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 1-6 of 6 results.

A306700 Decimal expansion of the constant S_2 = Sum_{j>=1} prime(2*j)!/prime(2*j + 1)!.

Original entry on oeis.org

0, 5, 1, 6, 6, 6, 6, 2, 2, 8, 8, 4, 2
Offset: 0

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Author

Marco Ripà, Mar 05 2019

Keywords

Comments

The constant S_2 is connected to the gap between the j-th and (j+1)-th primes.
Together with the constant S_1 (see A306658), S_2 involves the prime gaps, since twin primes produce the heaviest terms of the summation in comparison to their next and previous addend.
On Mar 07 2019, the first 2445000000 prime numbers were used and from Rosser's theorem we obtain:
0.05166662288423 < S_2 < 0.05166662288424 + Sum_{j>=1222500000} 1/((2*j*log(2*j) + log(log(2*j)) - 1) * (2*j*log(2*j) + log(log(2*j)) - 2)) < 0.05166662288424 + 3.22757*10^(-13) < 0.05166662288457.

Examples

			S_2 = 0.0516666228842...
		

Crossrefs

Programs

  • Mathematica
    b = 0; Do[f = Prime[Range[n - 999999, n]]; Do[b += N[1/Product[k, {k, f[[i]] + 1, f[[i + 1]]}], 100], {i, 1, 1000000, 2}]; Print[n, ": ", N[b, 100]], {n, 1000001, 100000001, 1000000}]; b
  • PARI
    suminf(j=1, prime(2*j)!/prime(2*j + 1)!) \\ Michel Marcus, Apr 02 2019

Formula

Sum_{j>=1} prime(2*j)!/prime(2*j + 1)! = Sum_{j>=1} 1/(Product{k=prime(2*j) + 1, prime(2*j + 1)} k) = 1/(5*4) + 1/(11*10*9*8) + 1/(17*16*15*14) + ...

A306744 Decimal expansion of the constant S_1 + S_2 = Sum_{j>=1} prime(j)!/prime(j + 1)!.

Original entry on oeis.org

4, 1, 9, 2, 2, 2, 0, 6, 4, 9, 0, 3
Offset: 0

Views

Author

Marco Ripà, Mar 07 2019

Keywords

Comments

The constant S_1 + S_2 is related to the prime gaps, since twin primes produce the largest terms of the sum compared with neighboring terms.

Examples

			S_1 + S_2 = 0.419222064903...
		

Crossrefs

Cf. A000040, A306658 (S_1), A306700 (S_2), A306780.

Programs

  • Mathematica
    s = 0; p = 2; q = 3; While[p < 10^10, s = N[s + 1/Times @@ Range[p +1, q], 32]; p = q; q = NextPrime@ q]; Take[ RealDigits[s][[1]], 20] (* Robert G. Wilson v, Mar 23 2019 *)
  • PARI
    suminf(j=1, prime(j)!/prime(j + 1)!) \\ Michel Marcus, Apr 02 2019

Formula

S_1 + S_2 = Sum_{j>=1} prime(j)!/prime(j + 1)! = Sum_{j>=2} 1/(Product{k=prime(j - 1) + 1, prime(j)} k) = 1/3 + 1/(4*5) + 1/(6*7) + 1/(8*9*10*11) + ...

A306780 Decimal expansion of the constant S_1 - S_2 = Sum_{j>=1} (-1)^(j+1)*(prime(j)!/prime(j + 1)!).

Original entry on oeis.org

3, 1, 5, 8, 8, 8, 8, 1, 9, 3, 5, 0
Offset: 0

Views

Author

Keywords

Comments

The constant S_1 - S_2 is related to the prime gaps, since twin primes produce the largest terms of the algebraic sum compared with neighboring terms.

Examples

			S_1 - S_2 = 0.315888819350...
		

Crossrefs

Cf. A000040, A306658 (S_1), A306700 (S_2), A306744 (S_1 + S_2).

Programs

  • PARI
    sumalt(j=1, (-1)^(j+1)*(prime(j)!/prime(j + 1)!)) \\ Michel Marcus, Apr 02 2019  \\ Needs default(realprecision, 10^4)  Jinyuan Wang, May 19 2019

Formula

S_1 - S_2 = Sum_{j>=1} (-1)^(j+1)*(prime(j)!/prime(j + 1)!) = Sum_{j>=2} (-1)^j/(Product{k=prime(j - 1) + 1, prime(j)} k) = 1/3 - 1/(4*5) + 1/(6*7) - 1/(8*9*10*11) + ...

A307383 Decimal expansion of the constant S_1* + S_2* = Sum_{j>=1} prime((j + 1) - 1)!/prime((j + 2) - 1)!.

Original entry on oeis.org

1, 3, 0, 4, 5, 6, 2, 5, 9, 8, 3, 5
Offset: 0

Views

Author

Keywords

Comments

The constant S_1* + S_2* is related to the prime gaps, since twin primes produce the largest terms of the sum compared with neighboring terms.
On Apr 06 2019, the first 4200000000 prime numbers were used in order to calculate S_1* and S_2* and using Rosser's theorem we get: 0.13045626983537 < S_1* + S_2* < 0.13045626983578.

Examples

			S_1* + S_2* = 0.130456269835...
		

Crossrefs

Cf. A000040, A306658 (S_1) A306700 (S_2), A306744 (S_1 + S_2).

Formula

S_1* + S_2* = Sum_{j>=1} prime((j + 1) - 1)!/prime((j + 2) - 1)! = Sum_{j>=1} 1/(Product{k=prime(j + 1), prime((j + 2) - 1)} k) = 1/(3*4) + 1/(5*6) + 1/(7*8*9*10) + 1/(11*12) + ...

A307384 Decimal expansion of the constant S_1* = Sum_{j>=1} prime((2*j) - 1)!/prime((2*j + 1) - 1)!.

Original entry on oeis.org

0, 8, 5, 1, 6, 1, 9, 1, 0, 9, 8, 5
Offset: 0

Views

Author

Keywords

Comments

Together with the constant S_2* and S_1* + S_2* (see A307383), S_1* involves the prime gaps, since twin primes produce the heaviest terms of the summation in comparison to their next and previous addend.
On Apr 06 2019, the first 4200000000 prime numbers were used and using Rosser's theorem we get: 0.08516191098523 < S_1* < 0.08516191098543.

Examples

			S_1* = 0.085161910985...
		

Crossrefs

Cf. A000040, A306658 (S_1) A306700 (S_2), A306744 (S_1 + S_2), A307383 (S_1* + S_2*).

Formula

S_1* = Sum_{j>=1} prime(2*j - 1)!/prime((2*j + 1) - 1)! = Sum_{j>=1} 1/(Product{k=prime(2*j), prime(2*j + 1)} k) = 1/(3*4) + 1/(7*8*9*10) + 1/(13*14*15*16) + 1/(19*20*21*22) +...

A307385 Decimal expansion of the constant S_2* = Sum_{j>=1} prime((2*j + 1) - 1)!/prime((2*j + 2) - 1)!.

Original entry on oeis.org

0, 4, 5, 2, 9, 4, 3, 4, 8, 8, 5, 0
Offset: 0

Views

Author

Keywords

Comments

The constant S_2* is related to the prime gaps, since twin primes produce the largest terms of the sum compared with neighboring terms.
On Apr 06 2019, the first 4200000000 prime numbers were used in order to calculate S_1* and S_2* and using Rosser's theorem we get: 0.04529434885014 < S_1* + S_2* < 0.04529434885035.

Examples

			S_2* = 0.045294348850...
		

Crossrefs

Cf. A000040, A306658 (S_1) A306700 (S_2), A306744 (S_1 + S_2), A307383 (S_1* + S_2*), A307384 (S_1*).

Formula

S_2* = Sum_{j>=1} prime((2*j + 1) - 1)!/prime((2*j + 2) - 1)! = Sum_{j>=1} 1/(Product{k=prime(2*j + 1), prime((2*j + 2) - 1)} k) = 1/(5*6) + 1/(11*12) + 1/(17*18) + 1/(23*24*25*26*27*28) +...
Showing 1-6 of 6 results.