A110260 -id:A110260 - cp's OEIS frontend

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

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

A110258 Denominators 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, 4, 64, 256, 16384, 65536, 1048576, 4194304, 1073741824, 4294967296, 68719476736, 274877906944, 17592186044416, 70368744177664, 1125899906842624, 4503599627370496, 4611686018427387904

Offset: 1

Paul D. Hanna, Jul 18 2005

Limit A110257(n)/a(n) = limit A110255(2*n-1)/A110256(2*n-1) = 4/Pi.

Apart from offset, identical to A056982.

			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
Index to divisibility sequences

See A056982 for another version of this sequence.

Cf. A110257 (numerators), A110255/A110256 (continued fraction), A110259/A110260.

{a(n)=denominator(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)=4^(2*n-vecsum(binary(n-1))-2) \\ Charles R Greathouse IV, Apr 09 2012

a(n) = 4^A005187(n-1).

a(n) = A110256(2*n-1).

Edited by N. J. A. Sloane, Jun 05 2007

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).

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

A110258 Denominators in the coefficients that form the odd-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula