A110258 -id:A110258 - cp's OEIS frontend

A056982 a(n) = 4^A005187(n). The denominators of the Landau constants.

1, 4, 64, 256, 16384, 65536, 1048576, 4194304, 1073741824, 4294967296, 68719476736, 274877906944, 17592186044416, 70368744177664, 1125899906842624, 4503599627370496, 4611686018427387904, 18446744073709551616, 295147905179352825856, 1180591620717411303424

Offset: 0

Eric W. Weisstein

Also equal to A046161(n)^2.
Let W(n) = Product_{k=1..n} (1- 1/(4*k^2)), the partial Wallis product with lim n -> infinity W(n) = 2/Pi; a(n) = denominator(W(n)). The numerators are in A069955.
Equivalently, denominators in partial products of the following approximation to Pi: Pi = Product_{n >= 1} 4*n^2/(4*n^2-1). Numerators are in A069955.
Denominator of h^(2n) in the Kummer-Gauss series for the perimeter of an ellipse.
Denominators of coefficients in hypergeometric([1/2,-1/2],[1],x). The numerators are given in A038535. hypergeom([1/2,-1/2],[1],e^2) = L/(2*Pi*a) with the perimeter L of an ellipse with major axis a and numerical eccentricity e (Maclaurin 1742). - Wolfdieter Lang, Nov 08 2010
Also denominators of coefficients in hypergeometric([1/2,1/2],[1],x). The numerators are given in A038534. - Wolfdieter Lang, May 29 2016
Also denominators of A277233. - Wolfdieter Lang, Nov 16 2016
A277233(n)/a(n) are the Landau constants. These constants are defined as G(n) = Sum_{j=0..n} g(j)^2, where g(n) = (2*n)!/(2^n*n!)^2 = A001790(n)/A046161(n). - Peter Luschny, Sep 27 2019

J.-P. Delahaye, Pi - die Story (German translation), Birkhäuser, 1999 Basel, p. 84. French original: Le fascinant nombre Pi, Pour la Science, Paris, 1997.
O. J. Farrell and B. Ross, Solved Problems in Analysis, Dover, NY, 1971; p. 77.

Charles R Greathouse IV, Table of n, a(n) for n = 0..500
B. Gourevitch, L'univers de Pi
Edmund Landau, Abschätzung der Koeffzientensumme einer Potenzreihe, Arch. Math. Phys. 21 (1913), 42-50. [Accessible in the USA through the Hathi Trust Digital Library.]
Edmund Landau, Abschätzung der Koeffzientensumme einer Potenzreihe (Zweite Abhandlung), Arch. Math. Phys. 21 (1913), 250-255. [Accessible in the USA through the Hathi Trust Digital Library.]
Cristinel Mortici, Sharp bounds of the Landau constants, Math. Comp. 80 (2011), pp. 1011-1018.
G. N. Watson, The constants of Landau and Lebesgue, Quart. J. Math. Oxford Ser. 1:2 (1930), pp. 310-318.
Eric Weisstein's World of Mathematics, Gauss-Kummer Series
Eric Weisstein's World of Mathematics, Ellipse
Index to divisibility sequences

Apart from offset, identical to A110258.
Cf. A005187, A046161, A056981, A069955.
Equals (1/2)*A038533(n), A038534, A277233.
Cf. A001790/A046161.

A056982 := n -> denom(binomial(1/2, n))^2:
seq(A056982(n), n=0..19); # Peter Luschny, Apr 08 2016
# Alternatively:
G := proc(x) hypergeom([1/2,1/2], [1], x)/(1-x) end: ser := series(G(x), x, 20):
[seq(coeff(ser,x,n), n=0..19)]: denom(%); # Peter Luschny, Sep 28 2019

Table[Power[4, 2 n - DigitCount[2 n, 2, 1]], {n, 0, 19}] (* Michael De Vlieger, May 30 2016, after Harvey P. Dale at A005187 *)
G[x_] := (2 EllipticK[x])/(Pi (1 - x));
CoefficientList[Series[G[x], {x, 0, 19}], x] // Denominator (* Peter Luschny, Sep 28 2019 *)

a(n)=my(s=n); while(n>>=1, s+=n); 4^s \\ Charles R Greathouse IV, Apr 07 2012

a(n) = (denominator(binomial(1/2, n)))^2. - Peter Luschny, Sep 27 2019

Edited by N. J. A. Sloane, Feb 18 2004, Jun 05 2007

A001068 a(n) = floor(5*n/4), numbers that are congruent to {0, 1, 2, 3} mod 5.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21, 22, 23, 25, 26, 27, 28, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 42, 43, 45, 46, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 63, 65, 66, 67, 68, 70, 71, 72, 73, 75, 76, 77, 78, 80, 81, 82, 83, 85, 86, 87, 88

Offset: 0

N. J. A. Sloane

From M. F. Hasler, Oct 21 2008: (Start)
Also, for n>0, the 4th term (after [0,n,3n]) in the continued fraction expansion of arctan(1/n). (Observation by V. Reshetnikov)
Proof:
arctan(1/n) = (1/n) / (1 + (1/n)^2/( 3 + (2/n)^2/( 5 + (3/n)^2/( 7 + ...)...)
            = 1 / ( n + 1/( 3n + 4/( 5n + 9/( 7n + 25/(...)...)
            = 1 / ( n + 1/( 3n + 1/( 5n/4 + (9/4)/( 7n + 25/(...)...),
and the term added to 5n/4, (9/4)/(7n+...) = (1/4)*9/(7n+...) is less than 1/4 for all n>=2. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Paul Erdős, Some recent problems and results in graph theory, Discr. Math., Vol. 164, No. 1-3 (1997), pp. 81-85.
Wikipedia, Continued fraction for arctangent.
R. Witula, P. Lorenc, M. Rozanski and M. Szweda, Sums of the rational powers of roots of cubic polynomials, Zeszyty Naukowe Politechniki Slaskiej, Seria: Matematyka Stosowana z. 4, Nr. kol. 1920, (2014), pp. 17-34.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A001622, A002265, A030308, A047203, A110255, A110256, A110257, A110258, A110259, A110260, A177704 (first differences), A249547.

[Floor(5*n/4): n in [0..80]]; // Vincenzo Librandi, Nov 13 2011

A001068:=n->floor(5*n/4); seq(A001068(k), k=0..100); # Wesley Ivan Hurt, Nov 07 2013

Table[Floor[5*n/4],{n,0,120}] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)
#+{0,1,2,3}&/@(5*Range[0,20])//Flatten (* or *) Complement[Range[0,103],5*Range[20]-1] (* Harvey P. Dale, Dec 03 2023 *)

a(n)=5*n\4 /* or, cf. comment: */
a(n)=contfrac(atan(1/n))[4] \\ M. F. Hasler, Oct 21 2008

contfrac( arctan( 1/n )) = 0 + 1/( n + 1/( 3n + 1/( a(n) + 1/(...)))). - M. F. Hasler, Oct 21 2008
a(n) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=1, b(1)=2 and b(k)=5*2^(k-2) for k>1. - Philippe Deléham, Oct 17 2011.
From Bruno Berselli, Oct 17 2011:  (Start)
G.f.: x*(1+x+x^2+2*x^3)/((1+x)*(1-x)^2*(1+x^2)).
a(n) = (10*n+2*(-1)^((n-1)n/2)+(-1)^n-3)/8.
a(-n) = -A047203(n+1). (End)
From Wesley Ivan Hurt, Sep 17 2015: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>4.
a(n) = n + floor(n/4). (End)
a(n) = n + A002265(n). - Robert Israel, Sep 17 2015
E.g.f.: (sin(x) + cos(x) + (5*x - 2)*sinh(x) + (5*x - 1)*cosh(x))/4. - Ilya Gutkovskiy, May 06 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = log(5)/4 + sqrt(5)*log(phi)/10 + sqrt(5-2*sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 10 2021

More terms from James Sellers, Sep 19 2000

A110255 Numerators in the fractional coefficients that form the partial quotients of the continued fraction representation of the inverse tangent of 1/x.

Original entry on oeis.org

1, 3, 5, 28, 81, 704, 325, 768, 20825, 311296, 83349, 1507328, 1334025, 3145728, 5337189, 130023424, 1366504425, 7516192768, 5466528925, 12884901888, 87470372561, 2954937499648, 349899121845, 12919261626368, 22394407746529

Offset: 1

Paul D. Hanna, Jul 18 2005

Limit a(2*n-1)/A110256(2*n-1) = limit A110257(n)/A110258(n) = 4/Pi.

Limit a(2*n)/A110256(2*n) = limit A110259(n)/A110260(n) = Pi.

			arctan(1/x) = 1/x - 1/(3*x^3) + 1/(5*x^5) - 1/(7*x^7) +-...
= [0; x, 3*x, 5/4*x, 28/9*x, 81/64*x, 704/225*x, 325/256*x,
768/245*x, 20825/16384*x, 311296/99225*x, 83349/65536*x,
1507328/480249*x, 1334025/1048576*x, 3145728/1002001*x,...]
= 1/(x + 1/(3*x + 1/(5/4*x + 1/(28/9*x + 1/(81/64*x +...))))).
The coefficients of x in the even-indexed partial quotients converge to Pi:
{3, 28/9, 704/225, 768/245, 311296/99225, ...}.
The coefficients of x in the odd-indexed partial quotients converge to 4/Pi:
{1, 5/4, 81/64, 325/256, 20825/16384, ...}.

Paul D. Hanna, Table of n, a(n) for n = 1..400

Cf. A110256 (denominators), A110257/A110258 (odd-indexed), A110259/A110260 (even-indexed).

{a(n)=numerator(subst((contfrac( sum(k=0,n,(-1)^k/x^(2*k+1)/(2*k+1)),n+1))[n+1],x,1))}

A110256 Denominators in the fractional coefficients that form the partial quotients of the continued fraction representation of the inverse tangent of 1/x.

Original entry on oeis.org

1, 1, 4, 9, 64, 225, 256, 245, 16384, 99225, 65536, 480249, 1048576, 1002001, 4194304, 41409225, 1073741824, 2393453205, 4294967296, 4102737925, 68719476736, 940839860961, 274877906944, 4113258565689, 17592186044416

Offset: 1

Paul D. Hanna, Jul 18 2005

Limit A110255(2*n-1)/a(2*n-1) = limit A110257(n)/A110258(n) = 4/Pi.

Limit A110255(2*n)/a(2*n) = limit A110259(n)/A110260(n) = Pi.

			arctan(1/x) = 1/x - 1/(3*x^3) + 1/(5*x^5) - 1/(7*x^7) +-...
= [0; x, 3*x, 5/4*x, 28/9*x, 81/64*x, 704/225*x, 325/256*x,
768/245*x, 20825/16384*x, 311296/99225*x, 83349/65536*x,
1507328/480249*x, 1334025/1048576*x, 3145728/1002001*x,...]
= 1/(x + 1/(3*x + 1/(5/4*x + 1/(28/9*x + 1/(81/64*x +...))))).
The coefficients of x in the even-indexed partial quotients converge to Pi:
{3, 28/9, 704/225, 768/245, 311296/99225, ...}.
The coefficients of x in the odd-indexed partial quotients converge to 4/Pi:
{1, 5/4, 81/64, 325/256, 20825/16384, ...}.

Paul D. Hanna, Table of n, a(n) for n = 1..400

Cf. A110255 (numerators), A110257/A110258 (odd-indexed), A110259/A110260 (even-indexed).

Cf. A095175. [From R. J. Mathar, Aug 18 2008]

{a(n)=denominator(subst((contfrac( sum(k=0,n,(-1)^k/x^(2*k+1)/(2*k+1)),n+1))[n+1],x,1))}

A110257 Numerators in the coefficients that form the odd-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x.

Original entry on oeis.org

1, 5, 81, 325, 20825, 83349, 1334025, 5337189, 1366504425, 5466528925, 87470372561, 349899121845, 22394407746529, 89580335298125, 1433319858545625, 5733391194015525, 5871086572691471625

Offset: 1

Paul D. Hanna, Jul 18 2005

Limit a(n)/A110258(n) = limit A110255(2*n-1)/A110256(2*n-1) = 4/Pi.

			arctan(1/x) = 1/x - 1/(3*x^3) + 1/(5*x^5) - 1/(7*x^7) +-...
= [0; x, 3*x, 5/4*x, 28/9*x, 81/64*x, 704/225*x, 325/256*x, 768/245*x, 20825/16384*x, 311296/99225*x, 83349/65536*x, 1507328/480249*x, 1334025/1048576*x, 3145728/1002001*x,...]
= 1/(x + 1/(3*x + 1/(5/4*x + 1/(28/9*x + 1/(81/64*x +...))))).
The coefficients of x in the even-indexed partial quotients converge to Pi: {3, 28/9, 704/225, 768/245, 311296/99225, ...}.
The coefficients of x in the odd-indexed partial quotients converge to 4/Pi: {1, 5/4, 81/64, 325/256, 20825/16384, ...}.

Paul D. Hanna, Table of n, a(n) for n = 1..200

Cf. A110258 (denominators), A110255/A110256 (continued fraction), A110259/A110260.

a := n -> (4*n+1)*binomial(2*n,n)^2/4^(add(i,i=convert(n,base,2)));
seq(a(n), n=0..16);  # Peter Luschny, Mar 23 2014

a[n_] := (4n+1) Binomial[2n, n]^2 / 4^DigitCount[n, 2, 1];
Array[a, 16] (* Jean-François Alcover, Jun 13 2019, from Maple *)

{a(n)=numerator(subst((contfrac( sum(k=0,2*n+1,(-1)^k/x^(2*k+1)/(2*k+1)),2*n+2))[2*n],x,1))}

a(n) = A110255(2*n-1).

a(n) = (4*n+1)*A002894(n)/4^A000120(n). - Peter Luschny, Mar 23 2014

A110259 Numerators in the coefficients that form the even-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x.

Original entry on oeis.org

3, 28, 704, 768, 311296, 1507328, 3145728, 130023424, 7516192768, 12884901888, 2954937499648, 12919261626368, 52776558133248, 774056185954304, 66428094503714816, 31525197391593472, 308982963234634989568

Offset: 1

Paul D. Hanna, Jul 18 2005

Lim_{n->infinity} a(n)/A110260(n) = lim_{n->infinity} A110255(2*n)/A110256(2*n) = Pi.

The continued fraction expansion arctan(z) = z/(1 + z^2/(3 + 4*z^2/(5 + 9*z^2/(7 + ...)))) is due to Lambert - see Roegel, Section 1.2.1. - Peter Bala, Dec 02 2021

			arctan(1/x) = 1/x - 1/(3*x^3) + 1/(5*x^5) - 1/(7*x^7) +- ...
arctan(1/x) = [0; x, 3*x, (5/4)*x, (28/9)*x, (81/64)*x, (704/225)*x, (325/256)*x, (768/245)*x, (20825/16384)*x, (311296/99225)*x, (83349/65536)*x, (1507328/480249)*x, (1334025/1048576)*x, (3145728/1002001)*x, ...]
arctan(1/x) = 1/(x + 1/(3*x + 1/(5/4*x + 1/(28/9*x + 1/(81/64*x + ...))))).
The coefficients of x in the even-indexed partial quotients converge to Pi: {3, 28/9, 704/225, 768/245, 311296/99225, ...}.
The coefficients of x in the odd-indexed partial quotients converge to 4/Pi: {1, 5/4, 81/64, 325/256, 20825/16384, ...}.
From _Peter Bala_, Dec 02 2021: (Start)
Making use of the expansion 2*arcsin(sqrt(x)/2)^2 = Sum_{n >= 1} x^n/ (n^2*binomial(2*n,n)) we calculate
3 + Pi = Sum_{n >= 1} (2^n)*n/binomial(2*n,n);
28 + 9*Pi = Sum_{n >= 3} (2^n)*n*(n-1)*(n-2)/binomial(2*n,n);
704 + 225*Pi = Sum_{n >= 5} (2^n)*n*(n-1)*...*(n-4)/binomial(2*n,n);
45*(768 + 245*Pi) = Sum_{n >= 7} (2^n)*n*(n-1)*...*(n-6)/binomial(2*n,n);
9*(311296 + 99225*Pi) = Sum_{n >= 9} (2^n)*n*(n-1)*...*(n-8)/ binomial(2*n,n).
It appears that Sum_{n >= 2*k+1} (2^n)*n*(n-1)*...*(n-2*k)/binomial(2*n,n) = N(2*k) + D(2*k)*Pi, where the ratios N(2*k)/D(2*k) are equal to the even-indexed partial quotients of Lambert's continued fraction representation of the inverse tangent of 1/x. (End)

Paul D. Hanna, Table of n, a(n) for n = 1..200
Denis Roegel, Lambert’s proof of the irrationality of Pi: Context and translation, [Research Report] LORIA, 2020, hal-02984214.

Cf. A110260 (denominators), A110255/A110256 (continued fraction), A110257/A110258.

{a(n)=numerator(subst((contfrac( sum(k=0,2*n+2,(-1)^k/x^(2*k+1)/(2*k+1)),2*n+2))[2*n+1],x,1))}

a(n) = A110255(2*n).

A110260 Denominators in the coefficients that form the even-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x.

Original entry on oeis.org

1, 9, 225, 245, 99225, 480249, 1002001, 41409225, 2393453205, 4102737925, 940839860961, 4113258565689, 16802526820625, 246430431820125, 21147754404155625, 10036045423404225, 98363281194784809225

Offset: 1

Paul D. Hanna, Jul 18 2005

Limit A110259(n)/a(n) = limit A110255(2*n)/A110256(2*n) = Pi.

			arctan(1/x) = 1/x - 1/(3*x^3) + 1/(5*x^5) - 1/(7*x^7) +-...
= [0; x, 3*x, 5/4*x, 28/9*x, 81/64*x, 704/225*x, 325/256*x,
768/245*x, 20825/16384*x, 311296/99225*x, 83349/65536*x,
1507328/480249*x, 1334025/1048576*x, 3145728/1002001*x,...]
= 1/(x + 1/(3*x + 1/(5/4*x + 1/(28/9*x + 1/(81/64*x +...))))).
The coefficients of x in the even-indexed partial quotients converge to Pi:
{3, 28/9, 704/225, 768/245, 311296/99225, ...}.
The coefficients of x in the odd-indexed partial quotients converge to 4/Pi:
{1, 5/4, 81/64, 325/256, 20825/16384, ...}.

Paul D. Hanna, Table of n, a(n) for n = 1..200

Cf. A110259 (numerators), A110255/A110256 (continued fraction), A110257/A110258.

{a(n)=denominator(subst((contfrac( sum(k=0,2*n+2,(-1)^k/x^(2*k+1)/(2*k+1)),2*n+2))[2*n+1],x,1))}

a(n) = A110256(2*n).

A381198 a(n) = denominator( [(xyz*u)^n] 1/sqrt(1 - (x + y + z + u(xy + xz + yz))) ).

Original entry on oeis.org

1, 4, 64, 256, 16384, 65536, 1048576, 4194304, 1073741824, 4294967296, 68719476736, 274877906944, 17592186044416, 70368744177664, 1125899906842624, 4503599627370496, 4611686018427387904, 18446744073709551616, 295147905179352825856, 1180591620717411303424, 75557863725914323419136

Offset: 0

Stefano Spezia, Feb 16 2025

Apparently a duplicate of A110258 and A056982. - R. J. Mathar, Feb 18 2025

S. Hassani, J.-M. Maillard, and N. Zenine, On the diagonals of rational functions: the minimal number of variables (unabridged version), arXiv:2502.05543 [math-ph], 2025. See p. 24.

Cf. A268554, A381197 (numerators).

a[n_]:=Denominator[SeriesCoefficient[1/Sqrt[1-(x+y+z+u*(x*y+x*z+y*z))],{x,0,n},{y,0,n},{z,0,n},{u,0,n}]]; Array[a,15,0]

cp's OEIS Frontend

A056982 a(n) = 4^A005187(n). The denominators of the Landau constants.

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A001068 a(n) = floor(5*n/4), numbers that are congruent to {0, 1, 2, 3} mod 5.

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A110255 Numerators in the fractional coefficients that form the partial quotients of the continued fraction representation of the inverse tangent of 1/x.

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A110256 Denominators in the fractional coefficients that form the partial quotients of the continued fraction representation of the inverse tangent of 1/x.

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A110257 Numerators in the coefficients that form the odd-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x.

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A110259 Numerators in the coefficients that form the even-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x.

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A110260 Denominators in the coefficients that form the even-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x.

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A381198 a(n) = denominator( [(xyz*u)^n] 1/sqrt(1 - (x + y + z + u(xy + xz + yz))) ).