Jaume Oliver Lafont - cp's OEIS frontend

A157718 Greedy Egyptian fraction expansion of log(3).

1, 11, 130, 91827, 42593758221, 2068726045016880942060, 20697114911379630588051784011292634933847536, 832769470129253476302780470023395858447487389073547955500158020204885523374048803963217

Offset: 0

Jaume Oliver Lafont, Mar 04 2009

			log(3) = Sum_{n>=0} 1/a(n) = 1/1 + 1/11 + 1/130 + 1/91827 + 1/42593758221 + ...

Wikipedia, Greedy algorithm for Egyptian fractions

Cf. A058962, A154920, A157024, A002391, A118324.

x=log(3); for (k=1, 8, d=ceil(1/x); x=x-1/d; print(d,","))

A158366 Least k such that n! divides (n+k)!/(n+1)!.

Original entry on oeis.org

1, 2, 3, 4, 5, 4, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 40, 43, 44, 43, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 64, 67, 68, 69, 70, 71, 72

Offset: 1

Jaume Oliver Lafont, Mar 17 2009

nonn

Motivated by 6!=10!/7!.

It appears that for most n, a(n) = n. The sequence of n for which a(n) != n begins: 6, 28, 42, 45, 66, 77, 91, 110, 126 ... that is probably A120624. - Michel Marcus, Aug 21 2013

Cf. A000142.

for (n=1,100,k=1;while((n+k)!/(n+1)!%n!,k++);print1(k,","))

A166741 E.g.f.: exp(2*arcsin(x)).

Original entry on oeis.org

1, 2, 4, 10, 32, 130, 640, 3770, 25600, 199810, 1740800, 16983850, 181043200, 2122981250, 26794393600, 367275756250, 5358878720000, 84106148181250, 1393308467200000, 24643101417106250, 457005177241600000

Offset: 0

Jaume Oliver Lafont, Oct 21 2009

nonn

exp(2*arcsin(1)) is Aleksandr Gelfond's constant.

Wikipedia, Gelfond's constant

Cf. A006228, A039661, A166748.

seq(simplify(2^(n-1) * (cosh(Pi)*(1-(-1)^n) + sinh(Pi)*(1+(-1)^n)) * GAMMA((1/2)*n-I)*GAMMA((1/2)*n+I) / Pi), n=0..20); # Vaclav Kotesovec, Nov 06 2014

CoefficientList[Series[E^(2*ArcSin[x]), {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Aug 04 2014 *)
FullSimplify[Table[2^(n-1) * (E^(Pi)-(-1)^n*E^(-Pi)) * Gamma[n/2-I] * Gamma[n/2+I] / Pi,{n,0,20}]] (* Vaclav Kotesovec, Nov 06 2014 *)

for (n=0,25,print(polcoeff(exp(2*asin(x)),n)*n!,","))

a(n) ~ 2 * n^(n-1) * (exp(Pi) - (-1)^n/exp(Pi)) / exp(n). - Vaclav Kotesovec, Aug 04 2014
From Vaclav Kotesovec, Nov 06 2014: (Start)
a(n) = (n^2 - 4*n + 8)*a(n-2).
a(n) = 2^(n-1) * (exp(Pi)-(-1)^n*exp(-Pi)) * GAMMA(n/2-I) * GAMMA(n/2+I) / Pi.
(End)

A157625 Product of the composite numbers between n+1 and 2n, both inclusive.

Original entry on oeis.org

1, 4, 24, 48, 4320, 8640, 120960, 3628800, 7257600, 14515200, 6706022400, 13412044800, 8717829120000, 470762772480000, 941525544960000, 1883051089920000, 2112783322890240000, 147894832602316800000

Offset: 1

Jaume Oliver Lafont, Mar 03 2009

nonn

This function is very useful in a problem due to Paul Erdős recorded in A157017. - M. F. Hasler, Feb 26 2014

Cf. A073840, A157017, A144186 (product of primes between n+2 and 2n, both inclusive).

nn=20;With[{comps=Complement[Range[2nn],Prime[Range[PrimePi[2nn]]]]}, Table[ Times@@ Select[comps,#>n&&#<=2n&],{n,nn}]] (* Harvey P. Dale, Feb 18 2013 *)

PARI
```
a(n)=prod(i=n+1,2*n,if(isprime(i),1,i))
```

a(n) = n!*A000984(n)*A034386(n)/A034386(2n). - M. F. Hasler, Feb 26 2014

A166871 Permutation of the integers: 3 positives, 2 negatives.

Original entry on oeis.org

0, 1, 2, 3, -1, -2, 4, 5, 6, -3, -4, 7, 8, 9, -5, -6, 10, 11, 12, -7, -8, 13, 14, 15, -9, -10, 16, 17, 18, -11, -12, 19, 20, 21, -13, -14, 22, 23, 24, -15, -16, 25, 26, 27, -17, -18, 28, 29, 30, -19, -20, 31, 32, 33, -21, -22, 34, 35, 36, -23, -24, 37, 38, 39, -25, -26, 40, 41

Offset: 0

Jaume Oliver Lafont, Oct 22 2009

This sequence enumerates the denominators with sign in case p=3 and n=2 of:
log(p/n) = sum( i>=0, sum(p*i+1<=j<=p*(i+1),1/j) - sum(n*i+1<=j<=n*(i+1),1/j) )
Similar sequences can be constructed for the logarithm of any rational r=p/n (p,n>0), enumerating p positive integers and n negative integers every p+n terms.
Case p=2, n=1 is A166711.
Case p=1, n=1 is A001057.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Wikipedia, Riemann series theorem
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,2,0,0,0,0,-1)

Cf. A016578, A166711.

LinearRecurrence[{0,0,0,0,2,0,0,0,0,-1},{0,1,2,3,-1,-2,4,5,6,-3}, 100] (* G. C. Greubel, May 27 2016 *)

Sum_{k>0} 1/a(k) = log(3/2).

G.f.: x*(1+2*x+3*x^2-x^3-2*x^4+2*x^5+x^6-x^8)/((x-1)^2*(x^4+x^3+x^2+x+1)^2 ).

keyword frac removed Jaume Oliver Lafont, Nov 02 2009

A165896 a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)a(n-2)+a(n-1)a(n-3)+a(n-2)*a(n-3))/a(n-4) with four initial ones.

Original entry on oeis.org

1, 1, 1, 1, 6, 51, 3001, 9180001, 14050074147451, 3870680638643416483474006, 4992392071450646411005278674572370014340582601, 2715030052293379508289500941366397276374058263752394148988972928520177978202810359001

Offset: 0

Jaume Oliver Lafont, Sep 29 2009

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..15
S. Fomin, A. Zelevinsky, The Laurent Phenomenon, Adv. Appl. Math. 28 (2) (2002) 119-144. [_R. J. Mathar_, Oct 23 2009]
Sergey Fomin, Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241 [math.CO], 2001. [_R. J. Mathar_, Oct 23 2009]

Cf. A006720, A165903.

RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==1,a[n]==(a[n-1]^2+a[n-2]^2+a[n-3]^2+ a[n-1]a[n-2]+ a[n-1]a[n-3]+a[n-2]a[n-3])/a[n-4]},a,{n,13}] (* Harvey P. Dale, May 21 2012 *)

a(n)=if(n<4,1,(a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)*a(n-2)+a(n-1)*a(n-3)+a(n-2)*a(n-3))/a(n-4))

a(n) ~ 1/sqrt(10) * c^(t^n), where t = A058265 = 1.8392867552141611325518525646532866..., c = 1.2712241060822553131735186905646486868228186258439... . - Vaclav Kotesovec, May 06 2015

a(n) = 10*a(n-1)*a(n-2)*a(n-3)-a(n-1)-a(n-2)-a(n-3)-a(n-4). - Bruno Langlois, Aug 21 2016

A164916 Denominators of a BBP series for Pi/4.

Original entry on oeis.org

1, -8, -20, -24, 144, -384, -832, -896, 4352, -10240, -21504, -22528, 102400, -229376, -475136, -491520, 2162688, -4718592, -9699328, -9961472, 42991616, -92274688, -188743680, -192937984, 822083584, -1744830464, -3556769792

Offset: 0

Jaume Oliver Lafont, Aug 31 2009

From the BBP formula for Pi, the following expression for Pi/4 in unit numerators is obtained
Pi/4 = Sum((1/(8k+1)+1/(-2*(8k+4))+1/(-4*(8k+5))+1/(-4*(8k+6)))/16^k, k>=0)
Therefore a(n) such that
a(4*n) = (8*n+1)*16^n.
a(4*n+1) = -2*(8*n+4)*16^n.
a(4*n+2) = -4*(8*n+5)*16^n.
a(4*n+3) = -4*(8*n+6)*16^n.
has
Sum_{n >= 0} (1/a(n)) = Pi/4.
Using PARI/GP suminf(n=0,1/(2^(n-2)*(2*(-1+(-1)^n+(1-I)*(-I)^n+(1+I)*I^n)+(-3+3*(-1)^n+(4-I)*(-I)^n+(4+I)*I^n)*n)))= 0.7853981633974483096156608454...=Pi/4. - Alexander R. Povolotsky, Sep 01 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A048581, A066968, A154925, A154962, A156269, A157142.

CoefficientList[Series[(1 - 8*x - 20*x^2 - 24*x^3 + 112*x^4 - 128*x^5 - 192*x^6 - 128*x^7)/(1 - 16*x^4)^2, {x,0,50}], x] (* G. C. Greubel, Feb 25 2017 *)

x='x + O('x^50); Vec((1 - 8*x - 20*x^2 - 24*x^3 + 112*x^4 - 128*x^5 - 192*x^6 - 128*x^7)/(1 - 16*x^4)^2) \\ G. C. Greubel, Feb 25 2017

G.f.: (1-8*x-20*x^2-24*x^3+112*x^4-128*x^5-192*x^6-128*x^7)/(1-16*x^4)^2.

a(n)= 2^(n-2)*(2*(-1+(-1)^n+(1-I)*(-I)^n+(1+I)*I^n)+(-3+3*(-1)^n+(4-I)*(-I)^n+(4+I)*I^n)*n). - Alexander R. Povolotsky, Sep 01 2009

Comment section corrected by Jaume Oliver Lafont, Sep 03 2009

A157327 Egyptian fraction expansion for Pi/4 = arctan(1/2) + arctan(1/3) (Hutton 1776).

Original entry on oeis.org

2, 3, -24, -81, 160, 1215, -896, -15309, 4608, 177147, -22528, -1948617, 106496, 20726199, -491520, -215233605, 2228224, 2195382771, -9961472, -22082967873, 44040192, 219667417263, -192937984, -2165293113021, 838860800

Offset: 0

Jaume Oliver Lafont, Feb 27 2009

Sum_{n>=0} 1/a(n) = Pi/4.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
X. Gourdon and P. Sebah, The constant Pi. The classic period

Cf. A157142, A155988, A058962.

CoefficientList[Series[2 (1 - 4 x^2)/(1 + 4 x^2)^2 + 3 x (1 - 9 x^2)/(1 + 9 x^2)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 12 2012 *)

G.f.: 2*(1-4*x^2)/(1+4*x^2)^2 + 3*x*(1-9*x^2)/(1+9*x^2)^2.

A176696 E.g.f.: exp(6arcsin(x))-6arcsin(x).

Original entry on oeis.org

1, 0, 36, 216, 1440, 9936, 74880, 608040, 5391360, 51732000, 539136000, 6055025400, 73322496000, 950831881200, 13198049280000, 194943875747400, 3061947432960000, 50884296047208000, 894088650424320000

Offset: 0

Jaume Oliver Lafont, Apr 24 2010

nonn

Sum_{k>=0} a(k)/(2^k*k!) = e^Pi - Pi (A018938).

Cf. A018938, A166748.

With[{nn=20},CoefficientList[Series[Exp[6ArcSin[x]]-6ArcSin[x],{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Mar 11 2012 *)

a(n)=polcoeff(exp(6*asin(x))-6*asin(x),n)*n!

A167205 a(n) = (3^n+1)/(3-(-1)^n).

Original entry on oeis.org

1, 1, 5, 7, 41, 61, 365, 547, 3281, 4921, 29525, 44287, 265721, 398581, 2391485, 3587227, 21523361, 32285041, 193710245, 290565367, 1743392201, 2615088301, 15690529805, 23535794707, 141214768241, 211822152361, 1270932914165

Offset: 0

Jaume Oliver Lafont, Oct 30 2009

This sequence is (3^n + 1) divided by the highest possible power of 2, which is 4 for odd n and 2 for even n. It is never divisible by 8 or any higher power of 2, which implies Levi ben Gerson's observation that (3^n + 1 = 2^k) has no solution for n > 1. Cf. the comments and links to A235365. - Joe Slater, Apr 02 2017

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,10,0,-9)

Cf. A007051, A074476, A235365.

List([0..27],n->(3^n+1)/(3-(-1)^n)); # Muniru A Asiru, Mar 05 2018

a:=n->(3^n+1)/(3-(-1)^n): seq(a(n),n=0..27); # Muniru A Asiru, Mar 05 2018

CoefficientList[Series[(1+x-5x^2-3x^3)/((1+x)(1-x)(1+3x)(1-3x)), {x,0,30}],x] (* or *) LinearRecurrence[{0,10,0,-9},{1,1,5,7},30] (* Harvey P. Dale, Apr 25 2011 *)

a(n) = (3^n+1)/(3-(-1)^n); \\ Altug Alkan, Mar 05 2018

a(n) = 10*a(n-2) - 9*a(n-4).
G.f.: (1 + x - 5*x^2 - 3*x^3)/((1+x)*(1-x)*(1+3*x)*(1-3*x)).
a(n) = numerator((1/4)^n + (3/4)^n), n > 0.

cp's OEIS Frontend

User: Jaume Oliver Lafont

A157718 Greedy Egyptian fraction expansion of log(3).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

A158366 Least k such that n! divides (n+k)!/(n+1)!.

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

A166741 E.g.f.: exp(2*arcsin(x)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A157625 Product of the composite numbers between n+1 and 2n, both inclusive.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

Formula

A166871 Permutation of the integers: 3 positives, 2 negatives.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A165896 a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)*a(n-2)+a(n-1)*a(n-3)+a(n-2)*a(n-3))/a(n-4) with four initial ones.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A164916 Denominators of a BBP series for Pi/4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A157327 Egyptian fraction expansion for Pi/4 = arctan(1/2) + arctan(1/3) (Hutton 1776).

Views

Author

Keywords

Comments

Links

A165896 a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-1)a(n-2)+a(n-1)a(n-3)+a(n-2)*a(n-3))/a(n-4) with four initial ones.

A176696 E.g.f.: exp(6arcsin(x))-6arcsin(x).