A258969 -id:A258969 - cp's OEIS frontend

A258880 E.g.f. satisfies: A(x) = Integral 1 + A(x)^3 dx.

1, 6, 540, 184680, 157600080, 270419925600, 816984611467200, 3971317527112003200, 29097143353353192480000, 305823675529741700675520000, 4435486895868663971869188480000, 86036822683997062842122964537600000, 2175352015640142857526698650779456000000

Offset: 0

Paul D. Hanna, Jun 13 2015

nonn

Note: Sum_{n>=0} (-1)^n*x^(3*n+1)/(3*n+1) = log( (1+x)/(1-x^3)^(1/3) )/2 + Pi*sqrt(3)/18 - atan( (1-2*x)*sqrt(3)/3 )*sqrt(3)/3.

			E.g.f.: A(x) = x + 6*x^4/4! + 540*x^7/7! + 184680*x^10/10! + 157600080*x^13/13! + 270419925600*x^16/16! +...
where Series_Reversion(A(x)) =  x - x^4/4 + x^7/7 - x^10/10 + x^13/13 - x^16/16 +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..150
Guo-Niu Han, Jing-Yi Liu, Divisibility properties of the tangent numbers and its generalizations, arXiv:1707.08882 [math.CO], 2017. See Table for k = 3 p. 8.

Cf. A000182, A000831, A258878, A258901, A258925, A258927, A259112, A259113, A258969.

terms = 13;
A[_] = 0;
Do[A[x_] = Integrate[1 + A[x]^3, x] + O[x]^k // Normal, {k, 1, 3 terms}];
DeleteCases[CoefficientList[A[x], x] Range[0, 3 terms - 2]!, 0] (* Jean-François Alcover, Jul 25 2018 *)

{a(n) = local(A=x); A = serreverse( sum(m=0,n, (-1)^m * x^(3*m+1)/(3*m+1) ) +O(x^(3*n+2)) ); (3*n+1)!*polcoeff(A,3*n+1)}
for(n=0,20,print1(a(n),", "))

/* E.g.f. A(x) = Integral 1 + A(x)^3 dx.: */
{a(n) = local(A=x); for(i=1,n+1, A = intformal( 1 + A^3 + O(x^(3*n+2)) )); (3*n+1)!*polcoeff(A,3*n+1)}
for(n=0,20,print1(a(n),", "))

E.g.f.: Series_Reversion( Integral 1/(1+x^3) dx ).
E.g.f.: Series_Reversion( Sum_{n>=0} (-1)^n * x^(3*n+1)/(3*n+1) ).
a(n) ~ 3^(15*n/2 + 17/4) * n^(3*n+1) / (exp(3*n) * (2*Pi)^(3*n+3/2)). - Vaclav Kotesovec, Jun 15 2015

A193534 Decimal expansion of (1/3) * (Pi/sqrt(3) - log(2)).

Original entry on oeis.org

3, 7, 3, 5, 5, 0, 7, 2, 7, 8, 9, 1, 4, 2, 4, 1, 8, 0, 3, 9, 2, 2, 8, 2, 0, 4, 5, 3, 9, 4, 6, 5, 9, 7, 2, 1, 4, 0, 2, 8, 5, 5, 3, 7, 1, 2, 4, 4, 1, 6, 1, 7, 7, 3, 8, 1, 6, 4, 0, 1, 6, 4, 1, 9, 6, 4, 9, 0, 9, 8, 5, 3, 0, 5, 2, 2, 1, 9, 7, 2, 2, 6, 9, 2, 7, 5, 3, 8, 8, 7, 0, 7, 1, 8, 8, 0, 4

Offset: 0

Alonso del Arte, Jul 29 2011

The formulas for this number and the constant in A113476 are exactly the same except for one small, crucial detail: the infinite sum has a denominator of 3i + 2 rather than 3i + 1, while in the closed form, log(2)/3 is subtracted from rather than added to (Pi * sqrt(3))/9.

Understandably, the typesetter for Spiegel et al. (2009) set the closed formula for this number incorrectly (as being the same as for A113476, compare equation 21.16 on the same page of that book).

			0.373550727891424180392282045394659721402855371244161773816401641964909853052219...

L. B. W. Jolley, Summation of Series, Dover, 1961, eq. (80), page 16.
J. Rivaud, Analyse, Séries, équations différentielles, Mathématiques supérieures et spéciales, Premier cycle universitaire, Vuibert, 1981, Exercice 3, p. 132.
Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill, 2009, p. 135, equation 21.18.

Gheorghe Coserea, Table of n, a(n) for n = 0..2015
Eric W. Weisstein, Euler's Series Transformation.

Cf. A073010, A193535, A024396, A113476, A196548, A258969.

evalf((Psi(5/6)-Psi(1/3))/6, 120); # Vaclav Kotesovec, Jun 16 2015

RealDigits[(Pi Sqrt[3])/9 - (Log[2]/3), 10, 100][[1]]

(Pi/sqrt(3)-log(2))/3 \\ Charles R Greathouse IV, Jul 29 2011

default(realprecision, 98);
eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(3*n+2)))), "3..-2")) \\ Gheorghe Coserea, Oct 06 2015

Equals Sum_{k >= 0} (-1)^k/(3k + 2) = 1/2 - 1/5 + 1/8 - 1/11 + 1/14 - 1/17 + ... (see A016789).
From Peter Bala, Feb 20 2015: (Start)
Equals (1/2) * Integral_{x = 0..1} 1/(1 + x^(3/2)) dx.
Generalized continued fraction: 1/(2 + 2^2/(3 + 5^2/(3 + 8^2/(3 + 11^2/(3 + ... ))))) due to Euler. For a sketch proof see A024396. (End)
Equals (Psi(5/6)-Psi(1/3))/6. - Vaclav Kotesovec, Jun 16 2015
Equals Integral_{x = 1..infinity} 1/(1 + x^3) dx. - Robert FERREOL, Dec 23 2016
Equals (1/2)*Sum_{n >= 0} n!*(3/2)^n/(Product_{k = 0..n} 3*k + 2) = (1/2)*Sum_{n >= 0} n!*(3/2)^n/A008544(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(3*k + 2)). - Peter Bala, Dec 01 2021
From Bernard Schott, Jan 28 2022: (Start)
Equals Integral_{x = 0..1} x/(1+ x^3) dx (see Rivaud reference).
Equals 3 * A196548. (End)
From Peter Bala, Mar 03 2024: (Start)
Equals (1/2)*hypergeom([2/3, 1], [5/3], -1).
Gauss's continued fraction: 1/(2 + 2^2/(5 + 3^2/(8 + 5^2/(11 + 6^2/(14 + 8^2/(17 + 9^2/(20 + 11^2/(23 + 12^2/(26 + ... ))))))))). (End)

A258994 E.g.f.: A'(x) = 1 + A(x)^6, with A(0)=1.

Original entry on oeis.org

1, 2, 12, 192, 4272, 124992, 4531392, 195869952, 9832326912, 562125837312, 36056880110592, 2564230500421632, 200237330428342272, 17032391106795159552, 1567547894591436275712, 155196096043697480466432, 16447362605632117421309952, 1857733260790463501532659712

Offset: 0

Vaclav Kotesovec, Jun 16 2015

nonn

In general, for k>1, if e.g.f. satisfies A'(x) = 1 + A(x)^k, with A(0)=1, then a(n) ~ n! * d^(n + 1/(k-1)) / ((k-1)^(1/(k-1)) * Gamma(1/(k-1)) * n^(1-1/(k-1))), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(k*j-1).

			A(x) = 1 + 2*x + 12*x^2/2! + 192*x^3/3! + 4272*x^4/4! + 124992*x^5/5! + ...
A'(x) = 2 + 12*x + 96*x^2 + 712*x^3 + 5208*x^4 + 188808*x^5/5 + ...
1 + A(x)^6 = 2 + 12*x + 96*x^2 + 712*x^3 + 5208*x^4 + 188808*x^5/5 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..50

Cf. A000831 (k=2), A258969 (k=3), A258970 (k=4), A258971 (k=5), A258927.

nmax=20; Subscript[a,0]=1; egf=Sum[Subscript[a,k]*x^k,{k,0,nmax+1}]; Table[Subscript[a,k]*k!,{k,0,nmax}] /.Solve[Take[CoefficientList[Expand[1+egf^6-D[egf,x]],x],nmax]==ConstantArray[0,nmax]][[1]]

{a(n) = local(A=1); A = 1 + serreverse( intformal( 1/(1 + (1+x)^6 +x*O(x^n)) )); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Jun 16 2015

a(n) ~ n! * d^(n+1/5) / (5^(1/5) * Gamma(1/5) * n^(4/5)), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(6*j-1) = 6/(Pi - sqrt(3)*log(2+sqrt(3))) = 6.97224737278326506475991855023425659249063565...

E.g.f.: 1 + Series_Reversion( Integral 1/(1 + (1+x)^6) dx ). - Paul D. Hanna, Jun 16 2015

A258970 E.g.f.: A'(x) = 1 + A(x)^4, with A(0)=1.

Original entry on oeis.org

1, 2, 8, 80, 1088, 19328, 422912, 10987520, 330555392, 11300913152, 432717037568, 18344259092480, 852932666851328, 43157160112160768, 2360748463307620352, 138821061188696145920, 8732741520836633034752, 585172975239737913638912, 41612642758392039581155328

Offset: 0

Vaclav Kotesovec, Jun 15 2015

nonn

			A(x) = 1 + 2*x + 8*x^2/2! + 80*x^3/3! + 1088*x^4/4! + 19328*x^5/5! + ...
A'(x) = 2 + 8*x + 40*x^2 + 544*x^3/3 + 2416*x^4/3 + 52864*x^5/15 + ...
1 + A(x)^4 = 2 + 8*x + 40*x^2 + 544*x^3/3 + 2416*x^4/3 + 52864*x^5/15 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..100

Cf. A000831, A258969, A258971, A258994, A258901.

nmax=20; Subscript[a,0]=1; egf=Sum[Subscript[a,k]*x^k, {k,0,nmax+1}]; Table[Subscript[a,k]*k!, {k,0,nmax}] /.Solve[Take[CoefficientList[Expand[1+egf^4-D[egf,x]],x],nmax]==ConstantArray[0,nmax]][[1]]

{a(n) = local(A=1); A = 1 + serreverse( intformal( 1/(1 + (1+x)^4 +x*O(x^n)) )); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Jun 16 2015

a(n) ~ n! * d^(n+1/3) / (3^(1/3) * GAMMA(1/3) * n^(2/3)), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(4*j-1) = 4*sqrt(2)/(Pi + log(3-2*sqrt(2))) = 4.10260201986929...

E.g.f.: 1 + Series_Reversion( Integral 1/(1 + (1+x)^4) dx ). - Paul D. Hanna, Jun 16 2015

A258971 E.g.f.: A'(x) = 1 + A(x)^5, with A(0)=1.

Original entry on oeis.org

1, 2, 10, 130, 2330, 54770, 1591690, 55065250, 2209888250, 100922263250, 5167670934250, 293215490277250, 18260340583516250, 1238269550334211250, 90824251513716786250, 7164531681653318001250, 604824006980892825496250, 54406894886223009690031250

Offset: 0

Vaclav Kotesovec, Jun 15 2015

nonn

In general, for k>1, if e.g.f. satisfies A'(x) = 1 + A(x)^k, with A(0)=1, then a(n) ~ n! * d^(n + 1/(k-1)) / ((k-1)^(1/(k-1)) * Gamma(1/(k-1)) * n^(1-1/(k-1))), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(k*j-1).

			A(x) = 1 + 2*x + 10*x^2/2! + 130*x^3/3! + 2330*x^4/4! + 54770*x^5/5! + ...
A'(x) = 2 + 10*x + 65*x^2 + 1165*x^3/3 + 27385*x^4/12 + 159169*x^5/12 + ...
1 + A(x)^5 = 2 + 10*x + 65*x^2 + 1165*x^3/3 + 27385*x^4/12 + 159169*x^5/12 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..80

Cf. A000831 (k=2), A258969 (k=3), A258970 (k=4), A258994 (k=6), A258925.

nmax=20; Subscript[a,0]=1; egf=Sum[Subscript[a,k]*x^k, {k,0,nmax+1}]; Table[Subscript[a,k]*k!, {k,0,nmax}] /.Solve[Take[CoefficientList[Expand[1+egf^5-D[egf,x]],x],nmax]==ConstantArray[0,nmax]][[1]]

{a(n) = local(A=1); A = 1 + serreverse( intformal( 1/(1 + (1+x)^5 +x*O(x^n)) )); n!*polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Jun 16 2015

a(n) ~ n! * d^(n+1/4) / (4^(1/4) * Gamma(1/4) * n^(3/4)), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(5*j-1) = 40*sqrt(5-sqrt(5)) / (8*sqrt(2)*Pi + sqrt(5+sqrt(5)) * ((9-5*sqrt(5))*log(2) + (sqrt(5)-5)*log(7+3*sqrt(5)))) = 5.53569595526739362969262739469167643400611216649309306882558956...

E.g.f.: 1 + Series_Reversion( Integral 1/(1 + (1+x)^5) dx ). - Paul D. Hanna, Jun 16 2015

A227250 Number of binary labeled trees with two-colored vertices that have n leaves and avoid the easiest to avoid 6-pattern set.

Original entry on oeis.org

1, 2, 6, 42, 390, 4698, 69174

Offset: 1

Vladimir Dotsenko, Jul 04 2013

There are two six-pattern sets that are the easiest to avoid, they are identified with one another by either swapping colors (black <-> white) or passing to complements (the latter implies that the compositional inverse e.g.f. F(x) of the sequence in question is -F(-x)). One of them is (in operation notation, with b/w encoding black/white vertices) {b(b(1,2),3), b(b(1,3),2), b(1,b(2,3)), b(w(1,3),2), b(1,w(2,3)), w(b(1,2),3)}, the other is {w(w(1,2),3), w(w(1,3),2), w(1,w(2,3)), w(b(1,3),2), w(1,b(2,3)), b(w(1,2),3)}.

Conjecture: E.g.f. (for offset 0) satisfies A'(x) = 1 + A(x)^3, with A(0)=1. The next terms are 1203498, 24163110, 549811962, 13982486166, 393026414922, ... - Vaclav Kotesovec, Jun 15 2015

V. Dotsenko, Pattern avoidance in labelled trees, Séminaire Lotharingien de Combinatoire, B67b (2012), 27 pp.

Cf. A258880, A258969.

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A258880 E.g.f. satisfies: A(x) = Integral 1 + A(x)^3 dx.

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A193534 Decimal expansion of (1/3) * (Pi/sqrt(3) - log(2)).

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A258994 E.g.f.: A'(x) = 1 + A(x)^6, with A(0)=1.

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A258970 E.g.f.: A'(x) = 1 + A(x)^4, with A(0)=1.

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A258971 E.g.f.: A'(x) = 1 + A(x)^5, with A(0)=1.

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A227250 Number of binary labeled trees with two-colored vertices that have n leaves and avoid the easiest to avoid 6-pattern set.

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