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 1-3 of 3 results.

A001539 a(n) = (4*n+1)*(4*n+3).

Original entry on oeis.org

3, 35, 99, 195, 323, 483, 675, 899, 1155, 1443, 1763, 2115, 2499, 2915, 3363, 3843, 4355, 4899, 5475, 6083, 6723, 7395, 8099, 8835, 9603, 10403, 11235, 12099, 12995, 13923, 14883, 15875, 16899, 17955, 19043, 20163, 21315, 22499, 23715, 24963, 26243, 27555, 28899
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

Sequence arises from reading the line from 3, in the direction 3, 35, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
log(sqrt(2)+1)/sqrt(2) = 0.62322524... = 2/3 - 2/35 + 2/99 - 2/195 + 2/323, ... = (1 - 1/3) + (1/7 - 1/5) + (1/9 - 1/11) + (1/15 - 1/13) + (1/17 - 1/19) + (1/23 - 1/21) + ... - Gary W. Adamson, Mar 01 2009
Numbers k such that k+1 is a square and k+5 is divisible by 8. - Bruno Berselli, Sep 27 2017
The concatenation of 8*A000217(n) and 99 is a term of the sequence. Example: for A000217(5) = 15, 8*15 = 120 and 12099 = a(27). In general, a(5*n+2) = 800*A000217(n) + 99. - Bruno Berselli, Sep 29 2017

Links

Crossrefs

Bisection of A000466.
Cf. A000217, A000290, A001533, A001538, A004767, A016286, A016813, A016826, A157142, A133766, A154633.

Programs

  • Mathematica
    CoefficientList[Series[(3 + 26 x + 3 x^2)/(1 - x)^3, {x, 0, 41}], x] (* or *) Table[(4 n + 1) (4 n + 3), {n, 0, 41}] (* Michael De Vlieger, Sep 29 2017 *)
  • Maxima
    makelist((4*n+1)*(4*n+3), n, 0, 30); /* Martin Ettl, Nov 12 2012 */
  • PARI
    a(n)=(4*n+1)*(4*n+3) \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = A016826(n) - 1 = (A001533(n)+5)/4 = (A001538(n)+16)/9.
Sum_{k>=0} 1/a(k) = Pi/8. - Benoit Cloitre, Aug 20 2002
G.f.: (3 + 26*x + 3*x^2)/(1 - x)^3. - Jaume Oliver Lafont, Mar 07 2009
a(n) = 32*n + a(n-1) for n > 0, a(0)=3. - Vincenzo Librandi, Nov 12 2010
a(n) = a(m) + 16*(n-m)*(n+m+1). The previous formula is obtained for m = n-1. - Bruno Berselli, Sep 29 2017
From Amiram Eldar, Feb 19 2023: (Start)
a(n) = A016813(n)*A004767(n).
Product_{n>=0} (1 - 1/a(n)) = sqrt(2)*cos(Pi/(2*sqrt(2))).
Product_{n>=0} (1 + 1/a(n)) = sqrt(2). (End)
From Elmo R. Oliveira, Oct 23 2024: (Start)
E.g.f.: exp(x)*(3 + 16*x*(2 + x)).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A133766 a(n) = (4*n+1)*(4*n+3)*(4*n+5).

Original entry on oeis.org

15, 315, 1287, 3315, 6783, 12075, 19575, 29667, 42735, 59163, 79335, 103635, 132447, 166155, 205143, 249795, 300495, 357627, 421575, 492723, 571455, 658155, 753207, 856995, 969903, 1092315, 1224615, 1367187, 1520415, 1684683, 1860375, 2047875, 2247567, 2459835
Offset: 0

Views

Author

Miklos Kristof, Jan 02 2008

Keywords

References

  • L. B. W. Jolley, Summation of Series, Dover, 1961.

Links

Crossrefs

Cf. A001539, A154633.

Programs

  • Maple
    seq((4*n+1)*(4*n+3)*(4*n+5),n=0..40);
  • Mathematica
    Table[c=4n;(c+1)(c+3)(c+5),{n,0,30}] (* or *) LinearRecurrence[{4,-6,4,-1},{15,315,1287,3315},30] (* Harvey P. Dale, May 06 2012 *)
  • PARI
    a(n)=(4*n+1)*(4*n+3)*(4*n+5) \\ Charles R Greathouse IV, Oct 16 2015

Formula

G.f.: 3*(5 + 85*x + 39*x^2 - x^3)/(1-x)^4 .
E.g.f: (15 + 300*x + 336*x^2 + 64*x^3)*exp(x) .
Sum_{n>=0} 4/a(n) = (Pi-2)/4. [Jolley, eq. 238]
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - Harvey P. Dale, May 06 2012
Sum_{n>=0} (-1)^n/a(n) = 1/8 + (log(2*sqrt(2)+3) - Pi)/(16*sqrt(2)). - Amiram Eldar, Feb 27 2022

A166107 A sequence related to the Madhava-Gregory-Leibniz formula for Pi.

Original entry on oeis.org

2, -10, 46, -334, 982, -10942, 140986, -425730, 7201374, -137366646, 410787198, -9473047614, 236302407090, -710245778490, 20563663645710, -638377099140510, 1912749274005030, -67020067316087550, 2477305680740159850
Offset: 0

Views

Author

Johannes W. Meijer, Oct 06 2009, Feb 26 2013, Mar 02 2013

Keywords

Comments

The EG1 matrix is defined in A162005. The first column of this matrix leads to the function PLS(z) = sum(2*eta(2*m-1)*z^(2*m-2), m=1..infinity) = 2*log(2) - Psi(z) - Psi(-z) + Psi(z/2) + Psi(-z/2). The values of this function for z=n+1/2 are related to Pi in a curious way.
Gauss's digamma theorem leads to PLS(z=n+1/2) = (-1)^n*4*sum((-1)^(k+1)/(2*k-1), k=1..n) + 2/(2*n+1). Now we define PLS(z=n+1/2) = a(n)/p(n) with a(n) the sequence given above and for p(n) we choose the esthetically nice p(n) = (2*n-1)!!/(floor((n-2)/3)*2+1)!!, n=>0. For even values of n the limit(a(2*n)/p(2*n), n=infinity) = Pi and for odd values of n the limit(a(2*n+1)/p(2*n+1), n=infinity) = - Pi. We observe that the a(n)/p(n) formulas resemble the partial sums of the Madhava-Gregory-Leibniz series for Pi = 4*(1-1/3+1/5-1/7+ ...), see the examples. The 'extra term' that appears in the a(n)/p(n) formulas, i.e., 2/(2*n+1), speeds up the convergence of abs(a(n)/p(n)) significantly. The first appearance of a digit in the decimal expansion of Pi occurs here for n: 1, 3, 9, 30, 74, 261, 876, 3056, .., cf. A126809. [Comment modified by the author, Oct 09 2009]

Examples

			The first few values of a(n)/p(n) are: a(0)/p(0) = 2/1; a(1)/p(1) = - 4*(1) + 2/3 = -10/3; a(2)/p(2) = 4*(1-1/3) + 2/5 = 46/15; a(3)/p(3) = - 4*(1-1/3+1/5) + 2/7 = - 334/105; a(4)/p(4)= 4*(1-1/3+1/5-1/7) + 2/9 = 982/315; a(5)/p(5) = - 4*(1-1/3+1/5-1/7+1/9) + 2/11 = -10942/3465; a(6)/p(6) = 4*(1-1/3+1/5-1/7+1/9-1/11) + 2/13 = 140986/45045; a(7)/p(7) = - 4*(1-1/3+1/5-1/7+1/9-1/11+1/13) + 2/15 = - 425730/135135.

Links

Crossrefs

Cf. A162005, A001147, A130823, A025547, A220747, A000796, A157142, A126809, A133766, A133767, A154633.

Programs

  • Maple
    A166107 := n -> A220747 (n)*((-1)^n*4*sum((-1)^(k+1)/(2*k-1), k=1..n) + 2/(2*n+1)): A130823 := n -> floor((n-1)/3)*2+1: A220747 := n -> doublefactorial(2*n+1) / doublefactorial(A130823(n)): seq(A166107(n), n=0..20);

Formula

a(n) = p(n)*(-1)^n*4*sum((-1)^(k+1)/(2*k-1), k=1..n) + 2/(2*n+1) with
p(n) = doublefactorial(2*n+1)/doublefactorial(floor((n-1)/3)*2+1) = A220747(n)
PLS(z) = 2*log(2) - Psi(z) - Psi(-z) + Psi(z/2) + Psi(-z/2)
PLS(z=n+1/2) = a(n)/p(n) = (-1)^n*4*sum((-1)^(k+1)/(2*k-1), k=1..n) + 2/(2*n+1)
PLS(z=2*n+5/2) - PLS(z=2*n+1/2) = 2/(4*n+5) - 4/(4*n+3) + 2/(4*n+1) which leads to:
Pi = 2 + 16 * sum(1/((4*n+5)*(4*n+3)*(4*n+1)), n=0 .. infinity).
PLS (z=2*n +7/2) - PLS(z=2*n+3/2) = 2/(4*n+7) - 4/(4*n+5) + 2/(4*n+3) which leads to:
Pi = 10/3 - 16*sum(1/((4*n+7)*(4*n+5)*(4*n+3)), n=0 .. infinity).
The combination of these two formulas leads to:
Pi = 8/3 + 48* sum(1/((4*n+7)*(4*n+5)*(4*n+3)*(4*n+1)), n=0 .. infinity).
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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