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.

Previous Showing 21-26 of 26 results.

A281974 Squarefree numbers excluding twin primes.

Original entry on oeis.org

1, 2, 6, 10, 14, 15, 21, 22, 23, 26, 30, 33, 34, 35, 37, 38, 39, 42, 46, 47, 51, 53, 55, 57, 58, 62, 65, 66, 67, 69, 70, 74, 77, 78, 79, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 102, 105, 106, 110, 111, 113, 114, 115, 118, 119, 122, 123
Offset: 1

Views

Author

Dimitris Valianatos, Feb 03 2017

Keywords

Examples

			30 is a term because: 30 = 2*3*5 is squarefree and is not a member of a twin primes pair.

Crossrefs

Cf. A005117 (squarefree numbers), A001097 (twin primes), A065421 (Brun's constant).

Programs

  • Mathematica
    Select[Range@ 123, SquareFreeQ@ # && ! And[PrimeQ@ #, Total@ Boole@ PrimeQ[# + {-2, 2}] >= 1] &] (* Michael De Vlieger, Feb 04 2017 *)
  • PARI
    {
for(n=1,1000,
     mb=moebius(n);
     if(mb<>0,
         if(isprime(n),
             if(!isprime(n-2)&&!isprime(n+2),
                 print1(n", ")
               ),
              ;print1(n", ")
           )
       )
    )
}

A306759 Decimal expansion of the sum of reciprocals of Brazilian primes, also called the Brazilian primes constant.

Original entry on oeis.org

3, 3, 1, 7, 5, 4, 4, 6, 6
Offset: 0

Views

Author

Bernard Schott, Mar 08 2019

Keywords

Comments

The name "constant of Brazilian primes" is used in the article "Les nombres brésiliens" in link, théorème 4, page 36. Brazilian primes are in A085104.
Let S(k) be the sum of reciprocals of Brazilian primes < k. These values below come from different calculations by Jon, Michel, Daniel and Davis.
q S(10^q)
== ========================
1 0.1428571428571428571... (= 1/7)
2 0.2889927283868234859...
3 0.3229022355626914481...
4 0.3295236806353669357...
5 0.3312171311946179843...
6 0.3316038696349217289...
7 0.3317139158654747333...
8 0.3317434191078170412...
9 0.3317513267394988538...
10 0.3317535651668937256...
11 0.3317542057931842329...
12 0.3317543906772274268...
13 0.3317544444033188051...
14 0.3317544601136967527...
15 0.3317544647354485208...
16 0.3317544661014868080...
17 0.3317544665073451951...
18 0.3317544666282877863...
19 0.3317544666644601817...
20 0.3317544666753095766...
According to the Goormaghtigh conjecture, there are only two Brazilian primes which are twice Brazilian: 31 = (111)_5 = (11111)_2 and 8191 = (111)_90 = (1111111111111)_2. The reciprocals of these two numbers are counted only once in the sum.

Examples

			1/7 + 1/13 + 1/31 + 1/43 + 1/73 + 1/127 + 1/157 + ... = 0.33175...

References

  • Daniel Lignon, Dictionnaire de (presque) tous les nombres entiers, Ellipses, Paris, 2012, page 175.

Links

Crossrefs

Cf. A085104 (Brazilian primes), A002383 (Brazilian primes (111)_b), A225148 (Brazilian primes of the form (b^q-1)/(b-1) with q prime >= 5).
Cf. A173898 (sum of the reciprocals of the Mersenne primes), A065421 (Brun's constant).

Programs

  • PARI
    brazil(N, L=List())=forprime(K=3, #binary(N+1)-1, for(n=2, sqrtnint(N-1, K-1), if(isprime((n^K-1)/(n-1)),listput(L, (n^K-1)/(n-1))))); Set(L);
brazilcons(lim,nbd) = r=brazil(10^lim); x=sum(M=1, #r, 1./r[M]);for(n=1, nbd, print1(floor(x*10^n)%10, ", "));\\ Davis Smith, Mar 10 2019
  • PARI
    cons(lim)=my(v=List(), t, k); for(n=2, sqrt(lim), t=1+n; k=1; while((t+=n^k++)<=lim, if(isprime(t), listput(v, t)))); v = vecsort(Vec(v), , 8); sum(k=1, #v, 1./v[k]); \\ Michel Marcus, Mar 11 2019

Formula

Equals Sum_{n>=1} 1/A085104(n).

A316189 Decimal expansion of Sum(1/p + 1/q) as (p, q) runs through the twin m^2 + 1 primes.

Original entry on oeis.org

3, 5, 7, 7, 4, 5, 1, 4, 7
Offset: 0

Views

Author

Michel Lagneau, Jun 26 2018

Keywords

Comments

Or decimal expansion of (1/5 + 1/17) + Sum_{i>=0} (1/p(i) + 1/q(i)) where p(i) and q(i) are primes of the form p(i) = m^2 + 1 = (10*i+4)^2 + 1 and q(i) = (m + 2)^2 + 1 = (10*i + 6)^2 + 1 (for m > 1, m == 4 (mod 10)). See A096012.
The sum is convergent; it must be less than 0.81459657... (see A172168).
Conjecture: the series of all twin m^2 + 1 prime reciprocals converges to 0.357745147...
It is probable that a(9) = 1.
A good approximation to the constant is (2*log(7/3)/log(17))^2 = 0.35774506... which agrees with the constant through the first 6 significant digits.

Examples

			0.3577451... = (1/5 + 1/17) + (1/17 + 1/37) + (1/197 + 1/257) + ...

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, pp. 94-98.
  • J. W. L. Glaisher, On the Sums of Inverse Powers of the Prime Numbers, Quart. J. Math. 25, 347-362, 1891.

Crossrefs

Cf. A002496, A005574, A065421, A085548, A096012, A172168, A206328.

Programs

  • Mathematica
    s=N[1/5+1/17,20];Do[p=(10*k+4)^2+1;q=(10*k+6)^2+1;If[PrimeQ[p]&&PrimeQ[q],s=s+1/p+1/q],{k,0,10^7}];Print[N[s,20]]

Formula

Equals (1/5 + 1/17) + Sum_{n>=1} (1/(A096012(n)^2 + 1) + 1/(A096012(n) + 2)^2 + 1).

A328498 Decimal expansion of Sum_{(p, q) runs through the twin primes} ((p mod 4) - 2) * (1/p + 1/q).

Original entry on oeis.org

1, 8, 3, 5, 0, 0, 3, 8
Offset: 0

Views

Author

Dimitris Valianatos, Oct 17 2019

Keywords

Comments

Because Viggo Brun's constant is 1.90216058... it is easy to find these 2 new constants:
(1/3 + 1/5) + (1/11 + 1/13) + (1/59 + 1/61) + (1/71 + 1/73) + ... = 1.04283048...
(1/5 + 1/7) + (1/17 + 1/19) + (1/29 + 1/31) + (1/41 + 1/43) + ... = 0.85933010...

Examples

			(1/3 + 1/5) - (1/5 + 1/7) + (1/11 + 1/13) - (1/17 + 1/19) - (1/29 + 1/31) + ... = 0.183500386... +- 0.000000002

Crossrefs

Cf. A001359, A006512, A065421.

Programs

  • PARI
    my(p = 3, s = 0.0); forprime(n = 5, 10^16, if(n-p == 2, if(p%4 == 3, s+= 1/p + 1/n, s+= -1/p - 1/n)); p = n); s

A331370 Decimal expansion of Sum_{(p1, p2) is twin prime pair} log(p2 / p1).

Original entry on oeis.org

1, 8, 7, 2, 1, 7, 8, 8
Offset: 1

Views

Author

Dimitris Valianatos, May 03 2020

Keywords

Comments

This constant is the difference between Brun's constant and the A331369 constant.
This sum converges because log(p_(k+1)/p_k) < 1/p_k + 1/p_(k+1) for every k >= 1 and (p_k, p_(k+1)) twin prime pair.

Examples

			1.8721788...

Crossrefs

Cf. A077800, A065421, A331369, A001359, A006512.

Formula

Equals A065421 - A331369.

A347359 Decimal expansion of Product_{p in A077800} (1 - 1/p).

Original entry on oeis.org

1, 2, 9, 3, 3, 7, 1, 7
Offset: 0

Views

Author

Kenneth H. Hicks, Aug 29 2021

Keywords

Comments

Note that A077800 is the sequence of twin primes with 5 repeated. The sequence of twin primes is A001097.
Related to Brun's constant (A065421) and the twin prime constant (A005597).
It is well known that the product of 1-1/p over all primes p is zero (it is related to the Riemann zeta function). Also the sum of 1/p diverges, whereas the sum of 1/p2 for p2 in the sequence A077800 converges to Brun's constant, regardless of whether there are an infinite number of twin primes or not. Similarly, the product in the present sequence also converges.
The repeated value of 1/5 is used in the calculation of Brun's constant (A065421) and we follow that convention here. The first two pairs of twin primes are (3,5) and (5,7), so the 4 initial terms in the product are (1-1/3)*(1-1/5)*(1-1/5)*(1-1/7).
This constant converges very slowly, similar to the convergence of Brun's constant. For example, for all twin primes below 1 billion, the product only reaches the value of 0.1469... Details on the error term in the convergence of the above product will be given in a forthcoming paper.

Examples

			0.12933717...

References

  • K. Hicks and K. Ward, Series and Product Relations Made from Primes, Pi Mu Epsilon Journal, Vol. 15, No. 3, Fall 2020, pp. 161-169.

Links

Crossrefs

Cf. A001097, A005597, A065421, A077800.

Extensions

Offset corrected by N. J. A. Sloane, Sep 20 2021
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