A054765 -id:A054765 - cp's OEIS frontend

A155728 INVERTi transform of A054765: (1, 3, 19, 160, 1744, ...).

1, 2, 14, 121, 1383, 19108, 309708, 5751027, 120357325, 2803145494, 71926499002, 2016492639229, 61338391284387, 2012321446421976, 70833707268623448, 2663117961930477847, 106515148705020928105, 4516063573152118802282, 202328834841437929100838

Offset: 1

Gary W. Adamson and Alexander R. Povolotsky, Jan 25 2009

This sequence convolved with A054765 prefaced with a 1: (1, 1, 3, 19, 160, ...) = (1, 3, 19, 160, ...), equivalent to row sums of triangle A155729 = A054765.

			We write (1, 3, 19, 160, ...) in reverse: (..., 19, 3, 1), top row.
Bottom row = (1, 2, ...), so that the format for a(3) = 14 becomes: ...3, 1 = A054765: (1, 3, 19, 160, ...). ..., 1, 2 for current format, take dot product = (3*1 + 1*2) = 5, then subtract from next term in A054765, getting (19 - 5) = 14. So a(3) = 14.
Continuing with analogous operations, we get (1, 2, 14, 121, 1383, ...).

Cf. A054765, A155729.

INVERTi transform of A054765 starting with offset 1: (1, 3, 19, 160, 1774, 23184, ...).

Corrected by R. J. Mathar, Apr 04 2012

More terms from Alois P. Heinz, Mar 31 2016

A054798 Let N(k) and D(k) be the sequences defined in A054765 and A012244; write N(k)* D(k+j ) - N(k+j)D(k) = (-1)^(k+1)(k!)^2*P(k) where P(k) is a polynomial in k of degree j-1; sequence gives coefficients of expansion of P(k) in powers of k for j=1,2,3,...

Original entry on oeis.org

1, 2, 3, 5, 20, 19, 12, 90, 214, 160, 29, 348, 1497, 2718, 1744, 70, 1225, 8236, 26453, 40336, 23184, 169, 4056, 39114, 193184, 512813, 689512, 364176, 408, 12852, 167884, 1174860, 4737628, 10955304, 13372072, 6598656, 985, 39400, 669078, 6282340, 35554929, 123708580, 257200712, 290478120, 135484416, 2378, 117711, 2519024, 30514946, 229958030, 1114357079, 3459179856, 6602445344, 6991966752, 3108695040

Offset: 1

N. J. A. Sloane, May 26 2000

			For j=1,2,3 the polynomials P(k) are 1, 3 + 2 k, 19 + 20 k + 5 k^2.
1;
2,3,
5,20,19,
12,90,214,160,
29,348,1497,2718,1744,
70,1225,8236,26453,40336,23184,
169,4056,39114,193184,512813,689512,364176,
408,12852,167884,1174860,4737628,10955304,13372072,6598656,
985,39400,669078,6282340,35554929,123708580,257200712,290478120,135484416,
2378,117711,2519024,30514946,229958030,1114357079,3459179856,6602445344,6991966752,3108695040,

K. S. Brown, Integer Sequences Related To Pi

Sign corrected by R. J. Mathar, Jul 13 2013

A012244 a(n+2) = (2n+3)a(n+1) + (n+1)^2a(n), a(0) = 1, a(1) = 1.

Original entry on oeis.org

1, 1, 4, 24, 204, 2220, 29520, 463680, 8401680, 172504080, 3958113600, 100370793600, 2787459998400, 84139894238400, 2742857884166400, 96034297911552000, 3594206259195552000, 143193586818810528000, 6050501147565883008000, 270263264589232282368000

Offset: 0

N. J. A. Sloane

a(n) is the number of n-letter words from an n-letter alphabet such that no letter appears more than twice. - Paul Boddington, Nov 17 2003

Alois P. Heinz, Table of n, a(n) for n = 0..393
K. S. Brown, Integer Sequences Related To Pi
D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions, arXiv:math/0501052 [math.CA], 2005.

Cf. A089975, A190392, A054765, A054766.

f := proc(n) option remember; if n <= 1 then 1 else (2*n-1)*f(n-1) +(n-1)^2*f(n-2); fi; end;

Range[0,20]! CoefficientList[Series[1/(1-2x-x^2)^(1/2), {x,0,20}], x]  (* Geoffrey Critzer, Dec 07 2011 *)

{a(n)=local(X=x+x^2*O(x^n));(n+1)!*polcoeff(serreverse(cos(X)+sin(X)-1),n+1)} \\ Paul D. Hanna, Aug 08 2012

E.g.f.: A(x) = (1 - 2*x - x^2)^(-1/2). - Paul Boddington, Nov 17 2003
a(n) = n!/2^n*A006139(n) = n!*Sum_{k=floor(n/2)..n} 2^(k-n)*C(n, k)*C(k, n-k). Sum_{n>=0} a(n)*x^n/n!^2 = exp(x)*BesselI(0, sqrt(2)*x). a(n) is the central coefficient of n!*(1+x+x^2/2)^n. - Vladeta Jovovic, Mar 22 2004
From Peter Bala, Aug 25 2011: (Start)
The function B(x) := int {t=0..x} A(t), obtained by integrating the generating function A(x), satisfies the autonomous differential equation d/dx(B(x)) = 1/(cos(B(x))-sin(B(x))). Compare with A190392.
Thus B(x), and hence A(x), can be found by inverting the function int {t=0..x} (cos(t)-sin(t)). By applying [Dominici, Theorem 4.1] the result can be expressed as
A(x) = 1 + sum {n>=1} D^n[1/(cos(t)-sin(t))](0)*x^n/n!, where the nested derivative D^n[f](x) of a function f(x) is defined recursively as D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0. Thus a(n) = D^n[1/(cos(t)-sin(t))](0). (End)
E.g.f. at offset 1: Series_Reversion(cos(x) + sin(x) - 1). - Paul D. Hanna, Aug 08 2012
a(n) ~ (1+sqrt(2))^(n+1/2) * n^n / (2^(1/4) * exp(n)). - Vaclav Kotesovec, Feb 18 2017

A054766 a(n+2) = (2n + 3)a(n+1) + (n + 1)^2*a(n), a(0) = 1, a(1) = 0.

Original entry on oeis.org

1, 0, 1, 5, 44, 476, 6336, 99504, 1803024, 37019664, 849418560, 21539756160, 598194037440, 18056575823040, 588622339549440, 20609136708249600, 771323264354361600, 30729606721005830400, 1298448658633614566400

Offset: 0

N. J. A. Sloane, May 26 2000

Numerators of the convergents of the generalized continued fraction expansion 4/Pi - 1 = [0; 1/3, 4/5, 9/7,..., n^2/(2*n + 1),...] = 1/(3 + 4/(5 + 9/(7 + ...))). The first 4 convergents are 1/3, 5/19, 44/160 and 476/1744.

Indranil Ghosh, Table of n, a(n) for n = 0..392
K. S. Brown, Integer Sequences Related To Pi

Cf. A012244, A054765.

RecurrenceTable[{a[n+2] == (2*n+3)*a[n+1] + (n+1)^2*a[n], a[0] == 1, a[1] == 0}, a, {n, 0, 25}] (* Vaclav Kotesovec, Feb 18 2017 *)
t={1,0};Do[AppendTo[t,(2(n-2)+3)*t[[-1]]+(n-1)^2*t[[-2]]],{n,2,18}];t (* Indranil Ghosh, Feb 25 2017 *)

a(n) ~ (1 - Pi/4) * (1 + sqrt(2))^(n + 1/2) * n^n / (2^(1/4) * exp(n)). - Vaclav Kotesovec, Feb 18 2017

More terms from James Sellers, May 27 2000
Definition expanded by Al Hakanson (hawkuu(AT)gmail.com), Dec 01 2008
Keyword frac added by Michel Marcus, Feb 25 2017

A155729 Triangle read by rows, M * Q; M = (T(n,k) = A155728(n-k+1)); Q = (A155728 * 0^(n-k)).

Original entry on oeis.org

1, 2, 1, 14, 2, 13, 121, 14, 6, 19, 1383, 121, 42, 38, 160, 19108, 1383, 363, 266, 320, 1744, 19108, 1383, 363, 266, 320, 1744, 309708, 19108, 4149, 2299, 2240, 3488, 23184, 2751027, 309708, 57324, 26277, 19360, 24416, 46368, 364176

Offset: 1

Gary W. Adamson & Alexander R. Povolotsky, Jan 25 2009

Row sums = A054765 starting with offset 1: (1, 3, 19, 160, 1744,...).
As a property of eigentriangles, sum of n-th row terms = rightmost term of next row.
A155728 = INVERTi transform of A054765: (1, 3, 19, 160, 1744,...).

			First few rows of the triangle =
1;
2, 1;
14, 2, 3;
121, 14, 6, 19;
1383, 121, 42, 38, 160;
19108, 1383, 363, 266, 320, 1744;
309708, 19108, 4149, 2299, 2240, 3488, 23184;
2751027, 309708, 57324, 26277, 19360, 24416, 46368, 364176;
...
Example: Row 4 = = (121, 14, 6, 19) termwise products of (121, 14, 2, 1) and (1, 1, 3, 19).

Cf. A054765, A155728

M = an infinite lower triangular matrix with A155728 in every column:
(1, 2, 14, 121, 1383, 19108, 309708,...).
Q = an infinite lower triangular matrix with A054765 prefaced with a 1:
(1, 1, 3, 19, 160, 1744,...) as the main diagonal and the rest zeros.

A345125 Numerator of 4/(1 + 1^2/(3 + 2^2/(5 + 3^2/(7 + ... + (n-1)^2/(2*n-1) )))).

Original entry on oeis.org

0, 4, 3, 19, 160, 1744, 644, 2529, 183296, 3763456, 4317632, 54743776, 1013549056, 30594128896, 35618973952, 10392576224, 3111643512832, 123968232030208, 48501417558016, 1083228572868608, 4080033616887808, 188557135970304, 3781715948011520

Offset: 0

Seiichi Manyama, Sep 16 2021

The limit of a(n)/A345259(n) is Pi.

			4/(1 + 1^2/(3 + 2^2/5)) = 19/6. So a(3) = 19.
0, 4, 3, 19/6, 160/51, 1744/555, 644/205, 2529/805, 183296/58345, ...

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Frits Beukers, A rational approach to Pi, Nieuw Archief voor de Wiskunde, December 2000, pp. 372-379.

Cf. A012244, A054765, A054766, A345259 (denominator).

nmax = 25; Join[{0}, Table[4/(1 + ContinuedFractionK[j^2, (2*j + 1), {j, 1, k}]), {k, 0, nmax}] // Numerator] (* Vaclav Kotesovec, Sep 16 2021 *)

a(n) = my(x=0); forstep(i=n, 2, -1, x = (i-1)^2/((2*i-1)+x);); if (n, numerator(4/(1+x)), numerator(x)); \\ Michel Marcus, Sep 16 2021

a(n)/A345259(n) = 4 * A054765(n)/A012244(n).

A345259 Denominator of 4/(1 + 1^2/(3 + 2^2/(5 + 3^2/(7 + ... + (n-1)^2/(2*n-1) )))).

Original entry on oeis.org

1, 1, 1, 6, 51, 555, 205, 805, 58345, 1197945, 1374345, 17425485, 322622685, 9738413685, 11337871545, 3308059755, 990466892415, 39460313827935, 15438480702645, 344802363740835, 1298715036217599, 60019600489849, 1203757572990973

Offset: 0

Seiichi Manyama, Sep 16 2021

The limit of A345125(n)/a(n) is Pi.

			4/(1 + 1^2/(3 + 2^2/5)) = 19/6. So a(3) = 6.
0, 4, 3, 19/6, 160/51, 1744/555, 644/205, 2529/805, 183296/58345, ...

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Frits Beukers, A rational approach to Pi, Nieuw Archief voor de Wiskunde, December 2000, pp. 372-379.

Cf. A012244, A054765, A054766, A345125 (numerator).

nmax = 25; Join[{1}, Table[4/(1 + ContinuedFractionK[j^2, (2*j + 1), {j, 1, k}]), {k, 0, nmax}] // Denominator] (* Vaclav Kotesovec, Sep 16 2021 *)

a(n) = my(x=0); forstep(i=n, 2, -1, x = (i-1)^2/((2*i-1)+x);); if (n, denominator(4/(1+x)), denominator(x)); \\ Michel Marcus, Sep 16 2021

A345125(n)/a(n) = 4 * A054765(n)/A012244(n).

cp's OEIS Frontend

A155728 INVERTi transform of A054765: (1, 3, 19, 160, 1744, ...).

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A054798 Let N(k) and D(k) be the sequences defined in A054765 and A012244; write N(k)* D(k+j ) - N(k+j)*D(k) = (-1)^(k+1)*(k!)^2*P(k) where P(k) is a polynomial in k of degree j-1; sequence gives coefficients of expansion of P(k) in powers of k for j=1,2,3,...

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A012244 a(n+2) = (2n+3)*a(n+1) + (n+1)^2*a(n), a(0) = 1, a(1) = 1.

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A054766 a(n+2) = (2*n + 3)*a(n+1) + (n + 1)^2*a(n), a(0) = 1, a(1) = 0.

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A155729 Triangle read by rows, M * Q; M = (T(n,k) = A155728(n-k+1)); Q = (A155728 * 0^(n-k)).

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A345125 Numerator of 4/(1 + 1^2/(3 + 2^2/(5 + 3^2/(7 + ... + (n-1)^2/(2*n-1) )))).

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A345259 Denominator of 4/(1 + 1^2/(3 + 2^2/(5 + 3^2/(7 + ... + (n-1)^2/(2*n-1) )))).

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A054798 Let N(k) and D(k) be the sequences defined in A054765 and A012244; write N(k)* D(k+j ) - N(k+j)D(k) = (-1)^(k+1)(k!)^2*P(k) where P(k) is a polynomial in k of degree j-1; sequence gives coefficients of expansion of P(k) in powers of k for j=1,2,3,...

A012244 a(n+2) = (2n+3)a(n+1) + (n+1)^2a(n), a(0) = 1, a(1) = 1.

A054766 a(n+2) = (2n + 3)a(n+1) + (n + 1)^2*a(n), a(0) = 1, a(1) = 0.