cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 12 results. Next

A016754 Odd squares: a(n) = (2n+1)^2. Also centered octagonal numbers.

Original entry on oeis.org

1, 9, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1225, 1369, 1521, 1681, 1849, 2025, 2209, 2401, 2601, 2809, 3025, 3249, 3481, 3721, 3969, 4225, 4489, 4761, 5041, 5329, 5625, 5929, 6241, 6561, 6889, 7225, 7569, 7921, 8281, 8649, 9025
Offset: 0

Views

Author

Keywords

Comments

The brown rat (rattus norwegicus) breeds very quickly. It can give birth to other rats 7 times a year, starting at the age of three months. The average number of pups is 8. The present sequence gives the total number of rats, when the intervals are 12/7 of a year and a young rat starts having offspring at 24/7 of a year. - Hans Isdahl, Jan 26 2008
Numbers n such that tau(n) is odd where tau(x) denotes the Ramanujan tau function (A000594). - Benoit Cloitre, May 01 2003
If Y is a fixed 2-subset of a (2n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007
Binomial transform of [1, 8, 8, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 8, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
All terms of this sequence are of the form 8k+1. For numbers 8k+1 which aren't squares see A138393. Numbers 8k+1 are squares iff k is a triangular number from A000217. And squares have form 4n(n+1)+1. - Artur Jasinski, Mar 27 2008
Sequence arises from reading the line from 1, in the direction 1, 25, ... and the line from 9, in the direction 9, 49, ..., in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
Equals the triangular numbers convolved with [1, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson & Alexander R. Povolotsky, May 29 2009
First differences: A008590(n) = a(n) - a(n-1) for n>0. - Reinhard Zumkeller, Nov 08 2009
Central terms of the triangle in A176271; cf. A000466, A053755. - Reinhard Zumkeller, Apr 13 2010
Odd numbers with odd abundance. Odd numbers with even abundance are in A088828. Even numbers with odd abundance are in A088827. Even numbers with even abundance are in A088829. - Jaroslav Krizek, May 07 2011
Appear as numerators in the non-simple continued fraction expansion of Pi-3: Pi-3 = K_{k>=1} (1-2*k)^2/6 = 1/(6+9/(6+25/(6+49/(6+...)))), see also the comment in A007509. - Alexander R. Povolotsky, Oct 12 2011
Ulam's spiral (SE spoke). - Robert G. Wilson v, Oct 31 2011
All terms end in 1, 5 or 9. Modulo 100, all terms are among { 1, 9, 21, 25, 29, 41, 49, 61, 69, 81, 89 }. - M. F. Hasler, Mar 19 2012
Right edge of both triangles A214604 and A214661: a(n) = A214604(n+1,n+1) = A214661(n+1,n+1). - Reinhard Zumkeller, Jul 25 2012
Also: Odd numbers which have an odd sum of divisors (= sigma = A000203). - M. F. Hasler, Feb 23 2013
Consider primitive Pythagorean triangles (a^2 + b^2 = c^2, gcd(a, b) = 1) with hypotenuse c (A020882) and respective even leg b (A231100); sequence gives values c-b, sorted with duplicates removed. - K. G. Stier, Nov 04 2013
For n>1 a(n) is twice the area of the irregular quadrilateral created by the points ((n-2)*(n-1),(n-1)*n/2), ((n-1)*n/2,n*(n+1)/2), ((n+1)*(n+2)/2,n*(n+1)/2), and ((n+2)*(n+3)/2,(n+1)*(n+2)/2). - J. M. Bergot, May 27 2014
Number of pairs (x, y) of Z^2, such that max(abs(x), abs(y)) <= n. - Michel Marcus, Nov 28 2014
Except for a(1)=4, the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 737", based on the 5-celled von Neumann neighborhood. - Robert Price, May 23 2016
a(n) is the sum of 2n+1 consecutive numbers, the first of which is n+1. - Ivan N. Ianakiev, Dec 21 2016
a(n) is the number of 2 X 2 matrices with all elements in {0..n} with determinant = 2*permanent. - Indranil Ghosh, Dec 25 2016
Engel expansion of Pi*StruveL_0(1)/2 where StruveL_0(1) is A197037. - Benedict W. J. Irwin, Jun 21 2018
Consider all Pythagorean triples (X,Y,Z=Y+1) ordered by increasing Z; the segments on the hypotenuse {p = a(n)/A001844(n), q = A060300(n)/A001844(n) = A001844(n) - p} and their ratio p/q = a(n)/A060300(n) are irreducible fractions in Q\Z. X values are A005408, Y values are A046092, Z values are A001844. - Ralf Steiner, Feb 25 2020
a(n) is the number of large or small squares that are used to tile primitive squares of type 2 (A344332). - Bernard Schott, Jun 03 2021
Also, positive odd integers with an odd number of odd divisors (for similar sequence with 'even', see A348005). - Bernard Schott, Nov 21 2021
a(n) is the least odd number k = x + y, with 0 < x < y, such that there are n distinct pairs (x,y) for which x*y/k is an integer; for example, a(2) = 25 and the two corresponding pairs are (5,20) and (10,15). The similar sequence with 'even' is A016742 (see Comment of Jan 26 2018). - Bernard Schott, Feb 24 2023
From Peter Bala, Jan 03 2024: (Start)
The sequence terms are the exponents of q in the series expansions of the following infinite products:
1) q*Product_{n >= 1} (1 - q^(16*n))*(1 + q^(8*n)) = q + q^9 + q^25 + q^49 + q^81 + q^121 + q^169 + ....
2) q*Product_{n >= 1} (1 + q^(16*n))*(1 - q^(8*n)) = q - q^9 - q^25 + q^49 + q^81 - q^121 - q^169 + + - - ....
3) q*Product_{n >= 1} (1 - q^(8*n))^3 = q - 3*q^9 + 5*q^25 - 7*q^49 + 9*q^81 - 11*q^121 + 13*q^169 - + ....
4) q*Product_{n >= 1} ( (1 + q^(8*n))*(1 - q^(16*n))/(1 + q^(16*n)) )^3 = q + 3*q^9 - 5*q^25 - 7*q^49 + 9*q^81 + 11*q^121 - 13*q^169 - 15*q^225 + + - - .... (End)

References

  • L. Lorentzen and H. Waadeland, Continued Fractions with Applications, North-Holland 1992, p. 586.

Crossrefs

Cf. A000447 (partial sums).
Cf. A348005, A379481 [= a(A048673(n)-1)].
Partial sums of A022144.
Positions of odd terms in A341528.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n) = 1 + Sum_{i=1..n} 8*i = 1 + 8*A000217(n). - Xavier Acloque, Jan 21 2003; Zak Seidov, May 07 2006; Robert G. Wilson v, Dec 29 2010
O.g.f.: (1+6*x+x^2)/(1-x)^3. - R. J. Mathar, Jan 11 2008
a(n) = 4*n*(n + 1) + 1 = 4*n^2 + 4*n + 1. - Artur Jasinski, Mar 27 2008
a(n) = A061038(2+4n). - Paul Curtz, Oct 26 2008
Sum_{n>=0} 1/a(n) = Pi^2/8 = A111003. - Jaume Oliver Lafont, Mar 07 2009
a(n) = A000290(A005408(n)). - Reinhard Zumkeller, Nov 08 2009
a(n) = a(n-1) + 8*n with n>0, a(0)=1. - Vincenzo Librandi, Aug 01 2010
a(n) = A033951(n) + n. - Reinhard Zumkeller, May 17 2009
a(n) = A033996(n) + 1. - Omar E. Pol, Oct 03 2011
a(n) = (A005408(n))^2. - Zak Seidov, Nov 29 2011
From George F. Johnson, Sep 05 2012: (Start)
a(n+1) = a(n) + 4 + 4*sqrt(a(n)).
a(n-1) = a(n) + 4 - 4*sqrt(a(n)).
a(n+1) = 2*a(n) - a(n-1) + 8.
a(n+1) = 3*a(n) - 3*a(n-1) + a(n-2).
(a(n+1) - a(n-1))/8 = sqrt(a(n)).
a(n+1)*a(n-1) = (a(n)-4)^2.
a(n) = 2*A046092(n) + 1 = 2*A001844(n) - 1 = A046092(n) + A001844(n).
Limit_{n -> oo} a(n)/a(n-1) = 1. (End)
a(n) = binomial(2*n+2,2) + binomial(2*n+1,2). - John Molokach, Jul 12 2013
E.g.f.: (1 + 8*x + 4*x^2)*exp(x). - Ilya Gutkovskiy, May 23 2016
a(n) = A101321(8,n). - R. J. Mathar, Jul 28 2016
Product_{n>=1} A033996(n)/a(n) = Pi/4. - Daniel Suteu, Dec 25 2016
a(n) = A014105(n) + A000384(n+1). - Bruce J. Nicholson, Nov 11 2017
a(n) = A003215(n) + A002378(n). - Klaus Purath, Jun 09 2020
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=0} a(n)/n! = 13*e.
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 3/e. (End)
Sum_{n>=0} (-1)^n/a(n) = A006752. - Amiram Eldar, Oct 10 2020
From Amiram Eldar, Jan 28 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = Pi/4 (A003881). (End)
From Leo Tavares, Nov 24 2021: (Start)
a(n) = A014634(n) - A002943(n). See Diamond Triangles illustration.
a(n) = A003154(n+1) - A046092(n). See Diamond Stars illustration. (End)
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k-1)) = 1/(9 - 3/(17 - 60/(33 - 315/(57 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*2^2 ))))).
3/2 - 2*log(2) = Sum_{k >= 1} 1/(k*a(k)*a(k-1)) = 1/(9 - 3/(17 - 60/(33 - 315/(57 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*2^2 - ... ))))).
Row 2 of A142992. (End)
From Peter Bala, Mar 26 2024: (Start)
8*a(n) = (2*n + 1)*(a(n+1) - a(n-1)).
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 1/2 - Pi/8 = 1/(9 + (1*3)/(8 + (3*5)/(8 + ... + (4*n^2 - 1)/(8 + ... )))). For the continued fraction use Lorentzen and Waadeland, p. 586, equation 4.7.9 with n = 1. Cf. A057813. (End)

Extensions

Additional description from Terrel Trotter, Jr., Apr 06 2002

A025547 Least common multiple of {1,3,5,...,2n-1}.

Original entry on oeis.org

1, 3, 15, 105, 315, 3465, 45045, 45045, 765765, 14549535, 14549535, 334639305, 1673196525, 5019589575, 145568097675, 4512611027925, 4512611027925, 4512611027925, 166966608033225, 166966608033225, 6845630929362225, 294362129962575675, 294362129962575675
Offset: 1

Views

Author

Keywords

Comments

This sequence coincides with the sequence f(n) = denominator of 1 + 1/3 + 1/5 + 1/7 + ... + 1/(2n-1) iff n <= 38. But a(39) = 6414924694381721303722858446525, f(39) = 583174972216520118520259858775. - T. D. Noe, Aug 04 2004 [See A350670(n-1).]
Coincides for n=1..42 with the denominators of a series for Pi*sqrt(2)/4 and then starts to differ. See A127676.
a(floor((n+1)/2)) = gcd(a(n), A051426(n)). - Reinhard Zumkeller, Apr 25 2011
A051417(n) = a(n+1)/a(n).

Crossrefs

Cf. A007509, A025550, A075135. The numerators are in A074599.
Cf. A003418 (LCM of {1..n}).

Programs

  • Haskell
    a025547 n = a025547_list !! (n-1)
    a025547_list = scanl1 lcm a005408_list
    -- Reinhard Zumkeller, Oct 25 2013, Apr 25 2011
    
  • Maple
    A025547:=proc(n) local i,t1; t1:=1; for i from 1 to n do t1:=lcm(t1,2*i-1); od: t1; end;
    f := n->denom(add(1/(2*k-1),k=0..n)); # a different sequence!
  • Mathematica
    a = 1; Join[{1}, Table[a = LCM[a, n], {n, 3, 125, 2}]] (* Zak Seidov, Jan 18 2011 *)
    nn=30;With[{c=Range[1,2*nn,2]},Table[LCM@@Take[c,n],{n,nn}]] (* Harvey P. Dale, Jan 27 2013 *)
  • PARI
    a(n)=lcm(vector(n,k,2*k-1)) \\ Charles R Greathouse IV, Nov 20 2012
    
  • Python
    # generates initial segment of sequence
    from math import gcd
    from itertools import accumulate
    def lcm(a, b): return a * b // gcd(a, b)
    def aupton(nn): return list(accumulate((2*i+1 for i in range(nn)), lcm))
    print(aupton(23)) # Michael S. Branicky, Mar 28 2022

A024199 a(n) = (2n-1)!! * Sum_{k=0..n-1}(-1)^k/(2k+1).

Original entry on oeis.org

0, 1, 2, 13, 76, 789, 7734, 110937, 1528920, 28018665, 497895210, 11110528485, 241792844580, 6361055257725, 163842638377950, 4964894559637425, 147721447995130800, 5066706567801827025, 171002070002301095250, 6548719685561840296125, 247199273204273879989500
Offset: 0

Views

Author

Keywords

Comments

(2*n + 1)!!/a(n+1), n>=0, is the n-th approximant for William Brouncker's continued fraction for 4/Pi = 1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + ... ))) See the C. Brezinski and J.-P. Delahaye references given under A142969 and A142970, respectively. The double factorials (2*n + 1)!! = A001147(n+1) enter. - Wolfdieter Lang, Oct 06 2008

Examples

			a(3) = (2*3 - 1)!! * Sum_{k=0..2} (-1)^k/(2k + 1) = 5!! * (1/(2*0 + 1) - 1/(2*1 + 1) + 1/(2*2 + 1)) = 5*3*1*(1/1 - 1/3 + 1/5) = 15 - 5 + 3 = 13. Notice that the first factor always cancels the common denominator of the sum. - _Michael B. Porter_, Jul 22 2016
		

References

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

Crossrefs

From Johannes W. Meijer, Nov 12 2009: (Start)
Cf. A007509 and A025547.
Equals first column of A167584.
Equals row sums of A167591.
Equals first right hand column of A167594.
(End)
Cf. A167576 and A135457.

Programs

  • Magma
    [0] cat [ n le 2 select (n) else 2*Self(n-1)+(2*n-3)^2*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 17 2015
  • Maple
    a := proc(n) option remember; if n=0 then 0 elif n=1 then 1 else 2*a(n-1)+(2*n-3)^2* a(n-2) fi end: seq(a(n), n=0..20); # Peter Luschny, Nov 16 2016 after N. J. A. Sloane
  • Mathematica
    f[k_] := (2 k - 1) (-1)^(k + 1)
    t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 22}]    (* A024199 signed *)
    (* Clark Kimberling, Dec 30 2011 *)
    RecurrenceTable[{a[n+1] == 2*a[n] + (2*n-1)^2*a[n-1],a[0] == 0, a[1] == 1},a,{n,0,20}] (* Vaclav Kotesovec, Mar 18 2014 *)
    CoefficientList[Series[Pi/4/Sqrt[1-2*x] - 1/2*Log[2*x+Sqrt[4*x^2-1]]/Sqrt[2*x-1], {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Mar 18 2014 *)

Formula

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) = 2k-1 for k = 1, 2, 3, ... (was previous definition). - Clark Kimberling
D-finite with recurrence a(0) = 0, a(1) = 1, a(n+1) = 2*a(n) + (2*n-1)^2*a(n-1). - N. J. A. Sloane, Jul 19 2002
a(n) + A024200(n) = A001147(n) = (2n-1)!!. - Max Alekseyev, Sep 23 2007
a(n)/A024200(n) -> Pi/(4-Pi) as n -> oo. - Max Alekseyev, Sep 23 2007
From Wolfdieter Lang, Oct 06 2008: (Start)
E.g.f. for a(n+1), n>=0: (sqrt(1-2*x)+arcsin(2*x)*sqrt(1+2*x)/2)/((1-4*x^2)^(1/2)*(1-2*x)). From the recurrence, solving (1-4*x^2)y''(x)-2*(8*x+1)*y'(x)-9*y=0 with inputs y(0)=1, y'(0)=2.
a(n+1) = A003148(n) + A143165(n), n>=0 (from the two terms of the e.g.f.). (End)
From Johannes W. Meijer, Nov 12 2009: (Start)
a(n) = (-1)^(n-1)*(2*n-3)!! + (2*n-1)*a(n-1) with a(0) = 0.
a(n) = (2*n-1)!!*sum((-1)^(k)/(2*k+1), k=0..n-1)
(End)
E.g.f.: Pi/4/sqrt(1-2*x) - 1/2*log(2*x+sqrt(4*x^2-1))/sqrt(2*x-1). - Vaclav Kotesovec, Mar 18 2014
a(n) ~ Pi * 2^(n-3/2) * n^n / exp(n). - Vaclav Kotesovec, Mar 18 2014
a(n) = (2*H(n+1/2)-Gamma(n+1/2))*2^(n-2)*sqrt(Pi) with H(x) the Hadamard factorial (see the link section). - Cyril Damamme, Jul 19 2015
a(n) = A135457(n) - (-1)^n A001147(n-1). - Cyril Damamme, Jul 19 2015
a(n) = (Pi + (-1)^n*(Psi(n/2 + 1/4) - Psi(n/2 + 3/4)))*Gamma(n+1/2)*2^(n-2)/sqrt(Pi). - Robert Israel, Jul 20 2015
a(n) = A167576(n) - A135457(n). - Cyril Damamme, Jul 22 2015
a(n)/A001147(n) -> Pi/4 as n -> oo. - Daniel Suteu, Jul 21 2016
From Peter Bala, Nov 15 2016: (Start)
Conjecture: a(n) = 1/2*Sum_{k = 0..2*n-1} (-1)^(n-k+1)*k!*(2*n - 2*k - 3)!!, where the double factorial of an odd integer (positive or negative) may be defined in terms of the gamma function as (2*N - 1)!! = 2^((N+1)/2)*Gamma(N/2 + 1)/sqrt(Pi).
E.g.f. 1/2*arcsin(2*x)/sqrt(1 - 2*x) = x + 2*x^2/2! + 13*x^3/3! + 76*x^4/4! + .... (End)

Extensions

Edited by N. J. A. Sloane, Jul 19 2002
New name from Cyril Damamme, Jul 19 2015

A167576 The first column of the ED3 array A167572.

Original entry on oeis.org

1, 5, 23, 167, 1473, 16413, 211479, 3192975, 54010305, 1030249845, 21566327895, 497334999735, 12405876372225, 335591130336525, 9716331072597975, 301633179343890975, 9941514351641143425, 348336799875365041125
Offset: 1

Views

Author

Johannes W. Meijer, Nov 10 2009

Keywords

Comments

Basically a(n) measures the difference between the Euler factorial n! and the Luschny factorial L(n) at half-integer values. For the Luschny factorial see the link. The formula given in the Maple section is a variant of a formula given by Cyril Damamme in A135457. - Peter Luschny, Jul 18 2015

Examples

			G.f. = x + 5*x^2 + 23*x^3 + 167*x^4 + 1473*x^5 + 16413*x^6 + ...
		

Crossrefs

Equals the first column of the ED3 array A167572.
Equals the first right hand column of A167583.
Other columns are A167577 and A167578.
Cf. A097801 (the 2*(-1)^n*(2*n-5)!! factor).
Cf. A007509 and A025547 (the sum((-1)^(k+n)/(2*k+1), k=0..n-1) factor).
Cf. A024199 and A135457.

Programs

  • Maple
    L := x -> (1+x*(Psi(1-x/2)-Psi(1/2-x/2)))/(-x)!:
    a := x -> (L(x-1/2)-(x-1/2)!)*2^(x-1)*sqrt(Pi):
    seq(simplify(a(n)),n=1..18); # Peter Luschny, Jul 18 2015
    a := proc(n) option remember: if n=1 then 1 else (2*n-1)*a(n-1)+2*(-1)^n*doublefactorial(2*n-5) fi: end: seq(a(n),n=1..18); # Johannes W. Meijer, Jul 20 2015
  • Mathematica
    a[ n_] := If[ n < 1, 0, (2 n - 3)!! ((-1)^n - I (4 n - 2) Sum[ I^k / k, {k, 1, 2 n - 1, 2}])]; (* Michael Somos, Jul 20 2015 *)
    a[ n_] := If[ n < 1, 0, (2 n - 3)!! ((-1)^n + (4 n - 2) Sum[ KroneckerSymbol[ -4, k]/ k, {k, 2 n - 1}])]; (* Michael Somos, Jan 31 2019 *)
  • PARI
    {a(n) = if( n<1, 0, prod(k=1, n-1, 2*k - 1) * ((-1)^n - (4*n - 2) * sum(k=1, n, (-1)^k / (2*k - 1))))}; /* Michael Somos, Jul 20 2015 */

Formula

a(n) = (-1)^n*(2*n-3)!!*(1 + (4*n-2)*Sum_{k=0..n-1} (-1)^(k+n)/(2*k+1)).
a(n) = (2*n-1)*a(n-1) + 2*(-1)^n*(2*n-5)!! with a(1) = 1.
a(n) = 4*a(n-1) + (4*n^2 - 16*n + 15)*a(n-2) with a(1) = 1 and a(2) = 5 [Superseeker].
0 = a(n)*a(n+1)*(-440*a(n+2) - 220*a(n+3) + 55*a(n+4)) + a(n)*a(n+2)*(536*a(n+2) - 118*a(n+3) - 4*a(n+4)) + a(n)*a(n+3)*(-4*a(n+3) + a(n+4)) + a(n+1)^2*(-220*a(n+2) - 32*a(n+3) + 8*a(n+4)) + a(n+1)*a(n+2)*(+71*a(n+2) + 4*a(n+3) - 2*a(n+4)) + a(n+2)^2*(-4*a(n+2) + a(n+3)) if n>0. - Michael Somos, Jul 19 2015
a(n) = (-1 + (n-1/2)*LerchPhi(-1,1,n+1/2) + (-n+1/2)*LerchPhi(-1,1,-n+1/2))/(1-2*n)!!. - Johannes W. Meijer, Jul 20 2015
a(n) = A024199(n) + A135457(n). - Cyril Damamme, Jul 22 2015
a(n) = ((-1)^n/(2*n - 1) + Pi/2 - (-1)^n LerchPhi(-1, 1, n + 1/2)) (2*n - 1)!!. - Michael Somos, Jan 31 2019

A167577 The second column of the ED3 array A167572.

Original entry on oeis.org

1, 11, 83, 741, 8169, 106107, 1592235, 27062325, 514246545, 10798366635, 248374594755, 6209158112325, 167651197407225, 4861802228946075, 150717766502187675, 4973638859450709525, 174078640829054894625
Offset: 1

Views

Author

Johannes W. Meijer, Nov 10 2009

Keywords

Crossrefs

Equals the second column of the ED3 array A167572.
Other columns are A167576 and A167578.
Cf. A007509 and A025547 (the sum((-1)^(k+n)/(2*k+1), k=0..n-1) factor).

Programs

  • Mathematica
    Table[(1/2)*(-1)^n*(2*n - 5)!!*((4*n^2 - 6*n - 2) + (16*n^3 - 24*n^2 - 4*n + 6)*Sum[(-1)^(k + n)/(2*k + 1), {k, 0, n - 1}]), {n, 1,50}] (* G. C. Greubel, Jun 16 2016 *)

Formula

a(n) = (1/2)*(-1)^n*(2*n-5)!!*((4*n^2-6*n-2)+(16*n^3-24*n^2-4*n+6)*sum((-1)^(k+n)/ (2*k+1), k=0..n-1)).

A167578 The third column of the ED3 array A167572.

Original entry on oeis.org

1, 17, 183, 2043, 26529, 398025, 6765975, 128556675, 2699661825, 62092533825, 1552309291575, 41912411683275, 1215458905032225, 37679245697871225, 1243414695550433175, 43519523831289457875, 1610222144582102522625
Offset: 1

Views

Author

Johannes W. Meijer, Nov 10 2009

Keywords

Crossrefs

Equals the third column of the ED3 array A167572.
Other columns are A167576 and A167577.
Cf. A007509 and A025547 (the sum((-1)^(k+n)/(2*k+1), k=0..n-1) factor).

Programs

  • Mathematica
    Table[(1/4)*(-1)^(n)*(2*n - 7)!!*((8*n^4 - 20*n^3 - 22*n^2 + 55*n + 12) + (32*n^5 - 80*n^4 - 80*n^3 + 200*n^2 + 18*n - 45)*(Sum[(-1)^(k + n)/(2*k + 1), {k, 0, n - 1}])), {n, 1, 50}] (* G. C. Greubel, Jun 16 2016 *)

Formula

a(n) = (1/4)*(-1)^(n)*(2*n-7)!!*((8*n^4-20*n^3-22*n^2+55*n+12)+(32*n^5-80*n^4-80*n^3+200*n^2+18*n-45)*(sum((-1)^(k+n)/(2*k+1),k=0..n-1))).

A167588 The second column of the ED4 array A167584.

Original entry on oeis.org

1, 6, 41, 372, 4077, 53106, 795645, 13536360, 257055705, 5400196830, 124170067665, 3104906420700, 83818724048325, 2431059231544650, 75354930324303525, 2486926158748693200, 87036225272850632625, 3220532233879435917750, 125594424461427237941625
Offset: 1

Views

Author

Johannes W. Meijer, Nov 10 2009

Keywords

Crossrefs

Equals the second column of the ED4 array A167584.
Other columns are A024199 and A167589.
Cf. A007509 and A025547 (the sum((-1)^(k+n)/(2*k+1), k=0..n-1) factor), A001147, A142970.

Programs

  • Mathematica
    Table[(1/2)*(-1)^(n)*(2*n - 3)!!*((n) + (4*n^2 - 1)*Sum[(-1)^(k + n)/(2*k + 1), {k, 0, n - 1}]), {n, 1, 50}] (* G. C. Greubel, Jun 17 2016 *)

Formula

a(n) = (1/2)*(-1)^(n)*(2*n-3)!!*(n+(4*n^2-1)*Sum_{k=0..n-1} ((-1)^(k+n)/(2*k+1))).
From Peter Bala, Nov 01 2016: (Start)
a(n) = (2*n + 1)!! * Sum_{k = 0..n-1} (-1)^(k-1)/((2*k - 1)*(2*k + 1)*(2*k + 3)).
a(n) ~ Pi * 2^(n-3/2) * ((n+1)/e)^(n+1).
E.g.f.: (4*x*sqrt(1 - 4*x^2) + 2*arcsin(2*x))/(8*(1 - 2*x)^(3/2)).
a(n) = 6*a(n-1) + (2*n - 5)*(2*n - 1)*a(n-2) with a(0) = 0, a(1) = 1.
The sequence b(n) := (2*n + 1)!! = (2*n + 2)!/((n + 1)!*2^(n+1)) satisfies the same recurrence with b(0) = 1 and b(1) = 3. This leads to the continued fraction representation a(n) = b(n)*[ 1/(3 - 3/(6 + 5/(6 + 21/(6 + ... + (2*n - 5)*(2*n - 1)/(6))))) ] for n >= 2.
As n -> infinity, a(n)/(A001147(n+1)) -> 1/2!*Pi/4 = 1/(3 - 3/(6 + 5/(6 + 21/(6 + ... + (2*n - 5)*(2*n - 1)/(6 + ...))))). Compare with the generalized continued fraction representation Pi = 3 + 1^2/(6 + 3^2/(6 + 5^2/(6 + ...))). See A142970. (End)

A167589 The third column of the ED4 array A167584.

Original entry on oeis.org

1, 10, 93, 1020, 13269, 198990, 3383145, 64276920, 1349846505, 31046064210, 776157686325, 20956154152500, 607730434609725, 18839602224969750, 621707822126431425, 21759750056864358000, 805111392478121276625
Offset: 1

Views

Author

Johannes W. Meijer, Nov 10 2009

Keywords

Crossrefs

Equals the third column of the ED4 array A167584.
Other columns are A024199 and A167588.
Cf. A007509 and A025547 (the sum((-1)^(k+n)/(2*k+1), k=0..n-1) factor), A001147.

Programs

  • Mathematica
    Table[(1/8)*(-1)^(n)*(2*n - 5)!!*((4*n^3 - 11*n) + (16*n^4 - 40*n^2 + 9)*(Sum[(-1)^(k + n)/(2*k + 1), {k, 0, n - 1}])), {n, 1, 50}] (* G. C. Greubel, Jun 17 2016 *)

Formula

a(n) = (1/8)*(-1)^n*(2*n-5)!!*((4*n^3-11*n)+(16*n^4-40*n^2+9)*(Sum_{k=0..n-1} (-1)^(k+n)/(2*k+1) ) ).
From Peter Bala, Nov 01 2016: (Start)
a(n) = 3*(2*n + 3)!! * Sum_{k = 0..n-1} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)*(2*k + 3)*(2*k + 5)).
a(n) ~ Pi*2^(n - 5/2)*((n + 2)/e)^(n + 2).
E.g.f.: (6*arcsin(2*x) + 4*x*sqrt(1 - 4*x^2)*(5 - 8*x^2))/(32*(1 - 2*x)^(5/2)).
a(n) = 10*a(n) + (2*n - 7)*(2*n + 1)*a(n-2) with a(0) = 0, a(1) = 1.
The sequence b(n) := (2*n + 3)!! = (2*n + 4)!/((n + 2)!*2^(n+2)) = A001147(n+2) satisfies the same recurrence with b(0) = 3 and b(1) = 15. This leads to the continued fraction representation a(n) = 1/3*b(n)*( 1/(5 - 15/(10 - 7/(10 + 9/(10 + 33/(10 + ... + (2*n - 7)*(2*n + 1)/(10)))))) ) for n >= 2.
As n -> infinity, 3*a(n)/(A001147(n+2)) -> 9/4!*Pi/4 = 1/(5 - 15/(10 - 7/(10 + 9/(10 + 33/(10 + ... + (2*n - 7)*(2*n + 1)/(10 + ...)))))). (End)

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

Views

Author

Wolfdieter Lang, Sep 15 2008

Keywords

Comments

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.

Examples

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

References

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

Formula

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.

A352395 Denominator of Sum_{k=0..n} (-1)^k / (2*k+1).

Original entry on oeis.org

1, 3, 15, 105, 315, 3465, 45045, 45045, 765765, 14549535, 14549535, 334639305, 1673196525, 5019589575, 145568097675, 4512611027925, 4512611027925, 4512611027925, 166966608033225, 166966608033225, 6845630929362225, 294362129962575675, 294362129962575675, 13835020108241056725
Offset: 0

Views

Author

Wolfdieter Lang, Apr 06 2022

Keywords

Comments

This is not the sequence A025547(n+1)_{n>=0}, because a(32) = 1420993851085122917681925 but A025547(33) = 18472920064106597929865025. Hence it is also not the sequence A350670.
The alternating sum Sum_{k=0..n} (-1)^k/(2*k+1) = (Psi(n + 3/2) - Psi((2*n - (-1)^n)/4 + 1) - log(2) + Pi/2)/2, with the Digamma function Psi(z).
Proof by subtracting twice the negative fractions from Sum_{k=0..n} 1/(2*k+1) = A350669(n)/A350670(n) (Abramowitz-Stegun, p. 258, eq. 6.3.4.), using Sum_{j=0..k} 1/(4*j + 3) = A074637((k+1)/4)/A074638(k+1) (Abramowitz-Stegun, p. 258, eqs. 6.3.6. with 6.3.5.) and, finally, replacing in the results for the even and odd n cases the formula for Psi(3/4) = -A200134.

Crossrefs

Cf. A007509 (numerators).

Programs

  • Mathematica
    Denominator @ Accumulate @ Table[(-1)^k/(2*k + 1), {k, 0, 25}] (* Amiram Eldar, Apr 08 2022 *)
  • PARI
    a(n) = denominator(sum(k=0, n, (-1)^k / (2*k+1))); \\ Michel Marcus, Apr 07 2022
    
  • Python
    from fractions import Fraction
    def A352395(n): return sum(Fraction(-1 if k % 2 else 1,2*k+1) for k in range(n+1)).denominator # Chai Wah Wu, May 18 2022

Formula

a(n) = denominator( (Psi(n + 3/2) - Psi((2*n - (-1)^n)/4 + 1) - log(2) + Pi/2)/2), for n >= 0, with the Digamma function. See the above comment.
a(n) = denominator(Pi/4 + (-1)^n * (Psi((n + 5/2)/2) - Psi((n + 3/2)/2))/4). - Vaclav Kotesovec, May 16 2022
Showing 1-10 of 12 results. Next