cp's OEIS Frontend

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.

Showing 1-4 of 4 results.

A369090 Expansion of e.g.f. A(x) satisfying A(x) = A( x^2*exp(x) ) / x, with A(0) = 0.

Original entry on oeis.org

1, 2, 9, 52, 425, 4206, 48307, 632360, 9444465, 159240250, 2983729331, 61300668012, 1367054727337, 32844312889766, 845234187028155, 23190947446000336, 675895337644401377, 20863665943202969586, 680448552777544884643, 23395823324931227353940, 846248620848062865320601
Offset: 1

Views

Author

Paul D. Hanna, Jan 26 2024

Keywords

Comments

Limit (a(n)/n!)^(1/n) = 1/w where w*exp(w) = 1 and w = LambertW(1) = 0.567143290409783872999968... (cf. A030178).

Examples

			E.g.f.: A(x) = x + 2*x^2/2! + 9*x^3/3! + 52*x^4/4! + 425*x^5/5! + 4206*x^6/6! + 48307*x^7/7! + 632360*x^8/8! + 9444465*x^9/9! + 159240250*x^10/10! + ...
RELATED SERIES.
The expansion of the logarithm of A(x)/x starts
log(A(x)/x) = x + 2*x^2/2! + 6*x^3/3! + 36*x^4/4! + 260*x^5/5! + 2190*x^6/6! + 21882*x^7/7! + 268856*x^8/8! + ... + A369091(n)*x^n/n! + ...
and equals the sum of all iterations of the function x^2*exp(x).
Let R(x) be the series reversion of A(x),
R(x) = x - 2*x^2/2! + 3*x^3/3! + 8*x^4/4! - 155*x^5/5! + 1464*x^6/6! - 7931*x^7/7! - 65360*x^8/8! + 2742345*x^9/9! + ...
then R(x) and e.g.f. A(x) satisfy:
(1) R( A(x) ) = x,
(2) R( x*A(x) ) = x^2 * exp(x).
GENERATING METHOD.
Let F(n) equal the n-th iteration of x^2*exp(x), so that
F(0) = x,
F(1) = x^2 * exp(x),
F(2) = x^4 * exp(2*x) * exp(x^2*exp(x)),
F(3) = x^8 * exp(4*x) * exp(2*x^2*exp(x)) * exp(F(2)),
F(4) = x^16 * exp(8*x) * exp(4*x^2*exp(x)) * exp(2*F(2)) * exp(F(3)),
F(5) = x^32 * exp(16*x) * exp(8*x^2*exp(x)) * exp(4*F(2)) * exp(2*F(3)) * exp(F(4)),
...
F(n+1) = F(n)^2 * exp(F(n))
...
Then the e.g.f. A(x) equals
A(x) = x * exp(F(0) + F(1) + F(2) + F(3) + ... + F(n) + ...).
equivalently,
A(x) = x * exp(x + x^2*exp(x) + x^4*exp(2*x)*exp(x^2*exp(x)) + x^8*exp(4*x)*exp(2*x^2*exp(x)) * exp(x^4*exp(2*x)*exp(x^2*exp(x))) + ...).

Links

Crossrefs

Cf. A369091, A369550 (a(n)/n), A030178.
Cf. A367390.

Programs

  • PARI
    {a(n) = my(A=x); for(i=0, #binary(n),
A = subst(A, x, x^2*exp(x +x^2*O(x^n)) )/x ); n! * polcoeff(H=A, n)}
for(n=1, 30, print1(a(n), ", "))

Formula

E.g.f. A(x) = Sum_{n>=1} a(n)*x^n/n! satisfies the following formulas.
(1) A(x) = A(x^2*exp(x)) / x.
(2) R(x*A(x)) = x^2*exp(x), where R(A(x)) = x.
(3) A(x) = x * exp( Sum_{n>=0} F(n) ), where F(0) = x, and F(n+1) = F(n)^2 * exp(F(n)) for n >= 0.
(4) A(x) = x * exp(L(x)), where L(x) = x + L(x^2*exp(x)) is the e.g.f. of A369091.

A367386 Expansion of g.f. A(x) satisfying A(x)^2 = (1+x) * A(x*A(x)) with A(0) = 0.

Original entry on oeis.org

1, 1, 1, 2, 3, 6, 12, 26, 56, 124, 278, 632, 1454, 3378, 7918, 18694, 44427, 106175, 255031, 615320, 1490588, 3624013, 8840006, 21628173, 53061676, 130508716, 321743567, 794907220, 1967848545, 4880622339, 12125865713, 30175562392, 75207082211, 187707922818, 469126856364
Offset: 1

Views

Author

Paul D. Hanna, Jan 08 2024

Keywords

Comments

Note that if F(x)^2 = (1+x) * F(x*F(x)) with F(0) = 1, then F(x) is the g.f. of A120056.

Examples

			G.f.: A(x) = x + x^2 + x^3 + 2*x^4 + 3*x^5 + 6*x^6 + 12*x^7 + 26*x^8 + 56*x^9 + 124*x^10 + 278*x^11 + 632*x^12 + 1454*x^13 + 3378*x^14 + 7918*x^15 + ...
where A(x)^2 = (1+x) * A(x*A(x)) as can be seen from the following expansions
A(x)^2 = x^2 + 2*x^3 + 3*x^4 + 6*x^5 + 11*x^6 + 22*x^7 + 46*x^8 + 100*x^9 + 221*x^10 + 496*x^11 + 1128*x^12 + 2592*x^13 + 6016*x^14 + 14080*x^15 + ...
A(x*A(x)) = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 15*x^7 + 31*x^8 + 69*x^9 + 152*x^10 + 344*x^11 + 784*x^12 + 1808*x^13 + 4208*x^14 + 9872*x^15 + ...
Let B(x) = x*A(x), then A(x) equals the infinite product involving successive iterations of B(x) starting with
A(x) = x*(1+x) * (1 + B(x)) * (1 + B(B(x))) * (1 + B(B(B(x)))) * (1 + B(B(B(B(x))))) * ...
which is equivalent to
A(x) = x*(1+x) * (1 + x*A(x)) * (1 + x*A(x) * A(x*A(x))) * (1 + x*A(x) * A(x*A(x)) * A(x*A(x) * A(x*A(x)))) * ...
RELATED SERIES.
Successive iterations of B(x) = x*A(x) begin
B(x) = x^2 + x^3 + x^4 + 2*x^5 + 3*x^6 + 6*x^7 + 12*x^8 + 26*x^9 + 56*x^10 + ...
B(B(x)) = x^4 + 2*x^5 + 4*x^6 + 9*x^7 + 18*x^8 + 39*x^9 + 85*x^10 + 191*x^11 + ...
B(B(B(x))) = x^8 + 4*x^9 + 12*x^10 + 34*x^11 + 89*x^12 + 228*x^13 + 575*x^14 + ...
B(B(B(B(x)))) = x^16 + 8*x^17 + 40*x^18 + 164*x^19 + 594*x^20 + 1984*x^21 + ...
B(B(B(B(B(x))))) = x^32 + 16*x^33 + 144*x^34 + 968*x^35 + 5412*x^36 + 26592*x^37 + ...
etc.
The coefficients in the iterations of x*A(x) form a table that begins
n=1: [1, 1, 1, 2, 3, 6, 12, 26, 56, 124, 278, 632, 1454, ...];
n=2: [1, 2, 4, 9, 18, 39, 85, 191, 433, 994, 2303, 5377, ...];
n=3: [1, 4, 12, 34, 89, 228, 575, 1441, 3595, 8943, 22215, ...];
n=4: [1, 8, 40, 164, 594, 1984, 6266, 19006, 55944, 160926, ...];
n=5: [1, 16, 144, 968, 5412, 26592, 118692, 491820, 1920852, ...];
n=6: [1, 32, 544, 6544, 62536, 505152, 3584008, 22917912, ...];
n=7: [1, 64, 2112, 47904, 839824, 12132480, 150360848, ...];
n=8: [1, 128, 8320, 366144, 12271904, 334108928, 7695888928, ...];
n=9: [1, 256, 33024, 2862208, 187499072, 9902461440, ...];
n=10: [1, 512, 131584, 22632704, 2931033216, 304847561728, ...];
etc.

Links

Crossrefs

Cf. A120056, A367387, A367390.

Programs

  • PARI
    {a(n) = my(A=x, V=[0,1]); for(i=1,n, V = concat(V,0); A = Ser(V);
V[#V] = polcoeff( subst(A,x,x*A) - A^2/(1+x), #V) ); V[n+1]}
for(n=1,40, print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n and B(x) = x*A(x) satisfy the following formulas.
(1) A(x)^2 = (1+x) * A(x*A(x)).
(2) A(x) = x*(1+x) * (1 + B(x)) * (1 + B(B(x))) * (1 + B(B(B(x)))) * (1 + B(B(B(B(x))))) * ..., an infinite product involving iterations of B(x) = x*A(x).
The iterations of B(x) = x*A(x) begin
(3.a) B(B(x)) = x*A(x)^3 / (1+x).
(3.b) B(B(B(x))) = x*A(x)^7 / ((1+x)^3 * (1 + x*A(x))).
(3.c) B(B(B(B(x)))) = x*A(x)^15 / ((1+x)^7 * (1 + x*A(x))^3 * (1 + x*A(x)^3/(1+x))).
(3.d) B(B(B(B(B(x))))) = x*A(x)^31 / ((1+x)^15 * (1+x*A(x))^7 * (1 + x*A(x)^3/(1+x))^3 * (1 + x*A(x)^7/((1+x)^3*(1+x*A(x))))).
The compositions of g.f. A(x) with the iterations of B(x) = x*A(x) begin
(4.a) A(B(x)) = A(x)^2 / (1+x).
(4.b) A(B(B(x))) = A(x)^4 / ((1+x)^2 * (1 + x*A(x))).
(4.c) A(B(B(B(x)))) = A(x)^8 / ((1+x)^4 * (1 + x*A(x))^2 * (1 + x*A(x)^3/(1+x))).
(4.d) A(B(B(B(B(x))))) = A(x)^16 / ((1+x)^8 * (1+x*A(x))^4 * (1 + x*A(x)^3/(1+x))^2 * (1 + x*A(x)^7/((1+x)^3*(1+x*A(x))))).

A367387 Expansion of g.f. A(x) satisfying A(x)^2 = A(x*A(x)) / (1-x) with A(0) = 0.

Original entry on oeis.org

1, 1, 2, 3, 7, 14, 34, 77, 193, 472, 1214, 3099, 8122, 21293, 56666, 151261, 407519, 1102006, 2998716, 8189515, 22467935, 61841586, 170818016, 473173219, 1314463002, 3660532769, 10218207713, 28584456783, 80124502593, 225011930357, 633003693094, 1783658958681, 5033641233827
Offset: 1

Views

Author

Paul D. Hanna, Jan 08 2024

Keywords

Comments

Note that if F(x)^2 = (1+x) * F(x*F(x)) with F(0) = 1, then F(x) is the g.f. of A088792.

Examples

			G.f.: A(x) = x + x^2 + 2*x^3 + 3*x^4 + 7*x^5 + 14*x^6 + 34*x^7 + 77*x^8 + 193*x^9 + 472*x^10 + 1214*x^11 + 3099*x^12 + 8122*x^13 + 21293*x^14 + 56666*x^15 + ...
where A(x)^2 = A(x*A(x)) / (1-x) as can be seen from the following expansions
A(x)^2 = x^2 + 2*x^3 + 5*x^4 + 10*x^5 + 24*x^6 + 54*x^7 + 133*x^8 + 320*x^9 + 809*x^10 + 2038*x^11 + 5278*x^12 + 13702*x^13 + 36144*x^14 + 95758*x^15 + ...
A(x*A(x)) = x^2 + x^3 + 3*x^4 + 5*x^5 + 14*x^6 + 30*x^7 + 79*x^8 + 187*x^9 + 489*x^10 + 1229*x^11 + 3240*x^12 + 8424*x^13 + 22442*x^14 + 59614*x^15 + ...
Let B(x) = x*A(x), then A(x) equals the infinite product involving successive iterations of B(x) starting with
A(x) = x/(1-x) / ( (1 - B(x)) * (1 - B(B(x))) * (1 - B(B(B(x)))) * (1 - B(B(B(B(x))))) * ...)
which is equivalent to
A(x) = x*(1-x) / ( (1 - x*A(x)) * (1 - x*A(x) * A(x*A(x))) * (1 - x*A(x) * A(x*A(x)) * A(x*A(x) * A(x*A(x)))) * ...).
RELATED SERIES.
Successive iterations of B(x) = x*A(x) begin
B(x) = x^2 + x^3 + 2*x^4 + 3*x^5 + 7*x^6 + 14*x^7 + 34*x^8 + 77*x^9 + ...
B(B(x)) = x^4 + 2*x^5 + 6*x^6 + 13*x^7 + 35*x^8 + 84*x^9 + 221*x^10 + ...
B(B(B(x))) = x^8 + 4*x^9 + 16*x^10 + 50*x^11 + 159*x^12 + 470*x^13 + ...
B(B(B(B(x)))) = x^16 + 8*x^17 + 48*x^18 + 228*x^19 + 974*x^20 + 3812*x^21 + ...
B(B(B(B(B(x))))) = x^32 + 16*x^33 + 160*x^34 + 1224*x^35 + 7900*x^36 + ...
etc.
The coefficients in the iterations of x*A(x) form a table that begins
n=1: [1, 1, 2, 3, 7, 14, 34, 77, 193, 472, 1214, 3099, ...];
n=2: [1, 2, 6, 13, 35, 84, 221, 556, 1464, 3801, 10107, ...];
n=3: [1, 4, 16, 50, 159, 470, 1397, 4033, 11656, 33284, ...];
n=4: [1, 8, 48, 228, 974, 3812, 14142, 50182, 172562, ...];
n=5: [1, 16, 160, 1224, 7900, 45096, 234764, 1136732, ...];
n=6: [1, 32, 576, 7568, 80568, 734672, 5938776, ...];
n=7: [1, 64, 2176, 52000, 977264, 15344032, 208985520, ...];
n=8: [1, 128, 8448, 382528, 13345504, 382081856, ...];
n=9: [1, 256, 33280, 2927744, 195986880, 10643805824, ...];
n=10: [1, 512, 132096, 22894848, 2998537088, 316503534848, ...];
etc.

Links

Crossrefs

Cf. A088792, A367386, A367390.

Programs

  • PARI
    {a(n) = my(A=x, V=[0,1]); for(i=1,n, V = concat(V,0); A = Ser(V);
V[#V] = polcoeff( subst(A,x,x*A) - (1-x)*A^2, #V) ); V[n+1]}
for(n=1,40, print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=1} a(n)*x^n and B(x) = x*A(x) satisfies the following formulas.
(1) A(x)^2 = A(x*A(x)) / (1-x).
(2) A(x) = x/(1-x) / ( (1 - B(x)) * (1 - B(B(x))) * (1 - B(B(B(x)))) * (1 - B(B(B(B(x))))) * ...), an infinite product involving iterations of B(x) = x*A(x).
The iterations of B(x) = x*A(x) begin
(3.a) B(B(x)) = x*(1-x) * A(x)^3.
(3.b) B(B(B(x))) = x*(1-x)^3 * (1 - x*A(x)) * A(x)^7.
(3.c) B(B(B(B(x)))) = x*(1-x)^7 * (1 - x*A(x))^3 * (1 - x*(1-x)*A(x)^3) * A(x)^15.
(3.d) B(B(B(B(B(x))))) = x*(1-x)^15 * (1 - x*A(x))^7 * (1 - x*(1-x)*A(x)^3)^3 * (1 - x*(1-x)^3*(1-x*A(x))*A(x)^7) * A(x)^31.
The compositions of g.f. A(x) with the iterations of B(x) = x*A(x) begin
(4.a) A(B(x)) = (1-x) * A(x)^2.
(4.b) A(B(B(x))) = (1-x)^2 * (1 - x*A(x)) * A(x)^4.
(4.c) A(B(B(B(x)))) = (1-x)^4 * (1 - x*A(x))^2 * (1 - x*(1-x)*A(x)^3) * A(x)^8.
(4.d) A(B(B(B(B(x))))) = (1-x)^8 * (1 - x*A(x))^4 * (1 - x*(1-x)*A(x)^3)^2 * (1 - x*(1-x)^3*(1-x*A(x))*A(x)^7) * A(x)^16.

A367391 Expansion of e.g.f. A(x) satisfying A(x)^2 = exp(x) * A(x*A(x)) with A(0) = 1.

Original entry on oeis.org

1, 1, 3, 28, 569, 19686, 1015357, 72213450, 6732370465, 794072741302, 115412704302581, 20251767162061986, 4220273910604275889, 1030325477950545779094, 291316596476686970503693, 94452315650030395608940066, 34815037905775665043220138561, 14478491178300336588521758911894
Offset: 0

Views

Author

Paul D. Hanna, Jan 08 2024

Keywords

Comments

Note that if F(x)^2 = exp(x) * F(x*F(x)) with F(0) = 0, then F(x) is the e.g.f. of A367390.

Examples

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 28*x^3/3! + 569*x^4/4! + 19686*x^5/5! + 1015357*x^6/6! + 72213450*x^7/7! + 6732370465*x^8/8! + 794072741302*x^9/9! + 115412704302581*x^10/10! + ...
where A(x)^2 = exp(x) * A(x*A(x)) as can be seen from the following expansions
A(x)^2 = 1 + 2*x + 8*x^2/2! + 74*x^3/3! + 1416*x^4/4! + 46742*x^5/5! + 2333836*x^6/6! + 162237574*x^7/7! + ...
A(x*A(x)) = 1 + x + 5*x^2/2! + 55*x^3/3! + 1161*x^4/4! + 40331*x^5/5! + 2073253*x^6/6! + 146835179*x^7/7! + ...
RELATED SERIES.
log(A(x)) = x + 2*x^2/2! + 21*x^3/3! + 460*x^4/4! + 16675*x^5/5! + 886926*x^6/6! + 64453095*x^7/7! + 6104710088*x^8/8! + 728774208459*x^9/9! + ...
Let B(x) = x*A(x), then log(A(x)) equals a sum over all iterations of B(x):
log(A(x)) = x/2 + B(x)/2^2 + B(B(x))/2^3 + B(B(B(x)))/2^4 + B(B(B(B(x))))/2^5 + ...
Successive iterations of B(x) = x*A(x) begin
B(x) = x + 2*x^2/2! + 9*x^3/3! + 112*x^4/4! + 2845*x^5/5! + 118116*x^6/6! + 7107499*x^7/7! + 577707600*x^8/8! + ...
B(B(x)) = x + 4*x^2/2! + 30*x^3/3! + 428*x^4/4! + 10760*x^5/5! + 430302*x^6/6! + 25021024*x^7/7! + ...
B(B(B(x))) = x + 6*x^2/2! + 63*x^3/3! + 1092*x^4/4! + 29625*x^5/5! + 1196658*x^6/6! + 68472705*x^7/7! + ...
B(B(B(B(x)))) = x + 8*x^2/2! + 108*x^3/3! + 2248*x^4/4! + 68200*x^5/5! + 2905524*x^6/6! + 168670432*x^7/7! + ...
etc.
where A(x) = exp(x/2 + B(x)/4 + B(B(x))/8 + B(B(B(x)))/16 + B(B(B(B(x))))/32 + ...).

Links

Crossrefs

Cf. A367390.

Programs

  • PARI
    {a(n) = my(A=1, V=[1]); for(i=1,n, V = concat(V,0); A = Ser(V);
V[#V] = polcoeff( subst(A,x,x*A) - exp(-x +x*O(x^(#V)))*A^2, #V-1) ); n!*V[n+1]}
for(n=0,20, print1(a(n),", "))

Formula

E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! and B(x) = x*A(x) satisfies the following formulas.
(1) A(x)^2 = exp(x) * A(x*A(x)).
Let B^n(x) denote the n-th iteration of B(x) = x*A(x), where B^(n+1)(x) = B( B^n(x) ) with B^0(x) = x, then
(2) log( A(x) ) = Sum_{n>=0} B^n(x) / 2^(n+1).
(3) B^n(x) = x*A(x)^(2^n - 1) / exp( Sum_{k=0..n-2} (2^(n-k-1) - 1) * B^k(x) ) for n > 1.
(3.a) B^2(x) = x*A(x)^3 / exp(x).
(3.b) B^3(x) = x*A(x)^7 / exp(3*x + B(x)).
(3.c) B^4(x) = x*A(x)^15 / exp(7*x + 3*B(x) + B^2(x)).
(3.d) B^5(x) = x*A(x)^31 / exp(15*x + 7*B(x) + 3*B^2(x) + B^3(x)).
(4) A( B^n(x) ) = A(x)^(2^n) / exp( Sum_{k=0..n-1} 2^(n-k-1) * B^k(x) ) for n > 0.
(4.a) A(B(x)) = A(x)^2 / exp(x).
(4.b) A(B^2(x)) = A(x)^4 / exp(2*x + B(x)).
(4.c) A(B^3(x)) = A(x)^8 / exp(4*x + 2*B(x) + B^2(x)).
(4.d) A(B^4(x)) = A(x)^16 / exp(8*x + 4*B(x) + 2*B^2(x) + B^3(x)).
Showing 1-4 of 4 results.

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