A115563 -id:A115563 - cp's OEIS frontend

A137245 Decimal expansion of Sum_{p prime} 1/(p * log p).

1, 6, 3, 6, 6, 1, 6, 3, 2, 3, 3, 5, 1, 2, 6, 0, 8, 6, 8, 5, 6, 9, 6, 5, 8, 0, 0, 3, 9, 2, 1, 8, 6, 3, 6, 7, 1, 1, 8, 1, 5, 9, 7, 0, 7, 6, 1, 3, 1, 2, 9, 3, 0, 5, 8, 6, 0, 0, 3, 0, 4, 9, 1, 9, 7, 8, 1, 3, 3, 9, 9, 7, 4, 4, 6, 7, 9, 4, 6, 9, 8, 6, 5, 4, 7, 0, 0, 4, 0, 3, 8, 5, 2, 5, 5, 8, 4, 7, 9, 8, 9, 8, 9, 4, 4

Offset: 1

R. J. Mathar, Mar 09 2008

Sum_{p prime} 1/(p^s * log p) equals this value here if s=1, equals A221711 if s=2, 0.22120334039... if s=3. See arXiv:0811.4739.
Erdős (1935) proved that for any sequence where no term divides another, the sum of 1/(x log x) is at most some constant C. He conjectures (1989) that C can be taken to be this constant 1.636..., that is, the primes maximize this sum. - Charles R Greathouse IV, Mar 26 2012 [The conjecture has been proved by Lichtman 2022. - Pontus von Brömssen, Jun 23 2022]
Note that sum 1/(p * log p) is almost (a tiny bit less than) 1 + 2/Pi = 1+A060294 = 1.63661977236758... (Why is it so close?) - Daniel Forgues, Mar 26 2012
Sum 1/(p * log p) is quite close to sum 1/n^2 = Pi^2/6 = 1.644934066... (Cf. David C. Ullrich, "Re: What is Sum(1/p log p)?" for why this is so; mentions A115563.) - Daniel Forgues, Aug 13 2012

			1.63661632335...

Henri Cohen, Number Theory, Volume II: Analytic and Modern Tools, GTM Vol. 240, Springer, 2007; see pp. 208-209.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 2.27.2, p. 186.

Karim Belabas and Henri Cohen, Computation of sum_{p prime} 1/(p^s log(p)), PARI/GP script, 2020.
Henri Cohen, High-precision computation of Hardy-Littlewood constants, (1998).
Henri Cohen, High-precision computation of Hardy-Littlewood constants. [pdf copy, with permission]
P. Erdős, Note on sequences of integers no one of which is divisible by any other, J. London Math. Soc. 10 (1935), pp. 126-128, [DOI].
P. Erdős, Some problems and results on combinatorial number theory, Graph theory and its applications: East and West (Jinan, 1986), Ann. New York Acad. Sci., 576, pp. 132-145, New York Acad. Sci., New York, 1989.
Brady Haran and Jared Duker Lichtman, Primes and Primitive Sets, Numberphile video (2022).
Jared Duker Lichtman, Almost primes and the Banks-Martin conjecture, arXiv:1909.00804 [math.NT], 2019.
Jared Duker Lichtman, A proof of the Erdős primitive set conjecture, arXiv:2202.02384 [math.NT], 2022.
Jared Duker Lichtman, Proving the Erdős Primitive Set Conjecture, Oxford Mathematics YouTube video, 2022.
Jared Duker Lichtman, A proof of the Erdős primitive set conjecture, YouTube video from Combinatorial and additive number theory conference (CANT), 2022.
R. J. Mathar, Twenty digits of some integrals of the prime zeta function, arXiv:0811.4739 [math.NT], 2008-2009, table in Section 2.4.
David C. Ullrich, Re: What is Sum(1/p log p)?, posting in newsgroup sci.math.research, 11 Feb 2006.

Cf. A000040, A060294, A221711 (p squared), A115563, A319231 (log squared), A319232 (p and log squared), A354952.

digits = 105;
precision = digits + 10;
tmax = 500; (* integrand considered negligible beyond tmax *)
kmax = 500; (* f(k) considered negligible beyond kmax *)
InLogZeta[k_] := NIntegrate[Log[Zeta[t]], {t, k, tmax}, WorkingPrecision -> precision, MaxRecursion -> 20];
f[k_] := With[{mu = MoebiusMu[k]}, If[mu == 0, 0, (mu/k^2)*InLogZeta[k]]];
s = 0;
Do[s = s + f[k]; Print[k, " ", s], {k, 1, kmax}];
RealDigits[s][[1]][[1 ;; digits]] (* Jean-François Alcover, Feb 06 2021, updated Jun 22 2022 *)

\\ See Belabas, Cohen link. Run as SumEulerlog(1) after setting the required precision.

default(realprecision, 200); s=0; for(k=1, 500, s = s + moebius(k)/k^2 * intnum(x=k,[[1], 1], log(zeta(x))); print(s)); \\ Vaclav Kotesovec, Jun 12 2022

Equals Sum_{n>=1} 1/(A000040(n)*log A000040(n)).

More terms from Hugo Pfoertner, Feb 01 2020

More terms from Vaclav Kotesovec, Jun 12 2022

A257812 Decimal expansion of Sum_{n>=2} (-1)^n/(n*log(n)).

Original entry on oeis.org

5, 2, 6, 4, 1, 2, 2, 4, 6, 5, 3, 3, 3, 1, 0, 4, 1, 0, 9, 3, 0, 6, 9, 6, 5, 0, 1, 4, 1, 1, 1, 3, 1, 4, 1, 3, 7, 2, 1, 7, 9, 0, 5, 9, 7, 8, 8, 7, 5, 5, 8, 5, 4, 0, 7, 4, 6, 9, 9, 5, 7, 0, 0, 8, 3, 3, 7, 8, 3, 2, 2, 3, 1, 3, 0, 2, 0, 8, 4, 4, 6, 9, 8, 4, 6, 3, 6, 2, 2, 7, 2, 9, 7, 3, 4, 6, 1, 5, 1, 7, 8, 8, 7, 6, 4, 9, 5, 5, 8

Offset: 0

Iaroslav V. Blagouchine, May 10 2015

This alternating series converges quite slowly. However, it can be efficiently computed via its integral representation (see formula below), which converges exponentially quickly. This formula and PARI were used to compute 1000 digits.

			0.5264122465333104109306965014111314137217905978875585...

Iaroslav V. Blagouchine, Table of n, a(n) for n = 0..1000

Cf. A099769, A115563, A257898, A257960, A257964, A257972, A257837.

evalf(sum((-1)^n/(n*log(n)), n=2..infinity), 120);
evalf(1/(4*log(2))+2*(Int((2*arctan(x)+x*log(4+4*x^2))/(sinh(2*Pi*x)*(log(4+4*x^2)^2+4*arctan(x)^2)*(x^2+1)), x=0..infinity)), 120);

NSum[(-1)^n/(n*Log[n]), {n, 2, Infinity}, AccuracyGoal -> 120, WorkingPrecision -> 200, Method -> AlternatingSigns]
1/(4*Log[2])+2*NIntegrate[(2*ArcTan[x]+x*Log[4+4*x^2])/((x^2+1)*Sinh[2*Pi*x]*(Log[4+4*x^2]^2+4*ArcTan[x]^2)), {x, 0,Infinity}, WorkingPrecision->120]

default(realprecision,120); sumalt(n=2, (-1)^n/(n*log(n))) \\ Vaclav Kotesovec, May 10 2015

allocatemem(50000000);
default(realprecision, 1200); 1/(4*log(2))+2*intnum(x=0, 1000, (2*atan(x)+x*log(4+4*x^2))/(sinh(2*Pi*x)*(log(4+4*x^2)^2+4*atan(x)^2)*(x^2+1)))

Equals 1/(4*log(2)) + 2*Integral_{x=0..oo} (2*arctan(x)+x*log(4+4*x^2))/(sinh(2*Pi*x)*(log(4+4*x^2)^2+4*arctan(x)^2)*(x^2+1)) dx.

A319231 Decimal expansion of Sum_{p = prime} 1/(p*log(p)^2).

Original entry on oeis.org

1, 5, 2, 0, 9, 7, 0, 4, 3, 9, 9, 3, 9, 5, 0, 0, 8, 6, 3, 4, 6, 1, 4, 2, 8, 6, 2, 8, 6, 1, 5, 5, 7, 9, 5, 2, 1, 9, 5, 6, 8, 4, 6, 1, 6, 7, 7, 6, 8, 3, 5, 0, 1, 1, 0, 6, 5, 5, 5, 2, 7, 5, 3, 5, 9, 6, 3, 4, 1, 0, 6, 4, 4, 3, 1, 0, 4, 1, 0, 4, 7, 2, 0, 6, 6, 3, 0, 7, 6, 1, 9, 5, 2, 2, 5, 2, 7, 5, 1, 3, 3, 4, 4, 6, 0

Offset: 1

R. J. Mathar, Sep 14 2018

Computed by expanding the formalism of arXiv:0811.4739 to double integrals over the Riemann zeta function.

			1/(2*A253191) + 1/(3*A175478) +1/(5*2.59029...) +1/(7*3.7865)+ ... = 1.52097043...

R. J. Mathar, Twenty digits of some integrals of the prime zeta function, arXiv:0811.4739 [math.NT], 2008-2018.

Cf. A137245, A115563, A221711, A319232, A354917.

digits = 105; precision = digits + 10;
tmax = 500; (* integrand considered negligible beyond tmax *)
kmax = 500; (* f(k) considered negligible beyond kmax *)
InLogZeta[k_] := NIntegrate[(t - k) Log[Zeta[t]], {t, k, tmax}, WorkingPrecision -> precision, MaxRecursion -> 20, AccuracyGoal -> precision];
f[k_] := With[{mu = MoebiusMu[k]}, If[mu == 0, 0, (mu/k^3)*InLogZeta[k]]];
s = 0;
Do[s = s + f[k]; Print[k, " ", s], {k, 1, kmax}];
RealDigits[s][[1]][[1 ;; digits]] (* Jean-François Alcover, Jun 21 2022, after Vaclav Kotesovec *)

default(realprecision, 200); s=0; for(k=1, 500, s=s+moebius(k)/k^3 * intnum(x=k,[[1], 1],(x-k)*log(zeta(x))); print(s)); \\ Vaclav Kotesovec, Jun 12 2022

More digits from Vaclav Kotesovec, Jun 12 2022

A319232 Decimal expansion of Sum_{p = prime} 1/(p*log p)^2.

Original entry on oeis.org

6, 3, 7, 0, 5, 6, 1, 8, 4, 0, 7, 4, 6, 7, 6, 4, 3, 3, 0, 5, 9, 9, 6, 8, 5, 8, 5, 0, 4, 7, 8, 5, 2, 7, 6, 9, 4, 5, 7, 9, 8, 9, 6, 0, 7, 7, 1, 9, 9, 5, 3, 3, 6, 7, 0, 9, 6, 0, 1, 3, 7, 1, 0, 7, 5, 5, 8, 8, 3, 1, 6, 0, 4, 3, 3, 2, 7, 1, 5, 1, 6, 8, 3, 6, 7, 5, 3, 8, 3, 5, 9, 6, 6, 1, 3, 3, 1, 8, 1, 3, 1, 3, 8, 2, 7, 5

Offset: 0

R. J. Mathar, Sep 14 2018

Obtained by expanding the formalism of arXiv:0811.4739 to double integrals over the Riemann zeta function.

			1/A016627^2 + 1/A016650^2 + 1/8.047189^2 + ... = 0.637056184074676....

R. J. Mathar, Twenty digits of some integrals of the prime zeta function, arXiv:0811.4739 (2008-2009).

Cf. A137245, A115563, A221711, A319231.

digits = 106; precision = digits + 10;
tmax = 500; (* integrand considered negligible beyond tmax *)
kmax = 300; (* f(k) considered negligible beyond kmax *)
InLogZeta[k_] := NIntegrate[(t - 2k) Log[Zeta[t]], {t, 2k, tmax}, WorkingPrecision -> precision, MaxRecursion -> 20, AccuracyGoal -> precision];
f[k_] := With[{mu = MoebiusMu[k]}, If[mu == 0, 0, (mu/k^3)*InLogZeta[k]]];
s = 0;
Do[s = s + f[k]; Print[k, " ", s], {k, 1, kmax}];
RealDigits[s][[1]][[1 ;; digits]] (* Jean-François Alcover, Jun 21 2022, after Vaclav Kotesovec *)

default(realprecision, 200); s=0; for(k=1, 300, s = s + moebius(k)/k^3 * intnum(x=2*k,[[1], 1], (x-2*k)*log(zeta(x))); print(s)); \\ Vaclav Kotesovec, Jun 12 2022

More terms from Vaclav Kotesovec, Jun 12 2022

A145419 Decimal expansion of Sum_{k>=2} 1/(k*(log k)^3).

Original entry on oeis.org

2, 0, 6, 5, 8, 8, 6, 5, 3, 8, 8, 8, 4, 1, 3, 5, 2, 5, 0, 9, 0, 3, 1, 4, 2, 2, 4, 1, 6, 4, 3, 7, 7, 3, 8, 1, 8, 0, 8, 6, 9, 7, 5, 2, 0, 6, 9, 3, 8, 3, 4, 7, 0, 7, 3, 4, 6, 3, 2, 4, 3, 6, 0, 2, 4, 1, 6, 8, 0, 7, 4, 0, 1, 3, 7, 7, 6, 5, 1, 5, 8, 6, 5, 5, 2, 6, 7, 3, 8, 2, 7, 3, 1, 4, 3, 0, 1, 3, 8, 8, 7, 7, 1, 8, 8

Offset: 1

R. J. Mathar, Feb 08 2009

Cubic analog of A115563. Evaluated by direct summation of the first 160 terms and accumulating the remainder with the 5 nontrivial terms in the Euler-Maclaurin expansion.

Theorem: Bertrand series Sum_{n>=2} 1/(n*log(n)^q) is convergent iff q > 1 (for q = 2, 4, 5 see respectively A115563, A145420, A145421). - Bernard Schott, Oct 23 2021

			2.0658865388841352509031422416437738180869752069383...

R. J. Mathar, The series limit of sum_k 1/[k log k (log log k)^2], arXiv:0902.0789 [math.NA], 2009-2021, constant D^(3).
Wikipédia, Série de Bertrand (in French).

Cf. A115563, A145420, A145421, A354917.

digits = 50; NSum[ 1/(n*Log[n]^3), {n, 2, Infinity}, NSumTerms -> 10000, WorkingPrecision -> digits + 10] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Feb 11 2013 *)
alfa = 3; maxiter = 20; nn = 10000; bas = Sum[1/(k*Log[k]^alfa), {k, 2, nn}] + 1/((alfa - 1)*Log[nn + 1/2]^(alfa - 1)); sub = 0; Do[sub = sub + 1/4^s/(2*s + 1)! * NSum[(D[1/(x*Log[x]^alfa), {x, 2 s}]) /. x -> k, {k, nn + 1, Infinity}, WorkingPrecision -> 120, NSumTerms -> 100000, PrecisionGoal -> 120, Method -> {"NIntegrate", "MaxRecursion" -> 100}]; Print[bas - sub], {s, 1, maxiter}] (* Vaclav Kotesovec, Jun 11 2022 *)

More terms from Jean-François Alcover, Feb 11 2013

More digits from Vaclav Kotesovec, Jun 11 2022

A145420 Decimal expansion of Sum_{k>=2} 1/(k*(log k)^4).

Original entry on oeis.org

2, 5, 5, 9, 1, 1, 9, 7, 4, 2, 9, 8, 6, 7, 3, 1, 4, 1, 8, 5, 7, 2, 0, 2, 0, 9, 7, 0, 3, 1, 0, 7, 6, 2, 9, 3, 3, 6, 1, 9, 1, 7, 8, 1, 5, 6, 3, 6, 6, 8, 7, 9, 4, 8, 7, 1, 7, 0, 6, 7, 9, 7, 0, 7, 9, 1, 4, 6, 5, 9, 0, 9, 8, 1, 6, 6, 1, 7, 1, 7, 6, 6, 5, 9, 3, 7, 9, 5, 9, 9, 2, 4, 9, 0, 3, 2, 1, 3, 8, 3, 5, 5, 4, 5, 8

Offset: 1

R. J. Mathar, Feb 08 2009

Quartic analog of A115563. Evaluated by direct summation of the first 160 terms and accumulating the remainder with the 5 nontrivial terms in the Euler-Maclaurin expansion.

Bertrand series Sum_{n>=2} 1/(n*log(n)^q) is convergent iff q > 1. - Bernard Schott, Jan 22 2022

			2.5591197429867314185720209703107629336191781563668...

R. J. Mathar, The series limit of sum_k 1/[k log k (log log k)^2], arXiv:0902.0789 [math.NA], 2009-2021, last sentence.
Wikipédia, Série de Bertrand (in French).

Cf. A115563 (q=2), A145419 (q=3), A145421 (q=5).

(* Computation needs a few minutes *) digits = 105; NSum[ 1/(n*Log[n]^4), {n, 2, Infinity}, NSumTerms -> 800000, WorkingPrecision -> digits + 5, Method -> {"EulerMaclaurin", Method -> {"NIntegrate", "MaxRecursion" -> 10}}] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Feb 12 2013 *)

More terms from Jean-François Alcover, Feb 12 2013

A168218 Decimal expansion of the Sum_{k=2..infinity} 1/(k^2*log(k)).

Original entry on oeis.org

6, 0, 5, 5, 2, 1, 7, 8, 8, 8, 8, 2, 6, 0, 0, 4, 4, 7, 6, 9, 9, 5, 4, 9, 0, 0, 5, 2, 0, 7, 2, 4, 0, 4, 4, 7, 3, 0, 3, 2, 3, 8, 8, 9, 8, 4, 5, 5, 0, 6, 5, 7, 8, 3, 3, 1, 1, 1, 4, 7, 5, 9, 0, 4, 2, 0, 6, 8, 9, 4, 1, 1, 9, 7, 8, 0, 8, 8, 6, 8, 1, 7, 6, 1, 1, 8, 3, 1, 2, 8, 4, 1, 9, 3, 0, 8, 9, 4, 0, 1, 9, 8, 6, 9, 9

Offset: 0

R. J. Mathar, Nov 20 2009

Also the value of the integral of the fractional part of the Riemann zeta function from 2 to infinity. - Alexander Bock, Apr 01 2014

			equals 1/(4*log(2))+1/(9*log(3))+1/(16*log(4))+ .... + = 0.605521788882600447699549005207240447303238898..

R. P. Boas, Jr, Partial sums of Infinite Series, and How They Grow, Am. Math. Monthly 84 (4) (1977) 237-235 [MR0440240].

Cf. A099769, A115563, A257812.

(* Computation needs a few minutes for 105 digits *) digits = 105; NSum[ 1/(n^2*Log[n]), {n, 2, Infinity}, NSumTerms -> 500000, WorkingPrecision -> digits + 5, Method -> {"EulerMaclaurin", Method -> {"NIntegrate", "MaxRecursion" -> 12}}] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Feb 11 2013 *)
RealDigits[NIntegrate[Zeta[x] - 1, {x, 2, Infinity}, WorkingPrecision->110], 10, 100] (* Alexander Bock, Apr 01 2014 *)

intnum(x=2,[oo,log(2)],zeta(x)-1) \\ Charles R Greathouse IV, Apr 01 2014

suminf(k=2,1/k^2/log(k)) \\ Charles R Greathouse IV, Apr 01 2014

Equals Integral_{x>=2} (zeta(x) - 1) dx.

More terms from Jean-François Alcover, Feb 11 2013

A097906 Decimal expansion of Sum_{i >= 1} i/prime(i)^2.

Original entry on oeis.org

1, 1, 4, 9, 0, 6, 4, 1, 7

Offset: 1

Pierre CAMI, Sep 04 2004, Dec 04 2014

By the prime number theorem the growth of n/prime(n)^2 is of the same order as 1/(n*log(n)^2). So the convergence of this series is as slow as the convergence of A115563. The current vague estimate is derived solely from the brute force accumulation of the partial sums (see Examples).
For the first 10000000 primes the sum = 1.0949524507.
For the first 20000000 primes the sum = 1.0970205319.
For the first 30000000 primes the sum = 1.0981565164.
For the first 40000000 primes the sum = 1.0989320390.
For the first 50000000 primes the sum = 1.0995170340.
For the first 60000000 primes the sum = 1.0999846910.
For the first 70000000 primes the sum = 1.1003730572.
For the first 80000000 primes the sum = 1.1007044057.
For the first 90000000 primes the sum = 1.1009928516.
For the first 100000000 primes the sum = 1.1012478922.
However, this does not take into account the non-negligible contribution from the tail.
From Charles R Greathouse IV, Dec 04 2014: (Start)
My current estimate of the constant is 1.1490642... based on a summation of the primes up to 10^10, 1.10463086777...
plus the integral from pi(10^10) + 1/2 to infinity of x/ali(x)^2 where ali(x) is the inverse logarithmic integral, which contributes 0.04443338335... (End)

			1.14906417...

Cf. A115563, A085548.

Corrected and extended by Charles R Greathouse IV, Dec 04 2014

Extended by David Broadhurst, Dec 07 2014

A145421 Decimal expansion of Sum_{k>=2} 1/(k*(log k)^5).

Original entry on oeis.org

3, 4, 2, 9, 8, 1, 6, 2, 6, 0, 0, 2, 3, 0, 5, 6, 0, 6, 5, 0, 2, 2, 4, 1, 1, 5, 8, 5, 6, 5, 5, 8, 6, 0, 5, 4, 0, 4, 5, 2, 3, 7, 6, 2, 0, 0, 1, 2, 0, 7, 1, 0, 3, 8, 9, 8, 4, 8, 2, 0, 0, 5, 2, 0, 9, 7, 4, 0, 4, 4, 4, 2, 8, 3, 5, 9, 4, 8, 1, 6, 1, 2, 1, 1, 8, 7, 4, 1, 9, 7, 2, 3, 8, 7, 3, 5, 3, 4, 5, 1, 6, 7, 7, 4, 2

Offset: 1

R. J. Mathar, Feb 08 2009

Quintic analog of A115563. Evaluated by direct summation of the first 160 terms and accumulating the remainder with the 5 nontrivial terms in the Euler-Maclaurin expansion.

Bertrand series Sum_{n>=2} 1/(n*log(n)^q) is convergent iff q > 1. - Bernard Schott, Feb 08 2022

			3.4298162600230560650224115856558605404523762001207...

Wikipédia, Série de Bertrand (in French).

Cf. A115563 (q=2), A145419 (q=3), A145420 (q=4).

(* Computation needs a few minutes *) digits = 105; NSum[ 1/(n*Log[n]^5), {n, 2, Infinity}, NSumTerms -> 1500000, WorkingPrecision -> digits + 10, Method -> {"EulerMaclaurin", Method -> {"NIntegrate", "MaxRecursion" -> 10}}] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Feb 12 2013 *)

More terms from Jean-François Alcover, Feb 12 2013

A118582 Decimal expansion of Sum_{k>=3} 1/(k log k (log log k)^2).

Original entry on oeis.org

3, 8, 4, 0, 6, 7, 6, 8, 0, 9, 2, 8, 2, 1, 7

Offset: 2

Robert G. Wilson v, May 16 2006

"The series [see title] converges to 38.43... so slowly that it requires 10^(3.14*10^86) terms to give two-decimal accuracy"

example 16 - "The series Sum_{k=3..inf} 1/(k log k (log log k)) diverges, but the partial sums exceed 10 only after a googolplex of terms have appeared"

			38.4067...

Daniel Zwillinger, Editor, CRC Standard Mathematical Tables and Formulae, 31st Edition, Chapman & Hall/CRC, Boca Raton, 1.3.9 Miscellaneous Sums and Series, example 15, page 42, 2003.

R. J. Mathar, The series limit of sum_k 1/[k log k (log log k)^2], arXiv:0902.0789 [math.NA], 2009-2021.
Eric Weisstein's World of Mathematics, Convergent Series

Cf. A115563.

(* Computation needs a few minutes *) digits = 15; m0 = 10^6; dm = 10^5; Clear[f]; f[m_] := f[m] = Sum[ 1/(k*Log[k]*Log[Log[k]]^2) // N[#, digits+2]&, {k, 3, m}] + 1/Log[Log[m + 1/2]] // RealDigits[#, 10, digits+2]& // First; f[m0]; f[m = m0 + dm]; While[f[m] != f[m - dm], m = m + dm]; f[m][[1 ;; digits]] (* Jean-François Alcover, Mar 07 2013 *)

Corrected the least significant digit and added 11 more digits. R. J. Mathar, Feb 03 2009

Name spelling and 3 least significant digits corrected by R. J. Mathar, Jul 07 2009

cp's OEIS Frontend

A137245 Decimal expansion of Sum_{p prime} 1/(p * log p).

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A257812 Decimal expansion of Sum_{n>=2} (-1)^n/(n*log(n)).

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A319231 Decimal expansion of Sum_{p = prime} 1/(p*log(p)^2).

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A319232 Decimal expansion of Sum_{p = prime} 1/(p*log p)^2.

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A145419 Decimal expansion of Sum_{k>=2} 1/(k*(log k)^3).

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A145420 Decimal expansion of Sum_{k>=2} 1/(k*(log k)^4).

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A168218 Decimal expansion of the Sum_{k=2..infinity} 1/(k^2*log(k)).

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