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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A210949 E.g.f. A(x) satisfies: A'(x) = 1/(1 - A(A(x))).

Original entry on oeis.org

1, 1, 4, 29, 309, 4383, 78121, 1684706, 42801222, 1255919755, 41918624013, 1572257236114, 65619165625383, 3022617826829288, 152615633802149416, 8397224009015443509, 500957609480739613321, 32261529179806961067634, 2234133327582388824135291
Offset: 1

Views

Author

Paul D. Hanna, Jul 22 2012

Keywords

Comments

An unsigned version of A067146.
Equals row sums of triangle A277410.
Is this a duplicate of A014622, which is related to f(f(x))f'(x)=-1 ? - R. J. Mathar, May 13 2025

Examples

			E.g.f.: A(x) = x + x^2/2! + 4*x^3/3! + 29*x^4/4! + 309*x^5/5! + 4383*x^6/6! +...
Let G(x) = Integral A(x) dx, then A(x) = x + G(A(x)) where
G(x) = x^2/2! + x^3/3! + 4*x^4/4! + 29*x^5/5! + 309*x^6/6! + 4383*x^7/7! +...
Also,
A(x) = x + G(x) + d/dx G(x)^2/2! + d^2/dx^2 G(x)^3/3! + d^3/dx^3 G(x)^4/4! +...
log(A(x)/x) = G(x)/x + d/dx G(x)^2/(2!*x) + d^2/dx^2 G(x)^3/(3!*x) + d^3/dx^3 G(x)^4/(4!*x) +...
By definition, A'(x) = 1/(1 - A(A(x))), where
A(A(x)) = x + 2*x^2/2! + 11*x^3/3! + 101*x^4/4! + 1313*x^5/5! + 22235*x^6/6! + 466356*x^7/7! + 11710760*x^8/8! +...

Links

Crossrefs

Cf. A277410, A067146, A279843, A279844, A279845, A280570, A280571, A280572, A280573, A280574, A280575.

Programs

  • Mathematica
    m = 20; A[_] = 0;
Do[A[x_] = InverseSeries[Integrate[1 - A[x], x] + O[x]^m], {m}];
CoefficientList[A[x], x] * Range[0, m - 1]! // Rest (* Jean-François Alcover, Sep 30 2019 *)
  • PARI
    {a(n)=local(A=x,G);for(i=1,n,G=intformal(A+x*O(x^n));A=serreverse(x-G));n!*polcoeff(A, n)}
  • PARI
    {a(n)=local(A=x,G);for(i=1,n,G=intformal(A+x*O(x^n));A=x+subst(G,x,A+x*O(x^n))); n!*polcoeff(A, n)}
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
{a(n)=local(A=x,G);for(i=0,n,G=intformal(A+x*O(x^n)); A=x+sum(m=1, n, Dx(m-1, G^m/m!))); n!*polcoeff(A, n)}
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
{a(n)=local(A=x,G);for(i=0,n,G=intformal(A+x*O(x^n)); A=x*exp(sum(m=1, n, Dx(m-1, G^m/x/m!)+x*O(x^n)))); n!*polcoeff(A, n)}
for(n=1, 25, print1(a(n), ", "))
  • PARI
    {a(n) = local(A=x); for(i=1,n, A = serreverse(intformal(1-A +x*O(x^n)))); n!*polcoeff(A,n)}
for(n=1, 25, print1(a(n), ", "))

Formula

E.g.f. A(x) satisfies:
(1) A(x) = Series_Reversion( Integral 1 - A(x) dx ).
(2) A''(x) = 1 / ( (1 - A(A(x)))^3 * (1 - A(A(A(x)))) ).
Let G(x) = Integral A(x) dx with G(0)=0, then the e.g.f. A(x) satisfies:
(3) A(x) = x + G(A(x)) or, equivalently, A(x - G(x)) = x.
(4) A(x) = x + Sum_{n>=1} d^(n-1)/dx^(n-1) G(x)^n / n!.
(5) A(x) = x*exp( Sum_{n>=1} d^(n-1)/dx^(n-1) G(x)^n/(n!*x) ).
a(n) = Sum_{k=0..n-1} A277410(n,k).

A277410 G.f. A(x,y) satisfies: A( x - y*G(x,y), y) = x + (1-y)*G(x,y) such that G(x,y) = Integral A(x,y) dx, where the coefficients T(n,k) of x^n*y^k form a triangle read by rows n>=1, for k=0..n-1.

Original entry on oeis.org

1, 1, 0, 1, 3, 0, 1, 13, 15, 0, 1, 38, 165, 105, 0, 1, 94, 1033, 2310, 945, 0, 1, 213, 4953, 26229, 36330, 10395, 0, 1, 459, 20370, 213511, 674520, 640710, 135135, 0, 1, 960, 76056, 1421225, 8559675, 18127935, 12588345, 2027025, 0, 1, 1972, 266334, 8283234, 85654979, 337805535, 515903850, 273544425, 34459425, 0, 1, 4007, 892542, 44013478, 729292193, 4822487682, 13506364410, 15631793100, 6529047525, 654729075, 0
Offset: 1

Views

Author

Paul D. Hanna, Oct 13 2016

Keywords

Comments

More generally, we have the following related identity.
Given functions F and G with F(0)=0, F'(0)=1, G(0)=0, G'(0)=0,
if F(x - y*G(x)) = x + (1-y)*G(x), then
(1) F(x) = x + G( y*F(x) + (1-y)*x ),
(2) y*F(x) + (1-y)*x = Series_Reversion(x - y*G(x)),
(3) F(x) = x + G(x + y*G(x + y*G(x + y*G(x +...)))),
(4) F(x) = x + Sum_{n>=1} y^(n-1) * d^(n-1)/dx^(n-1) G(x)^n / n!.
The g.f. of this sequence A(x,y) equals F(x) in the above when G(x) = Integral F(x) dx.

Examples

			G.f.: A(x,y) = x + x^2/2! + (3*y + 1)*x^3/3! + (15*y^2 + 13*y + 1)*x^4/4! + (105*y^3 + 165*y^2 + 38*y + 1)*x^5/5! + (945*y^4 + 2310*y^3 + 1033*y^2 + 94*y+ 1)*x^6/6! + (10395*y^5 + 36330*y^4 + 26229*y^3 + 4953*y^2 + 213*y + 1)*x^7/7! + (135135*y^6 + 640710*y^5 + 674520*y^4 + 213511*y^3 + 20370*y^2 + 459*y + 1)*x^8/8! + (2027025*y^7 + 12588345*y^6 + 18127935*y^5 + 8559675*y^4 + 1421225*y^3 + 76056*y^2 + 960*y + 1)*x^9/9! + (34459425*y^8 + 273544425*y^7 + 515903850*y^6 + 337805535*y^5 + 85654979*y^4 + 8283234*y^3 + 266334*y^2 + 1972*y + 1)*x^10/10! +...
such that A( x - y*G(x,y), y)  =  x + (1-y)*G(x,y)
also,
A(x,y) = x + G( y*A(x,y) + (1-y)*x, y)
where G(x,y) = Integral A(x,y).
...
This triangle of coefficients T(n,k) of x^n*y^k/n! in g.f. A(x,y) begins:
1;
1, 0;
1, 3, 0;
1, 13, 15, 0;
1, 38, 165, 105, 0;
1, 94, 1033, 2310, 945, 0;
1, 213, 4953, 26229, 36330, 10395, 0;
1, 459, 20370, 213511, 674520, 640710, 135135, 0;
1, 960, 76056, 1421225, 8559675, 18127935, 12588345, 2027025, 0;
1, 1972, 266334, 8283234, 85654979, 337805535, 515903850, 273544425, 34459425, 0;
1, 4007, 892542, 44013478, 729292193, 4822487682, 13506364410, 15631793100, 6529047525, 654729075, 0;
1, 8089, 2900353, 218797958, 5531376285, 57226590953, 264482764305, 555756298020, 505173143475, 170116046100, 13749310575, 0; ...
in which the diagonal equals A001147 (odd double factorials), and the row sums yield A210949.
...
APPLICATION.
Given F(x) such that
F(x - Integral p*F(x) dx) = x + Integral q*F(x) dx
then
F(x) = Sum_{n>=1} a(n)*x^n/n!
where
a(n) = Sum_{k=0..n-1} A277410(n,k) * p^k * (p+q)^(n-k-1) for n>=1.
EXAMPLES.
A210949(n) = Sum_{k=0..n-1} A277410(n,k).
A277403(n) = Sum_{k=0..n-1} A277410(n,k) * 2^(n-k-1).
A279843(n) = Sum_{k=0..n-1} A277410(n,k) * 3^(n-k-1).
A279844(n) = Sum_{k=0..n-1} A277410(n,k) * 2^k * 3^(n-k-1).
A279845(n) = Sum_{k=0..n-1} A277410(n,k) * 2^k.
A280570(n) = Sum_{k=0..n-1} A277410(n,k) * 4^(n-k-1).
A280571(n) = Sum_{k=0..n-1} A277410(n,k) * 3^k * 4^(n-k-1).
A280572(n) = Sum_{k=0..n-1} A277410(n,k) * 5^(n-k-1).
A280573(n) = Sum_{k=0..n-1} A277410(n,k) * 2^k * 5^(n-k-1).
A280574(n) = Sum_{k=0..n-1} A277410(n,k) * 3^k * 5^(n-k-1).
A280575(n) = Sum_{k=0..n-1} A277410(n,k) * 4^k * 5^(n-k-1).
...
COLUMN GENERATING FUNCTIONS.
From _Paul D. Hanna_, Nov 05 2016: (Start)
_Colin Barker_ observed that column 1 of this triangle (A277411) appears to have the o.g.f. x*(3*x-2*x^2) / ((1-x)^3*(1-2*x)).
This observation led to the following conjecture.
Let F(k,x) = o.g.f. of column k in this triangle,
then
F(k,x) = P(k,x) * x^(k+1) / Product_{j=0..k} (1 - (j+1)*x)^(2*(k-j)+1)
where P(k,x) is a polynomial in x with degree k*(k+1) for k>=0.
Example:
F(0,x) = x/(1-x) ;
F(1,x) = P(1,x)*x^2/((1-x)^3*(1-2*x)) ;
F(2,x) = P(2,x)*x^3/((1-x)^5*(1-2*x)^3*(1-3*x)) ;
F(3,x) = P(3,x)*x^4/((1-x)^7*(1-2*x)^5*(1-3*x)^3*(1-4*x)) ;
...
The polynomials P(k,x) begin:
P(0,x) = 1 ;
P(1,x) = 3*x - 2*x^2 ;
P(2,x) = 15*x - 45*x^2 - 2*x^3 + 106*x^4 - 92*x^5 + 24*x^6 ;
P(3,x) = 105*x - 840*x^2 + 504*x^3 + 16321*x^4 - 75880*x^5 + 154483*x^6 - 152077*x^7 + 39208*x^8 + 59000*x^9 - 60336*x^10 + 23328*x^11 - 3456*x^12 ;
P(4,x) = 945*x - 15645*x^2 + 32445*x^3 + 1255770*x^4 - 15120061*x^5 + 86803308*x^6 - 291640845*x^7 + 529758178*x^8 - 50236668*x^9 - 2553002523*x^10 + 7695202852*x^11 - 12713196156*x^12 + 13351222596*x^13 - 8752472980*x^14 + 2871967920*x^15 + 387984096*x^16 - 884504448*x^17 + 427064832*x^18 - 100694016*x^19 + 9953280*x^20 ;
P(5,x) = 10395*x - 305235*x^2 + 1299375*x^3 + 77300220*x^4 - 1834009998*x^5 + 21447595316*x^6 - 156933684108*x^7 + 721294719700*x^8 - 1490891586137*x^9 - 5868653004882*x^10 + 70213320019895*x^11 - 359261247450016*x^12 + 1234731543184308*x^13 - 3081038591203028*x^14 + 5553265322783926*x^15 - 6518085613542516*x^16 + 2256970375232288*x^17 + 9498116639867573*x^18 - 25485484994020128*x^19 + 37162639109810884*x^20 - 37419816866322296*x^21 + 27200926921683600*x^22 - 14055671260790656*x^23 + 4698364855901568*x^24 - 583485067952640*x^25 - 341605998065664*x^26 + 237336648708096*x^27 - 72380729917440*x^28 + 11910492979200*x^29 - 859963392000*x^30 ;
...
where the coefficient of x^(k*(k+1)) in P(k,x) equals A059332(k+1).
(End)

Links

Crossrefs

Cf. A210949 (row sums), A067146, A001147 (diagonal), A277411 (column 1), A277412 (diagonal).
Cf. A279843, A279844, A279845, A280570, A280571, A280572, A280573, A280574, A280575.

Programs

  • PARI
    {T(n, k) = my(A=x); for(i=1, n, A = x + subst(intformal(A +x*O(x^n)), x, y*A + (1-y)*x ) ); n!*polcoeff(polcoeff(A,n,x),k,y)}
for(n=1, 12, for(k=0, n-1, print1(T(n, k), ", ")); print(""))

Formula

Given g.f. A(x,y), define G(x,y) = Integral A(x,y) dx, then
(1) A(x,y) = x + G( y*A(x,y) + (1-y)*x, y),
(2) y*A(x,y) + (1-y)*x = Series_Reversion( x - y*G(x,y) ),
(3) y*x + (1-y)*B(x,y) = Series_Reversion( x + (1-y)*G(x,y) ), where B( A(x,y), y) = x.
(4) A(x,y) = x + Sum_{n>=1} y^(n-1) * d^(n-1)/dx^(n-1) G(x,y)^n / n!.
In formulas 2 and 3, the series reversion is taken with respect to variable x.

A144006 Triangle, read by rows of coefficients of x^n*y^k for k=0..n(n-1)/2 for n>=0, defined by e.g.f.: A(x,y) = 1 + Series_Reversion( Integral A(-x*y,y) dx ), with leading zeros in each row suppressed.

Original entry on oeis.org

1, 1, 1, 3, -1, 15, -10, 3, -1, 105, -105, 55, -30, 10, -3, 1, 945, -1260, 910, -630, 350, -168, 76, -30, 10, -3, 1, 10395, -17325, 15750, -12880, 9135, -5789, 3381, -1806, 910, -434, 196, -76, 30, -10, 3, -1, 135135, -270270, 294525, -275275, 228375
Offset: 0

Views

Author

Paul D. Hanna, Sep 10 2008

Keywords

Comments

Comment from Lucas Larsen, Aug 20 2024: (Start)
The nonzero entries in the n-th row appear to be the nonzero coefficients (up to sign) in the following:
Let c be a fixed point in (0,oo) and f a smooth function such that f(c) = c and f(f'(x)) = x in a neighborhood of c. Then the n-th derivative of f evaluated at c can be written as a Laurent polynomial in c with the (descending) coefficients in question.
For instance:
f'(c) = c
f''(c) = c^(-1)
f'''(c) = -c^(-4)
f''''(c) = 3c^(-7) + c^(-8)
(End)

Examples

			Triangle begins (without suppressing leading zeros):
1;
1;
0, 1;
0,0, 3, -1;
0,0,0, 15, -10, 3, -1;
0,0,0,0, 105, -105, 55, -30, 10, -3, 1;
0,0,0,0,0, 945, -1260, 910, -630, 350, -168, 76, -30, 10, -3, 1;
0,0,0,0,0,0, 10395, -17325, 15750, -12880, 9135, -5789, 3381, -1806, 910, -434, 196, -76, 30, -10, 3, -1;
0,0,0,0,0,0,0, 135135, -270270, 294525, -275275, 228375, -172200, 120960, -78519, 48006, -28336, 16065, -8609, 4461, -2166, 1018, -470, 196, -76, 30, -10, 3, -1; ...

Crossrefs

Cf. A144005, A067146.
Generates A014621, A014622 and A014623, which are related to Levine's sequence A011784.

Programs

  • PARI
    {T(n,k)=local(A=1+x*O(x^n)); for(i=0,n,A=1+serreverse(intformal(subst(A,x,-x*y))));n!*polcoeff(polcoeff(A,n,x),k,y)}
  • Python
    #This is only correct if the observation in the comment from 2024/08/20 is true.
def T(n,k):
    if 0 <= n <= 1:
        return 1 if k == 0 else 0
    c = {(-1,):1} #Polynomial in infinitely many variables (function iterates)
    for _ in range(n-1):
        cnext = {}
        for key, value in c.items():
            key += (0,)
            for i, ni in enumerate(key):
                term = tuple(nj-2 if j==i else nj-1 if j<=i+1 else nj
                             for j,nj in enumerate(key))
                cnext[term] = cnext.get(term,0) + value*ni
                if cnext[term] == 0:
                    del cnext[term]
        c = cnext
    pairs = {} #Reduction to single variable (evaluation at fixpoint)
    for key, value in c.items():
        s = -sum(key)
        pairs[s] = pairs.get(s,0) + value
    _, row = zip(*sorted(pairs.items())) #Coefficients
    if 0 <= k-n+1 < len(row): #Correcting number of leading 0s
        return (-1)**(n+k+1)*abs(row[k-n+1]) #Correcting signs
    else:
        return 0
# Lucas Larsen, Aug 22 2024

Formula

E.g.f. satisfies: A(x,y) = 1 + Series_Reversion[Integral A(-x*y,y) dx].
T(n,k) = [x^n*y^k] n!*A(x,y) for k=0..n(n-1)/2, n>=0.
Row sums equal A144005.
A067146(n) = Sum_{k=0..n(n-1)/2} (-1)^k*T(n,k).
This is a signed version of table A014621 because setting f((1+x)/y):=A(-x*y,y)/y for fixed y>0 implies f(f(x))*f'(x)=-1 and f(1/y)=1/y, as in the second formula of A014621. Therefore, the row sums form A014623 and the unsigned row sums form A014622. - Roland Miyamoto, Jun 03 2024
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