*x^3/3! - 16*x^4/4! - 384*x^5/5! - 2624*x^6/6! -...

Cf. A232690, A232692.

seq(n! * coeff(series((LambertW(-1, (8*x-3)*exp(-3))/(8*x-3))^(1/4), x, n+1), x, n), n=0..20) # Vaclav Kotesovec, Jan 05 2014

{a(n)=local(A=1+x);for(i=1,n,A=exp(1/A^2*intformal(A^6+x*O(x^n))));n!*polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

E.g.f.: (LambertW(-1, (8*x-3)*exp(-3))/(8*x-3))^(1/4). - Vaclav Kotesovec, Jan 05 2014

cp's OEIS Frontend

A333044 Exponential self-convolution of A232690.

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A232687 G.f. A(x) satisfies: the sum of the coefficients of x^k, k=0..n, in A(x)^n equals Sum_{k=0..n} C(n,k)^3 = A000172(n) (Franel numbers), for n>=0.

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A232692 E.g.f. satisfies: A(x) = exp( 1/A(x)^3 * Integral A(x)^8 dx ).

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A232691 E.g.f. satisfies: A(x) = exp( 1/A(x)^2 * Integral A(x)^6 dx ).

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Examples

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Programs

Maple

PARI

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