A126809 -id:A126809 - cp's OEIS frontend

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

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

Offset: 0

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

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]

			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.

Frits Beukers, A rational approach to Pi, Nieuw Archief voor de Wiskunde, December 2000, pp. 372-379.
Peter Borwein, The amazing number Pi, Nieuw Archief voor de Wiskunde, September 2000, pp. 254-258.
Johannes W. Meijer, A modified Madhava-Gregory-Leibniz formula for Pi, Mar 02 2013.
Pacific Institute for the Mathematical Sciences, Pi in the Sky magazine, 2000-2013.
Wislawa Szymborska, The admirable number Pi, Miracle Fair, 2002.
Eric. W. Weisstein, Gauss's Digamma Theorem., from Wolfram MathWorld.
Wikipedia, Madhava of Sangamagrama.

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

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);

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).

A274982 a(n) is the number of terms required in the Basel Problem, i.e., Sum_{m >= 1} 1/m^2, for the first appearance of n correct digits in the decimal expansion of Pi^2/6 to occur.

Original entry on oeis.org

1, 22, 203, 1071, 29354, 245891, 14959260, 14959260, 146023209, 1178930480, 20735515065, 121559317130, 4416249685106, 37826529360487, 155364605873808, 2291095531474075, 27417981382118579, 154501831890087986, 2116782166626093033, 13809261875873749757

Offset: 1

G. L. Honaker, Jr., Sep 23 2016

a(n) = round(1/(floor((1/6)Pi^2 * 10^(n-1))/10^(n-1))) for all n up to at least n=1000 (and it can be shown that this formula almost certainly holds for all n beyond that; see A126809 for a similar problem). - Jon E. Schoenfield, Nov 06 2016, Nov 12 2016

			a(2) = 22 because 22 terms (Sum_{m = 1..22} 1/m^2) are required for the first two decimal digits of Pi^2/6 to occur for the first time.

Jon E. Schoenfield, Table of n, a(n) for n = 1..1000
Ed Sandifer, How Euler Did It: Estimating the Basel Problem, MAA Online (2003).

Cf. A013661, A126809.

use ntheory ":all"; use bignum try=>"GMP"; my ($dig,$sum,$exp) = (0, 0, (Pi(40)**2)/6); $exp =~ s/\.//; for my $m (1 .. 1e9) { $sum += 1/($m*$m); (my $str = $sum) =~ s/\.//; print ++$dig, " $m\n" while length($str) > $dig && index($exp, substr($str,0,$dig+1)) == 0; } # Dana Jacobsen, Sep 29 2016

a(7)-a(11) from Dana Jacobsen, Oct 03 2016

A342584 Minimum k such that all partial sums of the Leibniz series 4/1 - 4/3 + 4/5 - ... with k or more terms give a value of Pi correct to n decimal digits.

Original entry on oeis.org

7, 25, 627, 2454, 136120, 376847, 2886750, 21546984, 278567575, 2437795018, 97974268952, 4836489478578, 4836489478578, 147895359776636, 308788493220129, 4193528956200935, 25999253094360135, 650467164953053602, 2161492060929047665, 26769019461318409710

Offset: 1

Ben Whitmore, Mar 16 2021

Let s(k) be the sum of the first k terms of the Leibniz series, and let eps(k) = |Pi-floor(Pi*10^(k-1))/10^(k-1)|. Then a(n) is either the minimum k such that |Pi-s(k)| < Min(eps(n), 1/10^(n-1)-eps(n)), or one less than the minimum such k.
a(n) is the minimum k such that the k-th and k+1-th partial sums agree with Pi in the first n decimal digits.
It is easy to calculate a(n) for large n because the partial sums of the Leibniz series can be expressed in terms of the digamma function, which can be computed efficiently using an asymptotic series.

			a(1) = 7 because the sum of the first 6 terms is approximately 2.9706, while the sum of the first 7 terms is approximately 3.2837, and all partial sums of more than 7 terms also have 3 as the first digit.
a(20) = 26769019461318409710 because the sum of the first 26769019461318409709 terms is approximately 3.1415926535897932385000000000000000000007642, while the sum of the first 26769019461318409710 terms is approximately 3.14159265358979323842528, which agrees with Pi in the first 20 digits, as do all partial sums with more terms.

Ben Whitmore, Table of n, a(n) for n = 1..500

Cf. A000796, A126809.

a[n_]:=Module[{eps,num,check,pi},
Block[{$MaxExtraPrecision=Infinity},
  pi[k_]:=If[EvenQ[k],
   Pi+PolyGamma[0,N[1/4,n+30]+k/2]-PolyGamma[0,3/4+k/2],
   pi[k+1]+4/(2k+1)
  ];
  eps=Min[#,10^-(n-1)-#]&[Abs[Pi-Floor[Pi 10^(n-1)]/10^(n-1)]];
  num=2Ceiling[k/2]/.FindRoot[PolyGamma[k/2+3/4]-PolyGamma[k/2+1/4]==eps,
   {k,10^(n-1)},WorkingPrecision->2n+30];
  check[a_]:=RealDigits[pi[a],10,n][[1,-1]]==RealDigits[Pi,10,n][[1,-1]];
  num-3+Position[check/@Range[num-3,num],False][[-1,1]]
]
];
a/@Range[20]

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

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A274982 a(n) is the number of terms required in the Basel Problem, i.e., Sum_{m >= 1} 1/m^2, for the first appearance of n correct digits in the decimal expansion of Pi^2/6 to occur.

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A342584 Minimum k such that all partial sums of the Leibniz series 4/1 - 4/3 + 4/5 - ... with k or more terms give a value of Pi correct to n decimal digits.

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