A024199 -id:A024199 - cp's OEIS frontend

A024217 a(n) = ( Product {k = 1..n} 3k - 2 ) ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 2) ).

1, 3, 25, 222, 3166, 47016, 951544, 19827408, 520029520, 13952218560, 449559799360, 14756761434240, 563961412362880, 21893890640563200, 968019931702297600, 43385863589508249600, 2178487766250586470400, 110704921777161066700800, 6222745685273069016064000

Offset: 1

Clark Kimberling

Original name was: s(1)*s(2)*...*s(n)(1/s(1) - 1/s(2) + ... + c/s(n)), where c=(-1)^(n+1) and s(k) = 3k-2 for k = 1,2,3,...

L. Euler, De fractionibus continuis observationes, The Euler Archive, Index Number 123, Section 5

Cf. A007559, A024199, A024383, A024396, A113476.

I:=[1,3]; [n le 2 select I[n] else 3*Self(n-1)+(3*n-5)^2*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2015

a[1] := 1: a[2] := 3: for n from 3 to 20 do a[n] := 3*a[n-1]+(3*n-5)^2*a[n-2] end do: seq(a[n], n = 1 .. 20); # Peter Bala, Feb 20 2015

Table[Product[3*k-2,{k,1,n}] * Sum[(-1)^(k+1)/(3*k-2),{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Feb 21 2015 *)

From Peter Bala, Feb 20 2015: (Start)
a(n) = A007559(n) * Sum {k = 1..n} (-1)^(k+1)/(3*k - 2).
Recurrence: a(n+1) = 3*a(n) + (3*n - 2)^2*a(n-1) with a(1) = 1 and a(2) = 3.
The triple factorial numbers A007559 also satisfy this second-order recurrence equation. This leads to the continued fraction representation a(n)/A007559(n) = 1/(1 + 1^2/(3 + 4^2/(3 + 7^2/(3 + ... + (3*n - 2)^2/(3 ))))).
Taking the limit as n -> infinity gives the generalized continued fraction: Sum {k >= 1} (-1)^(k+1)/(3*k - 2) = 1/(1 + 1^2/(3 + 4^2/(3 + 7^2/(3 + 10^2/(3 +  ... ))))) due to Euler.
The alternating sum has the value 1/3*( log(2) + Pi/sqrt(3) ) = A113476. Cf. A024396. (End)
a(n) ~ sqrt(2*Pi) * (sqrt(3)*Pi + 3*log(2)) * 3^(n-2) * n^(n-1/6) / (GAMMA(1/3) * exp(n)). - Vaclav Kotesovec, Feb 21 2015

New name from Peter Bala, Feb 20 2015

a(18)-a(19) from Vincenzo Librandi, Feb 21 2015

A135457 a(n) = (2n-1)!! * Sum_{k=0..n-2}(-1)^k/(2k+1).

Original entry on oeis.org

0, 3, 10, 91, 684, 8679, 100542, 1664055, 25991640, 532354635, 10455799410, 255542155155, 6044821114500, 171748491958575, 4751436512960550, 153911731348760175, 4874807783839316400, 177334729873063945875

Offset: 1

Benoit Cloitre, Dec 15 2007

nonn

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Peter Luschny, Is the Gamma function misdefined? Hadamard versus Euler - Who found the better Gamma function?

Cf. A167576 and A024199.

I:=[0,3,10]; [n le 3 select I[n] else 4*Self(n-1)+(4*n^2-12*n+1)*Self(n-2)-(8*n^2-48*n+70)*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jul 17 2015

a := n -> (2^(n-2)*GAMMA(n+1/2)*((-1)^n*(Psi(n/2+1/4)-Psi(n/2-1/4))+Pi))/sqrt(Pi);
seq(a(n), n=1..18); # Peter Luschny, Jul 18 2015

FullSimplify[Table[(2^(n-2)*(n-1/2)!*(Pi+2*(-1)^n*LerchPhi[-1,1,n-1/2]))/Sqrt[Pi],{n,1,20}]] (* Vaclav Kotesovec, Oct 11 2013 *)

a(n)=round((-1/4)*prod(i=1,n,2*i-3)*(Pi*(2*n-1)+2*(-1)^n*sum(k=0,1500,1.*k!/prod(i=1,k,(2*i+2*n-1)))))

a(n) = (-1/4)(Product_{i=1..n}(2i-3))((2n-1)Pi + 2(-1)^n*Sum_{k>=0}k!/ Product_{j=1..k}(2j+2n-1)). - Benoit Cloitre, Dec 15 2007
a(n+3) = 4*a(n+2) + (4n^2+12n+1)*a(n+1) - (8n^2-2)*a(n) with a(1)=0, a(2)=3, a(3)=10. - Benoit Cloitre, Dec 15 2007
a(n) ~ Pi * 2^(n-3/2) * n^n / exp(n). - Vaclav Kotesovec, Oct 11 2013
a(n+1) = (2n+1)*(a(n) - (-1)^n (2n-3)!!) with a(1)=0. - Cyril Damamme, Jul 16 2015
a(n) = (2^(n-2)*Gamma(n+1/2)*((-1)^n*(Psi(n/2+1/4)-Psi(n/2-1/4))+Pi))/sqrt(Pi). - Peter Luschny, Jul 18 2015
a(n) = A167576(n) - A024199(n). - Cyril Damamme, Jul 22 2015

Definition replaced by a simplified one by Cyril Damamme, Jul 18 2015

A142969 Numerators of approximants of a continued fraction for 4/Pi - 1 = (4 - Pi)/Pi.

Original entry on oeis.org

1, 2, 29, 52, 887, 8066, 11069, 143128, 3485197, 2792362, 78773861, 326941444, 1166735057, 28815727078, 1038855637093, 902109848368, 1031041592023, 33635927876926, 37917122954701, 1387635433109516, 66513954553071413, 59972573887236398, 3113073102662686381

Offset: 1

Wolfdieter Lang, Sep 15 2008

Denominators are A007509(n), n >= 1.
This results from William Brouncker's continued fraction for 4/Pi without the leading 1.
William Brouncker's result appears in John Wallis's "Arithmetica infinitorum" from 1655.

			Approximants a(n)/A007509(n): 1/2, 2/13, 29/76, 52/263, 887/2578, 8066/36979, ...

C. Brezinski, History of Continued Fractions and Padé approximants, Springer, 1991, ch. 3.

Wolfdieter Lang, Rationals and more.

a(n) = numerator(C(n)) with C(n) the n-th approximant to the continued fraction (1^2)(2+(3^2)/(2+(5^2)/(2+...

C(n) = Sum_{k=1..n} (-1)^(k+1)*(Product_{j=1..k} (2*k-1))^2/(q(k)*q(k-1)), with q(n) = A024199(n+1). Proof with Euler's conversion of continued fractions to alternating series. For this conversion see, e.g., the Brezinski reference, p. 98.

A254795 Numerators of the convergents of the generalized continued fraction 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))).

Original entry on oeis.org

2, 9, 54, 441, 4410, 53361, 747054, 12006225, 216112050, 4334247225, 95353438950, 2292816782025, 59613236332650, 1671463434096225, 50143903022886750, 1606276360166472225, 54613396245660055650, 1967688541203928475625, 74772164565749282073750

Offset: 0

Peter Bala, Feb 23 2015

Brouncker gave the generalized continued fraction expansion 4/Pi = 1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + ... ))). The sequence of convergents begins [1/1, 3/2, 15/13, 105/76, ... ]. The numerators of the convergents are in A001147, the denominators in A024199.

In extending Brouckner's result, Osler showed that 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))) = L^2/Pi, where L is the lemniscate constant A062539. The sequence of convergents to Osler's continued fraction begins [2/1, 9/4, 54/25, 441/200, 4410/2025, ...]. Here we list the (unreduced) numerators of these convergents. See A254796 for the sequence of denominators. See A254794 for the decimal expansion of L^2/Pi.

T. J. Osler, The missing fractions in Brouncker’s sequence of continued fractions for Pi

Cf. A008545, A001147, A024199, A142970, A254794, A254796.

a[0] := 2: a[1] := 9:
for n from 2 to 18 do a[n] := 4*a[n-1] + (2*n-1)^2*a[n-2] end do:
seq(a[n], n = 0 .. 18);

a(2*n-1) = ( A008545(n) )^2 = ( Product {k = 0..n-1} 4*k + 3 )^2.
a(2*n) = (4*n + 2)*( A008545(n) )^2 = (4*n + 2)*( Product {k = 0..n-1} 4*k + 3 )^2.
a(n) = 4*a(n-1) + (2*n - 1)^2*a(n-2) with a(0) = 2, a(1) = 9.
a(2*n) = (4*n + 2)*a(2*n-1); a(2*n+1) = (4*n + 4)*a(2*n) + a(2*n-1).

A024383 a(n) = s(1)s(2)...s(n)(1/s(1) - 1/s(2) + ... + c/s(n)), where c = (-1)^(n+1) and s(k) = 4*k - 3 for k = 1, 2, 3, ....

Original entry on oeis.org

1, 4, 41, 488, 8881, 176556, 4622745, 128838480, 4403082465, 157917434580, 6659489632905, 292097166060600, 14653855170875025, 759940716395000700, 44202442040567948025, 2645857155729629066400, 175060715455871850866625

Offset: 1

Clark Kimberling

Cf. A024199, A024217, A024396, A007696, A181048.

a := proc(n) option remember; if n = 0 then 1 elif n = 1 then 4 else 4*a(n-1) + (4*n - 3)^2*a(n-2) end if; end:
seq(a(n), n = 0..20);

Table[Product[4*k - 3, {k, 1, n}] * Sum[(-1)^(k+1)/(4*k - 3), {k, 1, n}], {n, 1, 20}] (* Vaclav Kotesovec, Jan 02 2020 *)

a(n) = prod(k=1, n, 4*k-3)*sum(k=1, n, (-1)^(k+1)/(4*k-3)); \\ Michel Marcus, Jul 06 2019

a(n) ~ (Pi^(3/2) + 2*sqrt(Pi)*log(1 + sqrt(2))) * 2^(2*n - 2) * n^(n - 1/4) / (Gamma(1/4) * exp(n)). - Vaclav Kotesovec, Jan 02 2020
From Peter Bala, Mar 21 2024: (Start)
a(n) = Product_{k = 0..n} (4*k + 1) * Sum_{k = 0..n} (-1)^k/(4*k + 1).
a(n) = 4*a(n-1) + (4*n - 3)^2*a(n-2) with a(0) = 1 and a(1) = 4.
b(n) := Product_{k = 0..n} (4*k + 1) = A007696(n+1) satisfies the same 3-term recurrence with b(0) = 1 and b(1) = 5, leading to the continued fraction expansion for the constant A181048 = Sum_{k >= 0} (-1)^k/(4*k + 1) = 1/(1 + 1^2/(4 + 5^2/(4 + 9^2/(4 + 13^2/(4 + ... ))))) due to Euler. (End)

More terms from Sean A. Irvine, Jul 06 2019

A143165 Expansion of the exponential generating function arcsin(2x)/(2(1-2*x)^(3/2)).

Original entry on oeis.org

0, 1, 6, 49, 468, 5469, 73362, 1138005, 19737000, 383284665, 8163588510, 190709475705, 4818820261500, 131650382056725, 3850053335966250, 120466494638624925, 4002649276431128400, 141156781966460192625, 5252646220794868029750, 206149276075766825426625

Offset: 0

Wolfdieter Lang, Sep 15 2008

Used in A024199(n+1) = A003148(n) + a(n).

Binomial convolution of [0,1^2,0,2^2,0,...,0,((2*k)!/k!)^2,0,...] (e.g.f. arcsin(2*x)/2) with the double factorials A001147.

			a(3) + A003148(3) = 49 + 27 = 76 = A024199(4).

Robert Israel, Table of n, a(n) for n = 0..402

Cf. A003148, A024199.

f:= gfun:-rectoproc({2*(n+1)*(3+2*n)^2*a(n)-(4*n^2+8*n+1)*a(n+1)-(2*(n+4))*a(n+2)+a(n+3)=0, a(0)=0,a(1)=1,a(2)=6},a(n),remember):
map(f, [$0 .. 30]); # Robert Israel, Feb 07 2018

With[{nn=20},CoefficientList[Series[ArcSin[2x]/(2(1-2x)^(3/2)),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Mar 18 2019 *)

x = 'x + O('x^40); concat(0, Vec(serlaplace(asin(2*x)/(2*(1-2*x)^(3/2))))) \\ Michel Marcus, Jun 18 2017

E.g.f.: arcsin(2*x)/(2*(1-2*x)^(3/2)).
a(n) = sum(binomial(n,2*k+1)*(4^k)*((2*k-1)!!)^2*(2*(n-2*k)-1)!!,k=0..floor(n/2)), with (2*n-1)!!:= A001147(n) (double factorials).
a(n) ~ Pi * 2^(n-1/2) * n^(n+1) / exp(n) * (1 - sqrt(2/(Pi*n))). - Vaclav Kotesovec, Mar 18 2014
2*(n+1)*(3+2*n)^2*a(n)-(4*n^2+8*n+1)*a(n+1)-(2*(n+4))*a(n+2)+a(n+3)=0. - Robert Israel, Feb 07 2018

A254796 Denominators of the convergents of the generalized continued fraction 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))).

Original entry on oeis.org

1, 4, 25, 200, 2025, 24300, 342225, 5475600, 98903025, 1978060500, 43616234025, 1046789616600, 27260146265625, 763284095437500, 22925783009390625, 733625056300500000, 24966177697226390625, 898782397100150062500, 34178697267502928765625

Offset: 0

Peter Bala, Feb 23 2015

The generalized continued fraction 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))) represents the constant L^2/Pi = 2.188439... = A254794, where L is the lemniscate constant A062539. See A254795 for the numerators of the convergents of the continued fraction.

			54/25 = 2.16, 441/200 = 2.205 etc approach 2.188..

T. J. Osler, The missing fractions in Brouncker’s sequence of continued fractions for Pi

Cf. A007696, A024199, A062539, A254794, A254795.

I:=[1,4]; [n le 2 select I[n] else 4*Self(n-1)+(2*n-3)^2*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 24 2015

a[0] := 1: a[1] := 4:
for n from 2 to 18 do a[n] := 4*a[n-1] + (2*n-1)^2*a[n-2] end do:
seq(a[n], n = 0 .. 18);

RecurrenceTable[{a[0] == 1, a[1] == 4, a[n] == 4 a[n - 1] + (2 n - 1)^2 a[n - 2]}, a, {n, 20}] (* Vincenzo Librandi, Feb 24 2015 *)

a(2*n) = A007696(n+1)^2 = ( Product {k = 0..n} 4*k + 1 )^2.
a(2*n-1) = 4*n*A007696(n)^2 = 4*n * ( Product {k = 0..n-1} 4*k + 1 )^2.
a(n) = 4*a(n-1) + (2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 4.
a(2*n+1) = 4*(n + 1)*a(2*n); a(2*n) = (4*n + 2)*a(2*n-1) + a(2*n-2).
Empirical e.g.f.: ((-Q(1/2, -3)-Q(-1/2, -3))*P(1/2, (2*x+3)/(2*x-1))+Q(1/2, (2*x+3)/(2*x-1))*(P(1/2, -3)+P(-1/2, -3)))/((1-2*x)^(3/2)*(-Q(-1/2, -3)*P(1/2, -3)+Q(1/2, -3)*P(-1/2, -3))) where P and Q are Legendre functions of the first and second kinds. - Robert Israel, Feb 24 2015

A072371 a(0) = 0, a(1) = 1, a(n+1) = 2a(n) + (2n-1)^2*a(n-1).

Original entry on oeis.org

1, 2, 5, 28, 181, 1734, 18129, 246072, 3555945, 62478090, 1152624285, 24859839060, 558026987805, 14266908838350, 377300685054825, 11155177913266800, 339620231957641425, 11399366438564677650, 392645165479000867125, 14749514218199731855500, 567030259977151650805125

Offset: 0

N. J. A. Sloane, Jul 19 2002

nonn

A. E. Jolliffe, Continued Fractions, in Encyclopaedia Britannica, 11th ed., pp. 30-33; see p. 31.

Cf. A024199, A024200.

f := proc(n) option remember; local a,b,t1,t2,t3,i,j,k; a := 1; b := 2; if n=0 then RETURN(a) elif n=1 then RETURN(b) else RETURN(2*f(n-1)+ (2*n-3)^2*f(n-2)); fi; end;

A072372 a(0) = 1, a(1) = 1, a(n) = 2a(n-1) + (2n-1)^2*a(n-2) for n > 1.

Original entry on oeis.org

1, 1, 11, 47, 633, 5073, 86739, 1030815, 21577905, 341061345, 8471746395, 167351545935, 4816256934825, 114227230079025, 3739505765645475, 103544112027750975, 3800753264840803425, 120361044527902418625, 4896644838485789032875, 174567559635669989163375

Offset: 0

N. J. A. Sloane, Jul 19 2002

nonn

A. E. Jolliffe, Continued Fractions, in Encyclopaedia Britannica, 11th ed., pp. 30-33; see p. 31.

Cf. A024199, A024200.

f := proc(n) local a, b, t1, t2, t3, i, j, k; option remember; a := 1; b := 1; if n = 0 then RETURN(a); elif n = 1 then RETURN(b); else RETURN(2*f(n - 1) + (2*n - 1)^2*f(n - 2)); fi; end: seq(f(n), n=0..20); # adapted to offset 0 by Georg Fischer, Dec 23 2019

RecurrenceTable[{a[0]==a[1]==1, a[n]==2*a[n-1] + (2n-1)^2*a[n-2]}, a, {n,0,20}]
(* Harvey P. Dale, Nov 15 2018, adapted to offset 0 by Georg Fischer, Dec 23 2019 *)

Definition adapted to offset 0 by Georg Fischer, Dec 23 2019

cp's OEIS Frontend

A024217 a(n) = ( Product {k = 1..n} 3*k - 2 ) * ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 2) ).

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A135457 a(n) = (2n-1)!! * Sum_{k=0..n-2}(-1)^k/(2k+1).

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A142969 Numerators of approximants of a continued fraction for 4/Pi - 1 = (4 - Pi)/Pi.

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A254795 Numerators of the convergents of the generalized continued fraction 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))).

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A024383 a(n) = s(1)*s(2)*...*s(n)*(1/s(1) - 1/s(2) + ... + c/s(n)), where c = (-1)^(n+1) and s(k) = 4*k - 3 for k = 1, 2, 3, ....

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A143165 Expansion of the exponential generating function arcsin(2*x)/(2*(1-2*x)^(3/2)).

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A254796 Denominators of the convergents of the generalized continued fraction 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))).

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A024217 a(n) = ( Product {k = 1..n} 3k - 2 ) ( Sum {k = 1..n} (-1)^(k+1)/(3*k - 2) ).

A024383 a(n) = s(1)s(2)...s(n)(1/s(1) - 1/s(2) + ... + c/s(n)), where c = (-1)^(n+1) and s(k) = 4*k - 3 for k = 1, 2, 3, ....

A143165 Expansion of the exponential generating function arcsin(2x)/(2(1-2*x)^(3/2)).

A072371 a(0) = 0, a(1) = 1, a(n+1) = 2a(n) + (2n-1)^2*a(n-1).

A072372 a(0) = 1, a(1) = 1, a(n) = 2a(n-1) + (2n-1)^2*a(n-2) for n > 1.