A135148 -id:A135148 - cp's OEIS frontend

A135147 A binomial recursion : a(n) = p(n) (see formula).

1, 4, 25, 188, 1671, 17190, 201125, 2638984, 38390179, 613363466, 10678267425, 201215691660, 4080450217247, 88609322165902, 2051573162708125, 50450534991347216, 1313219083705400475, 36072797094375866898, 1042811362801447763225, 31647646914322017237652, 1006032342980535954429463

Offset: 1

Benoit Cloitre, Nov 20 2007

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 1..400

Cf. A135148, A135149, A135150, A135074, A135075.

z[1]:= x; z[n_] := 1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n - 1}]; Table[ Coefficient[z[n], x], {n, 1, 20}] (* G. C. Greubel, Sep 28 2016 *)
z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[CoefficientList[Series[(1 - E^x)*(-1 - E^x + 2*x)/(2*(2 - E^x)^2), {x, 0, nmax}], x] * Range[0, nmax]!] (* Vaclav Kotesovec, Nov 25 2020 *)

r=1; s=2; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (2 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).
Limit_{n->oo} p(n)/q(n) = (3 - 2*log(2))/(2*log(2) - 1) = 4.17739889912417966161076...
a(n) ~ (3 - 2*log(2)) * n * n! / (8 * log(2)^(n+2)). - Vaclav Kotesovec, Nov 25 2020
E.g.f.: (1 - exp(x)) * (2*x - 1 - exp(x)) / (2*(2 - exp(x))^2). - Vaclav Kotesovec, Nov 25 2020

More terms from Amiram Eldar, Nov 25 2020

A135149 A binomial recursion: a(n) = p(n) (see formula).

Original entry on oeis.org

1, 5, 36, 304, 2973, 33156, 415962, 5803307, 89172846, 1496858836, 27258427263, 535299208890, 11277600621714, 253741796354921, 6072776118043704, 154050364873902628, 4128986249628307077, 116598919802471049936, 3460199566405679555310, 107659401911343963741971

Offset: 1

Benoit Cloitre, Nov 20 2007

nonn

Benoit Cloitre, Binomial recursions, Pi and log2, in preparation 2007.

Vaclav Kotesovec, Table of n, a(n) for n = 1..400

Cf. A135147, A135148, A135150, A135074, A135075.

z[1] := x; z[n_] := 1 + Sum[(3 + Binomial[n, k])*z[k], {k, 1, n - 1}]; Table[ Coefficient[z[n], x, 1], {n, 1, 10}] (* G. C. Greubel, Sep 28 2016 *)
z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(3 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[Simplify[CoefficientList[Series[E^(5*x/2)*(60*ArcSin[E^(x/2) / Sqrt[2]] - 22 - 15*Pi) / (150*(2 - E^x)^(5/2)) + (24*(-3 + 5*x) - 8*E^x*(-4 + 15*x) + 2*E^(2*x)*(31 + 15*x))/(150*(2 - E^x)^2), {x, 0, nmax}], x] * Range[0, nmax]!]] (* Vaclav Kotesovec, Nov 25 2020 *)

r=1; s=3; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (3 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).
Limit_{n->oo} p(n)/q(n) = (15*Pi - 22)/(52 - 15*Pi) = 5.1524450418835554775446337...
a(n) ~ 2 * (15*Pi - 22) * n^(3/2) * n! / (225 * sqrt(Pi) * log(2)^(n + 5/2)). - Vaclav Kotesovec, Nov 25 2020
E.g.f.: exp(5*x/2) * (60*arcsin(exp(x/2)/sqrt(2)) - 22 - 15*Pi) / (150*(2 - exp(x))^(5/2)) + (24*(-3 + 5*x) - 8*exp(x)*(-4 + 15*x) + 2*exp(2*x)*(31 + 15*x)) / (150*(2 - exp(x))^2). - Vaclav Kotesovec, Nov 25 2020

More terms from Vaclav Kotesovec, Nov 25 2020

A135150 A binomial recursion: a(n) = q(n) (see formula).

Original entry on oeis.org

0, 1, 7, 59, 577, 6435, 80731, 1126321, 17306899, 290514275, 5290386805, 103892269503, 2188786203451, 49246871008285, 1178620260610039, 29898497436003155, 801364442718809233, 22629823094599476315, 671564575318740405283, 20894818098241648524577, 680161672262047334987995

Offset: 1

Benoit Cloitre, Nov 20 2007

nonn

Benoit Cloitre, Binomial recursions, Pi and log2, in preparation 2007.

Vaclav Kotesovec, Table of n, a(n) for n = 1..400

Cf. A135147, A135148, A135149, A135074, A135075.

z[1] := x; z[n_] := 1 + Sum[(3 + Binomial[n, k])*z[k], {k, 1, n - 1}]; Table[
Coefficient[z[n], x, 0], {n, 1, 10}] (* G. C. Greubel, Sep 28 2016 *)
z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(3 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x, 0], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[CoefficientList[Series[((-24 + 44*E^x - 46*E^(2*x))/(2 - E^x)^2 - 15*x + E^(5*x/2)*(52 + 15*Pi - 60*ArcSin[E^(x/2)/Sqrt[2]])/(2*(2 - E^x)^(5/2)))/75, {x, 0, nmax}], x] * Range[0, nmax]!] (* Vaclav Kotesovec, Nov 25 2020 *)

r=1; s=3; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (3 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).
Limit_{n->oo} p(n)/q(n) = (15*Pi - 22)/(52 - 15*Pi) = 5.1524450418835554775446337...
a(n) ~ 2 * (52 - 15*Pi) * n^(3/2) * n! / (225 * sqrt(Pi) * log(2)^(n + 5/2)). - Vaclav Kotesovec, Nov 25 2020
E.g.f.: ((-24 + 44*exp(x) - 46*exp(2*x))/(2 - exp(x))^2 - 15*x + exp(5*x/2)*(52 + 15*Pi - 60*arcsin(exp(x/2)/sqrt(2))) /(2*(2 - exp(x))^(5/2)))/75. - Vaclav Kotesovec, Nov 25 2020

More terms from Vaclav Kotesovec, Nov 25 2020

A142980 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1) + (n + 1)^2a(n).

Original entry on oeis.org

1, 5, 29, 190, 1414, 11820, 110004, 1129200, 12686256, 154896480, 2043108000, 28958014080, 438997622400, 7088892491520, 121487996448000, 2202440792832000, 42113131054848000, 847071044402688000, 17880009683784192000, 395192695448291328000, 9127967350755133440000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 2 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.

Seiichi Manyama, Table of n, a(n) for n = 1..448

Cf. A024167, A135148, A142979, A142981, A142982.

p := n -> 2*n^2+ 2*n+1: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)

Module[{a, n}, RecurrenceTable[{a[n+2] == 5*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 5}, a, {n, 25}]] (* Paolo Xausa, Dec 12 2024 *)

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n^2 + 2*n + 1 = A001844(n) is the Ehrhart polynomial for the 2-dimensional cross polytope (a square).
Recurrence: a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n+1)^2*a(n).
The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 5, b(2) = 26.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + (n-1)^2/5)))), for n >= 2.
The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + n^2/(5 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4*k^4 + 1)) = log(2) - (1 - 1/2); the final equality is a result of Glaisher.
Thus a(n) ~ c*n^2*n! as n -> oo, where c = 2*log(2) - 1.
From Peter Bala, Dec 09 2024: (Start)
E.g.f.: A(x) = ((1 + x)^2 *log(1 + x) - 2*x^2)/(1 - x)^3 satisfies the differential equation 1 + (x + 5)*A(x) + (x^2 - 1)*A(x)' with A(0) = 0.
Sum_{k = 1..n} Stirling2(n, k)*a(k) = A135148(n+1). (End)

A132436 A binomial recursion: a(n) = p(n) (see comment).

Original entry on oeis.org

1, 1, 4, 20, 129, 1020, 9542, 103063, 1262134, 17279744, 261531315, 4335950346, 78146040374, 1521220672933, 31808447321848, 711019048106744, 16919695824732249, 427046133330613512, 11394750238551713066, 320486422239301377007, 9476411014096567341034

Offset: 1

Benoit Cloitre, Nov 20 2007

nonn

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (-1 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).

Vaclav Kotesovec, Table of n, a(n) for n = 1..400

Cf. A135147, A135148, A135149, A135150, A135074, A135075.

z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(-1 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[Simplify[CoefficientList[Series[1 + x + E^(x/2)*(2*ArcSin[E^(x/2)/Sqrt[2]] - 1 - Pi/2)/Sqrt[2 - E^x], {x, 0, nmax}], x] * Range[0, nmax]!]] (* Vaclav Kotesovec, Nov 25 2020 *)

r=1; s=-1; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Limit_{n->oo} p(n)/q(n) = (Pi-2)/(4-Pi) = 1.329896183162743847239353...
From Vaclav Kotesovec, Nov 25 2020: (Start)
a(n) ~ (Pi - 2) * n! / (2*sqrt(Pi*n) * log(2)^(n + 1/2)).
a(n) ~ (Pi - 2) * n^n / (sqrt(2) * exp(n) * log(2)^(n + 1/2)).
E.g.f.: 1 + x + exp(x/2)*(2*arcsin(exp(x/2)/sqrt(2)) - 1 - Pi/2) / sqrt(2 - exp(x)).
(End)

A132437 A binomial recursion: a(n) = q(n) (see comment).

Original entry on oeis.org

0, 1, 3, 15, 97, 767, 7175, 77497, 949047, 12993303, 196655437, 3260367539, 58761008087, 1143864229549, 23917992791139, 534642521054391, 12722568903456817, 321112383611040455, 8568150193087139231, 240986045600284560553, 7125677277725450247087

Offset: 1

Benoit Cloitre, Nov 20 2007

nonn

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (-1 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).

Vaclav Kotesovec, Table of n, a(n) for n = 1..400

Cf. A135147, A135148, A135149, A135150, A135074, A135075.

z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(-1 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x, 0], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
Rest[CoefficientList[Series[-2 - x + E^(x/2)*((4 + Pi)/2 - 2*ArcSin[E^(x/2) / Sqrt[2]]) / Sqrt[2 - E^x], {x, 0, 20}], x] * Range[0, 20]!] (* Vaclav Kotesovec, Nov 25 2020 *)

r=1; s=-1; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Limit_{n->oo} p(n)/q(n) = (Pi-2)/(4-Pi) = 1.329896183162743847239353...
From Vaclav Kotesovec, Nov 25 2020: (Start)
E.g.f.: -2-x + exp(x/2)*((4+Pi)/2 - 2*arcsin(exp(x/2)/sqrt(2))) / sqrt(2-exp(x)).
a(n) ~ (4 - Pi) * n! / (2*sqrt(Pi*n) * log(2)^(n + 1/2)).
a(n) ~ (4 - Pi) * n^n / (sqrt(2) * exp(n) * log(2)^(n + 1/2)). (End)

A298673 Inverse matrix of A135494.

Original entry on oeis.org

1, 1, 1, 4, 3, 1, 26, 19, 6, 1, 236, 170, 55, 10, 1, 2752, 1966, 645, 125, 15, 1, 39208, 27860, 9226, 1855, 245, 21, 1, 660032, 467244, 155764, 32081, 4480, 434, 28, 1, 12818912, 9049584, 3031876, 635124, 92001, 9576, 714, 36, 1, 282137824, 198754016, 66845340, 14180440, 2108085, 230097, 18690, 1110, 45, 1

Offset: 1

Tom Copeland, Jan 24 2018

Since this is the inverse matrix of A135494 with row polynomials q_n(t), first introduced in that entry by R. J. Mathar, and the row polynomials p_n(t) of this entry are a binomial Sheffer polynomial sequence, the row polynomials of the inverse pair are umbral compositional inverses, i.e., p_n(q.(t)) = q_n(p.(t)) = t^n. For example, p_3(q.(t)) = 4q_1(t) + 3q_2(t) + q_3(t) = 4t + 3(-t + t^2) + (-t -3t^2 +t^3) = t^3. In addition, both sequences possess the umbral convolution property (p.x) + p.(y))^n = p_n(x+y) with p_0(t) = 1.

This is the inverse of the Bell matrix generated by A153881; for the definition of the Bell matrix see the link. - Peter Luschny, Jan 26 2018

			Matrix begins as
     1;
     1;    1;
     4,    3,    1;
    26,   19,    6,    1;
   236,  170,   55,   10,    1;
  2752, 1966,  645,  125,   15,    1;

Peter Luschny, The Bell transform
Andrew Elvey Price and Alan D. Sokal, Phylogenetic trees, augmented perfect matchings, and a Thron-type continued fraction (T-fraction) for the Ward polynomials, arXiv:2001.01468 [math.CO], 2020.

Cf. A000311, A000169, A000670, A006351, A135494.

# The function BellMatrix is defined in A264428. Adds (1,0,0,0, ..) as column 0.
BellMatrix(n -> `if`(n=0, 1, -1), 9): MatrixInverse(%); # Peter Luschny, Jan 26 2018

BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
B = BellMatrix[Function[n, If[n == 0, 1, -1]], rows = 12] // Inverse;
Table[B[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 28 2018, after Peter Luschny *)

E.g.f.: e^[p.(t)x] = e^[t*h(x)] = exp[t*[(x-1)/2 + T{ (1/2) * exp[(x-1)/2] }], where T is the tree function of A000169 related to the Lambert function. h(x) = sum(j=1,...) A000311(j) * x^j / j! = exp[xp.'(0)], so the first column of this entry's matrix is A000311(n) for n > 0 and the second column of the full matrix for p_n(t) to n >= 0. The compositional inverse of h(x) is h^(-1)(x) = 1 + 2x - e^x.
The lowering operator is L = h^(-1)(D) = 1 + 2D - e^D with D = d/dt, i.e., L p_n(t) = n * p_(n-1)(t). For example, L p_3(t) = (D - D^2! - D^3/3! - ...) (4t + 6t^ + t^3) = 3 (t + t^2) = 3 p_2(t).
The raising operator is R = t * 1/[d[h^(-1)(D)]/dD] = t * 1/[2 - e^D)] = t (1 + D + 3D^2/2! + 13D^3/3! + ...). The coefficients of R are A000670. For example, R p_2(t) = t (1 + D + 3D^2/2!  + ...)  (t + t^2) = 4t + 3t^2 + t^3   = p_3(t).
The row sums are A006351, or essentially 2*A000311.
Conjectures from Mikhail Kurkov, Mar 01 2025: (Start)
T(n,k) = Sum_{j=0..n-k} binomial(n+j-1, k-1)*A269939(n-k, j) for 1 <= k <= n.
T(n,k) = A(n-1,k,0) for n > 0, k > 0 where A(n,k,q) = A(n-1,k,q+1) + 2*(q+1)!*Sum_{j=0..q} A(n-1,k,j)/j! for n >= 0, k > 0, q >= 0 with A(0,k,q) = Stirling1(q+1,k) for k > 0, q >= 0 (see A379458). In other words, T(n,k) = Sum_{j=0}^{n-1} A379460(n-j-1,j)*Stirling1(j+1,k) for n > 0, k > 0.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} b(j-1)*binomial(-k,j)*T(n,k+j-1)*(-1)^j for 1 <= k < n with T(n,n) = 1 where b(n) = 1 + 4*Sum_{i=1..n} A135148(i).
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} c(j-1)*binomial(n,j)*T(n-j+1,k) for 1 <= k < n with T(n,n) = 1 where c(n) = A000311(n+1) + (n-1)*A000311(n). (End)

cp's OEIS Frontend

A135147 A binomial recursion : a(n) = p(n) (see formula).

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A135149 A binomial recursion: a(n) = p(n) (see formula).

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A135150 A binomial recursion: a(n) = q(n) (see formula).

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A142980 a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n + 1)^2*a(n).

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A132436 A binomial recursion: a(n) = p(n) (see comment).

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A132437 A binomial recursion: a(n) = q(n) (see comment).

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A298673 Inverse matrix of A135494.

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A142980 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1) + (n + 1)^2a(n).