A251592 -id:A251592 - cp's OEIS frontend

A369301 Expansion of (1/x) * Series_Reversion( x * (1-x)^3 * (1-x^3)^3 ).

1, 3, 15, 94, 657, 4902, 38236, 308025, 2542965, 21401780, 182934144, 1583745114, 13858675065, 122379042879, 1089156646584, 9759520978270, 87975115569873, 797233088237190, 7258632128721117, 66367727370376632, 609132332475784548, 5610015849998778144

Offset: 0

Seiichi Manyama, Jan 18 2024

nonn

Cf. A369299, A369300.

Cf. A369270.

my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x)^3*(1-x^3)^3)/x)

a(n, s=3, t=3, u=3) = sum(k=0, n\s, binomial(t*(n+1)+k-1, k)*binomial((u+1)*(n+1)-s*k-2, n-s*k))/(n+1);

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} binomial(3*n+k+2,k) * binomial(4*n-3*k+2,n-3*k).

a(n) = (1/(n+1)) * [x^n] 1/( (1-x)^3 * (1-x^3)^3 )^(n+1). - Seiichi Manyama, Feb 14 2024

A220883 Triangle read by rows: row n gives coefficients of expansion of Product_{k = 1..n-1} ((n + 1)*x + k), starting with lowest power.

Original entry on oeis.org

1, 1, 3, 2, 12, 16, 6, 55, 150, 125, 24, 300, 1260, 2160, 1296, 120, 1918, 11025, 29155, 36015, 16807, 720, 14112, 103936, 376320, 716800, 688128, 262144, 5040, 117612, 1063692, 4934601, 12859560, 19013778, 14880348, 4782969, 40320, 1095840, 11812400, 67284000, 224490000, 453600000, 546000000, 360000000, 100000000, 362880, 11292336, 141896700, 963218080, 3943187325, 10190179923, 16741251450, 16953838770, 9646149645, 2357947691

Offset: 1

N. J. A. Sloane, Dec 29 2012

Related to Stirling numbers A008275, A008277.

			Triangle begins:
    1
    1     3
    2    12     16
    6    55    150    125
   24   300   1260   2160   1296
  120  1918  11025  29155  36015  16807
  720 14112 103936 376320 716800 688128 262144
  ...

Paolo Xausa, Table of n, a(n) for n = 1..11325 (rows 1..150 of the triangle, flattened).
Peter Bala, Fractional iteration of a series inversion operator
Jean-Christophe Novelli and Jean-Yves Thibon, Duplicial algebras and Lagrange inversion, arXiv preprint arXiv:1209.5959 [math.CO], 2012.

Cf. A220884, A008275, A008277, A056856, A260687.

seq(seq(coeff(mul((n+1)*t + k, k = 1..n-1), t, i), i = 0..n-1), n = 1 .. 10); # Peter Bala, Nov 16 2015
# Alternative:
T := (n, k) -> (-1)^(n-k)*(n+1)^(k-1)*Stirling1(n, k):
seq(print(seq(T(n, k), k=1..n)), n=1..8);
# Peter Luschny, Mar 20 2024

A220883[n_, k_] := (-1)^(n-k)*(n+1)^(k-1)*StirlingS1[n, k];
Table[A220883[n, k], {n, 10}, {k, n}] (* Paolo Xausa, Mar 19 2024 *)

From Peter Bala, Nov 16 2015: (Start)
E.g.f.: A(x,t) = x + (1 + 3*t)*x^2/2! + (1 + 4*t)*(2 + 4*t)*x^3/3! + ....
The function F(x,t) := 1 + t*A(x,t) has several nice properties:
F(x,t) = 1/x*Revert( x*(1 - x)^t ) = 1 + t*x + t*(1 + 3*t)*x^2/2! + t*(2 + 12*t + 16*t^2)*x^3/3! + ..., where Revert denotes the series reversion operator with respect to x.
F(x,t)*(1 - x*F(x,t))^t = 1.
F(x,t)^m = 1 + m*t*x + m*t*((m + 2)*t + 1)*x^2/2! + m*t*((m + 3)*t + 1)*((m + 3)*t + 2)*x^3/3! + m*t*((m + 4)*t + 1)*((m + 4)*t + 2)*((m + 4)*t + 3)*x^4/4! + ....
Log(F(x,t)) = t*x + t*(1 + 2*t)*x^2/2! + t*(1 + 3*t)*(2 + 3*t)*x^3/3! + t*(1 + 4*t)*(2 + 4*t)*(3 + 4*t)*x^4/4! + ... is the e.g.f for A056856.
F(x,t) = G(x,t)^t, where G(x,t) = 1 + x + (2 + 2*t)*x^2/2! + (2 + 3*t)*(3 + 3*t)*x^3/3! + (2 + 4*t)*(3 + 4*t)*(4 + 4*t)*x^4/4! + ... is the o.g.f. for A260687. (End)
T(n, k) = (-1)^(n-k)*(n+1)^(k-1)*Stirling1(n, k). - Peter Luschny, Mar 01 2021 [Corrected by Paolo Xausa, Mar 19 2024]

A260687 Triangular array with n-th row giving coefficients of polynomial Product_{k = 2..n} (k + n*t) for n >= 1.

Original entry on oeis.org

1, 2, 2, 6, 15, 9, 24, 104, 144, 64, 120, 770, 1775, 1750, 625, 720, 6264, 20880, 33480, 25920, 7776, 5040, 56196, 250096, 571095, 708295, 453789, 117649, 40320, 554112, 3127040, 9433088, 16486400, 16744448, 9175040, 2097152, 362880, 5973264, 41229324, 156498804

Offset: 1

Peter Bala, Nov 16 2015

Related to A220883 and A251592.

			Triangle begins
...1
...2      2
...6     15       9
..24    104     144      64
.120    770    1775    1750     625
.720   6264   20880   33480   25920    7776
5040  56196  250096  571095  708295  453789  117649
...

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998

P. Bala, Fractional iteration of a series inversion operator

A000142 (column 0), A000169 (main diagonal), A006675 (column 1). Cf. A001700, A025174, A056856, A163456, A220883, A224274, A251592.

seq(seq(coeff(mul(n*t + k, k = 2 .. n), t, i), i = 0..n-1), n = 1..10);

E.g.f. (with constant term 1 included): A(x,t) = [ 1/x*Revert( x*(1 - x)^t ) ]^(1/t) = Sum_{n >= 0} 1/(n*t + 1)*binomial(n*t + n,n)*x^n = 1 + x + (2 + 2*t)*x^2/2! + (2 + 3*t)*(3 + 3*t)*x^3/3! + (2 + 4*t)*(3 + 4*t)*(4 + 4*t)*x^4/4! + ..., where Revert denotes the series reversion operator with respect to x.
In the notation of the Bala link, A(x,t) = I^t(1/(1 - x)) where I^t is a fractional inversion operator.
A(x,t) = B_(1+t)(x), where B_t(x) is the e.g.f. for A251592 and is the generalized binomial series of Lambert. See Graham et al., Section 5.4 and Section 7.5.
A(x,t)^m = Sum_{n >= 0} m/(n*t + m)*binomial(n*t + n + m - 1,n)*x^n = 1 + m*x + m*(2*t + m + 1)*x^2/2! + m*(3*t + m + 1)*(3*t + m + 2)*x^3/3! + m*(4*t + m + 1)*(4*t + m + 2)*(4*t + m + 3)*x^4/4! + ....
A(x,t)^t = 1 + t*x + t(1 + 3*t)*x^2/2! + t*(1 + 4*t)*(2 + 4*t)*x^3/3! + t*(1 + 5*t)*(2 + 5*t)*(3 + 5*t)*x^4/4! + ... is the e.g.f for A220883 with an extra constant term 1 and an extra factor of t included.
t*log( A(x,t) ) = t*x + t*(1 + 2*t)*x^2/2! + t*(1 + 3*t)*(2 + 3*t)*x^3/3! + t*(1 + 4*t)*(2 + 4*t)*(3 + 4*t)*x^4/4! + ... is the e.g.f for A056856.
For n = 1,2,3,..., the sequence [x^n] A(x,t)^n = [1, (2*t + 3), (3*t + 4)*(3*t + 5)/2!, (4*t + 5)*(4*t + 6)*(4*t + 7)/3!, ...]. This sequence has the following specializations:
t = 0: [1, 3, 10, 35, 126, ...] = A001700 (with different offset).
t = 1: [1, 5, 28, 165, 1001, ...] = A025174.
t = 2: [1, 7, 55, 455, 3876, ...] = A224274.
t = 3: [1, 9, 91, 969, 10626, ...] = A163456.

A369266 Expansion of (1/x) * Series_Reversion( x * (1-x) / (1+x^3)^2 ).

Original entry on oeis.org

1, 1, 2, 7, 24, 84, 313, 1209, 4769, 19166, 78253, 323570, 1352122, 5701467, 24229122, 103663575, 446163435, 1930390329, 8391341664, 36630504952, 160509484616, 705750073063, 3112865367660, 13769327908980, 61066953746400, 271488240652950, 1209671359828154

Offset: 0

Seiichi Manyama, Jan 18 2024

nonn

Cf. A071969, A369268.

Cf. A369267, A370248.

my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x)/(1+x^3)^2)/x)

a(n, s=3, t=2, u=1) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial((u+1)*(n+1)-s*k-2, n-s*k))/(n+1);

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} binomial(2*n+2,k) * binomial(2*n-3*k,n-3*k).

a(n) = (1/(n+1)) * [x^n] ( 1/(1-x) * (1+x^3)^2 )^(n+1). - Seiichi Manyama, Feb 14 2024

A377223 Coefficients of the series whose 12th power is 1/x * series_reversion(x*E_6(x)), where E_6(x) is the Eisenstein series of weight 6.

Original entry on oeis.org

1, 42, 34020, 39770808, 54603156174, 82058923220904, 130685055490645992, 216707827984305135744, 370213729923354622242084, 647073665508052293475274898, 1151627718366568095339000345192, 2079918757332503030219456972007720, 3802403760868562402170776739039126584, 7022808067106759130277006634854345528104

Offset: 0

Peter Bala, Nov 08 2024

Let R = 1 + x*Z[[x]] denote the set of integral power series with constant term equal to 1. Let P_n = {g^n, g in R}. If f belongs to P_n then the power series 1/x * series_reversion(x*f(x)) is also in P_n. Apply Bala, Theorem 1, Corollary 2.

Here we take f to be the Eisenstein series E_6. See A013973. It is known that the 12th root f^(1/12) has integer coefficients (Heninger et al.). See A109817. It follows that the present sequence is integral.

Peter Bala, Fractional iteration of a series inversion operator
N. Heninger, E. M. Rains, and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.
N. Heninger, E. M. Rains, and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.

Cf. A013973, A109817, A377221, A377222.

with(numtheory):
Order := 30:
E_6 := 1 - 504*add(sigma[5](n)*x^n, n = 1..30):
solve(series(x*E_6, x) = y, x):
seq(coeftayl(series((%/y)^(1/12), y), y = 0, n), n = 0..20);

G.f.: A(x) = the 12-fold iterate I^12( 1/E_6(x)^(1/12) ), where the operator I : R -> R is defined by I(f(x)) = 1/x * series_reversion(x/f(x)), showing that the g.f. A(x) is integral.

A369208 Expansion of (1/x) * Series_Reversion( x * (1-x)^2 / (1+x^2) ).

Original entry on oeis.org

1, 2, 8, 38, 200, 1122, 6576, 39790, 246672, 1558658, 10001592, 64997814, 426922392, 2829624514, 18901301984, 127115260894, 859978039840, 5848754717314, 39964745880552, 274231943135686, 1888891689752680, 13055393137141282, 90517646431869328

Offset: 0

Seiichi Manyama, Jan 18 2024

nonn

Cf. A218045, A369263.
Cf. A052709, A109081.
Cf. A370242.

my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x)^2/(1+x^2))/x)

a(n, s=2, t=1, u=2) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial((u+1)*(n+1)-s*k-2, n-s*k))/(n+1);

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(n+1,k) * binomial(3*n-2*k+1,n-2*k).

a(n) = (1/(n+1)) *[x^n] ( 1/(1-x)^2 * (1+x^2) )^(n+1). - Seiichi Manyama, Feb 14 2024

A369226 Expansion of (1/x) * Series_Reversion( x * (1-x) / (1+x^2)^2 ).

Original entry on oeis.org

1, 1, 4, 13, 53, 220, 968, 4373, 20271, 95705, 458904, 2228220, 10934524, 54143848, 270189008, 1357428997, 6860264323, 34853234867, 177900211204, 911867479717, 4691701977973, 24222505191984, 125448280976224, 651555603531308, 3392951906596708, 17711433386188300

Offset: 0

Seiichi Manyama, Jan 18 2024

nonn

Cf. A052709, A369262.

Cf. A218045, A370243.

my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x)/(1+x^2)^2)/x)

a(n, s=2, t=2, u=1) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial((u+1)*(n+1)-s*k-2, n-s*k))/(n+1);

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(2*n+2,k) * binomial(2*n-2*k,n-2*k).

a(n) = (1/(n+1)) * [x^n] ( 1/(1-x) * (1+x^2)^2 )^(n+1). - Seiichi Manyama, Feb 14 2024

A369298 Expansion of (1/x) * Series_Reversion( x * (1-x)^2 * (1-x^3)^2 ).

Original entry on oeis.org

1, 2, 7, 32, 163, 884, 5011, 29342, 176092, 1077384, 6695093, 42140930, 268108170, 1721372836, 11138994028, 72573587520, 475674650717, 3134297846792, 20750020222815, 137953554890508, 920667400056250, 6165565645765092, 41419898169301995

Offset: 0

Seiichi Manyama, Jan 18 2024

nonn

Cf. A369297, A369300.

Cf. A369296, A370275.

my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x)^2*(1-x^3)^2)/x)

a(n, s=3, t=2, u=2) = sum(k=0, n\s, binomial(t*(n+1)+k-1, k)*binomial((u+1)*(n+1)-s*k-2, n-s*k))/(n+1);

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/3)} binomial(2*n+k+1,k) * binomial(3*n-3*k+1,n-3*k).

a(n) = (1/(n+1)) * [x^n] 1/( (1-x)^2 * (1-x^3)^2 )^(n+1). - Seiichi Manyama, Feb 14 2024

A377221 Coefficients of the series whose 8th power is 1/x * series_reversion(x * E_4(x)), where E_4(x) is the Eisenstein series of weight 4.

Original entry on oeis.org

1, -30, 10980, -5822040, 3623245710, -2467207358280, 1779938570782440, -1336872265001920320, 1034337566576031632100, -818707881037376263396710, 659829780447854309255690280, -539628866179308154664183513160, 446708428717281359928910138018680, -373580804664955058627213489276760840

Offset: 0

Peter Bala, Nov 07 2024

Let R = 1 + x*Z[[x]] denote the set of integer power series with constant term equal to 1. Define the operator T: R -> R by T(f(x)) = 1/x * series_reversion(x*f(x)). Let P_n = {g^n, g in R}. It follows from Bala, Theorem 1, Corollary 2, that if f belongs to P_n then T(f) is also in P_n.

Here we take f to be the Eisenstein series E_4, the theta series of the E_8 lattice. See A004009. It is known that the 8th root E_4^(1/8) has integer coefficients (Heninger et al.). It follows that the present sequence is integral.

			Let F(x) = 1/(E_4(x))^(1/8) = 1 - 30*x + 3780*x^2 - 616440*x^3 + 111056910*x^4 - 21135698280*x^5 + ...
Then
I(F(x))   = 1 - 30*x + 4680*x^2 - 983640*x^3 + 234828510*x^4 - 60324330780*x^5 + ...
I^2(F(x)) = 1 - 30*x + 5580*x^2 - 1431840*x^3 + 422752110*x^4 - 135277163280*x^5 + ...
I^3(F(x)) = 1 - 30*x + 6480*x^2 - 1961040*x^3 + 687787710*x^4 - 262396695780*x^5 + ...
I^4(F(x)) = 1 - 30*x + 7380*x^2 - 2571240*x^3 + 1042895310*x^4 - 461122928280*x^5 + ...
I^5(F(x)) = 1 - 30*x + 8280*x^2 - 3262440*x^3 + 1501034910*x^4 - 753933360780*x^5 + ...
I^6(F(x)) = 1 - 30*x + 9180*x^2 - 4034640*x^3 + 2075166510*x^4 - 1166342993280*x^5 + ...
I^7(F(x)) = 1 - 30*x + 10080*x^2 - 4887840*x^3 + 2778250110*x^4 - 1726904325780*x^5 + ...
I^8(F(x)) = 1 - 30*x + 10980*x^2 - 5822040*x^3 + 3623245710*x^4 - 2467207358280*x^5 + ... = the g.f. A(x).

Peter Bala, Fractional iteration of a series inversion operator
N. Heninger, E. M. Rains, and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.
N. Heninger, E. M. Rains, and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.

Cf. A004009, A108091, A377220, A377223.

with(numtheory):
Order := 30:
E_4 := 1 + 240*add(sigma[3](n)*x^n, n = 1..30):
solve(series(x*E_4, x) = y, x):
seq(coeftayl(series((%/y)^(1/8), y), y = 0, n), n = 0..20);

G.f.: A(x) = the 8-fold iterate I^8( 1/(E_4(x))^(1/8) ), where I : R -> R denotes the operator I(f(x)) = 1/x * series_reversion(x/f(x)), showing that the g.f. A(x) is integral.

cp's OEIS Frontend

A369301 Expansion of (1/x) * Series_Reversion( x * (1-x)^3 * (1-x^3)^3 ).

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A220883 Triangle read by rows: row n gives coefficients of expansion of Product_{k = 1..n-1} ((n + 1)*x + k), starting with lowest power.

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A260687 Triangular array with n-th row giving coefficients of polynomial Product_{k = 2..n} (k + n*t) for n >= 1.

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A369266 Expansion of (1/x) * Series_Reversion( x * (1-x) / (1+x^3)^2 ).

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A377223 Coefficients of the series whose 12th power is 1/x * series_reversion(x*E_6(x)), where E_6(x) is the Eisenstein series of weight 6.

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A369208 Expansion of (1/x) * Series_Reversion( x * (1-x)^2 / (1+x^2) ).

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A369226 Expansion of (1/x) * Series_Reversion( x * (1-x) / (1+x^2)^2 ).

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A369298 Expansion of (1/x) * Series_Reversion( x * (1-x)^2 * (1-x^3)^2 ).

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