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.

A274263 Integer part of the ratio of consecutive prime gaps.

Original entry on oeis.org

2, 1, 2, 0, 2, 0, 2, 1, 0, 3, 0, 0, 2, 1, 1, 0, 3, 0, 0, 3, 0, 1, 1, 0, 0, 2, 0, 2, 3, 0, 1, 0, 5, 0, 3, 1, 0, 1, 1, 0, 5, 0, 2, 0, 6, 1, 0, 0, 2, 1, 0, 5, 0, 1, 1, 0, 3, 0, 0, 5, 1, 0, 0, 2, 3, 0, 1, 0, 2, 1, 1, 0, 1, 0, 1, 1, 0, 2, 1, 0, 5, 0, 3, 0, 1, 1, 0, 0, 2, 3, 0, 0, 2, 0, 1, 2
Offset: 1

Views

Author

Andres Cicuttin, Jun 17 2016

Keywords

Comments

It seems that the distribution of the ratios of consecutive prime gaps exhibits a quite symmetric pattern, in the sense that the relative frequency of each ratio is similar to that of the inverse of that ratio (at least for the first 2*10^5 primes). This is more clearly seen by mean of a histogram of the logarithm of the ratios which is nearly symmetric and nearly centered around zero (see link).
Integer part of 2, 1, 2, 1/2, 2, 1/2, 2, 3/2, 1/3, 3, 2/3, 1/2,.... - R. J. Mathar, Jun 26 2016

Examples

			For n = 1 we have (prime(3)-prime(2))/(prime(2)-prime(1)) = (5-3)/(3-2) = 2 and its integer part is 2: a(1) = 2.
For n = 4 we have (prime(6)-prime(5))/(prime(5)-prime(4)) = (13-11)/(11-7) = 1/2 and its integer part is 0: a(4) = 0.
		

Crossrefs

Cf. A001223.

Programs

  • Maple
    A274264 := proc(n)
        A001223(n+1)/A001223(n) ;
        floor(%) ;
    end proc: # R. J. Mathar, Jun 26 2016
  • Mathematica
    Table[Floor[(Prime[j+2]-Prime[j+1])/(Prime[j+1]-Prime[j])],{j,1,200}];
    IntegerPart[#[[2]]/#[[1]]]&/@Partition[Differences[Prime[Range[200]]],2,1] (* Harvey P. Dale, Mar 07 2018 *)
  • PARI
    a(n) = (prime(n+2)-prime(n+1))\(prime(n+1)-prime(n)); \\ Michel Marcus, Jun 18 2016

Formula

a(n) = floor(A001223(n+1)/A001223(n)).

A274043 Number of squarefree integers congruent to {1, 2, 3} mod 8 <= 10^n.

Original entry on oeis.org

4, 28, 300, 3033, 30389, 303947, 3039643, 30396338, 303963527, 3039635535, 30396355364, 303963551074, 3039635509269, 30396355092700, 303963550926732, 3039635509266675, 30396355092702331, 303963550927021020
Offset: 1

Views

Author

Frank M Jackson, Jun 18 2016

Keywords

Comments

Empirically, the limit of a(n)/10^n tends to 3/Pi^2 (A104141) and implies that the asymptotic density of squarefree numbers congruent to {1, 2, 3} mod 8 is half that of the asymptotic density of all squarefree integers (A071172). When this sequence is compared with squarefree numbers congruent to {5, 6, 7} mod 8 (A274264) it contains slightly fewer squarefree integers at each of the sampling points, 10^n for n > 1. It can be argued heuristically that, as {1, 2, 3} mod 8 contains a square residue, its equivalence class should contain fewer squarefree numbers.
Also it has been shown, conditional on the Birch Swinnerton-Dyer conjecture, that all squarefree integers congruent to {5, 6, 7} mod 8 (A273929) are primitive congruent numbers (A006991). However, this property applies only sparsely to squarefree integers congruent to {1, 2, 3} mod 8 (A062695).

Crossrefs

Programs

  • Mathematica
    Table[Length@Select[Range[10^n], MemberQ[{1, 2, 3}, Mod[#, 8]]&&SquareFreeQ[#] &], {n, 1, 8}]

Extensions

a(10)-a(11) from Giovanni Resta, Jun 19 2016
a(12)-a(18) from Hiroaki Yamanouchi, Dec 25 2016

A274403 Number of primitive (squarefree) congruent numbers (A006991) <= 10^n.

Original entry on oeis.org

3, 36, 361, 3503, 34065, 332712, 3252966, 31925924
Offset: 1

Views

Author

Frank M Jackson, Jun 20 2016

Keywords

Comments

Conjecture: the limit of a(n)/10^n tends to 3/Pi^2 (A104141). This is based on the assumption, conditional on the Birch Swinnerton-Dyer conjecture, that all squarefree integers congruent to {5, 6, 7} mod 8 (A273929) are a subset of primitive congruent numbers (A006991) and have a natural density of 3/Pi^2. However, squarefree integers congruent to {1, 2, 3} mod 8 are conjecturally sparsely congruent numbers with a natural density of 0. It has been proved without the BSD conjecture that the natural density of congruent numbers is at least 55.9% the natural density of squarefree numbers congruent to {5, 6, 7} mod 8 (see A. Smith link).
The Mathematica program below is a slow implementation of the Tunnell criteria for determining congruent numbers. It will give counts for up to 10^5 in realistic time. Counts for 10^6 and 10^7 have been derived from tables generated by Giovanni Resta (see link).
From Jose Aranda, Jul 04 2024: (Start)
The C++ program I have written calculates a(8) = 31925924 in 75 minutes. The time grows almost exponentially.
Looking at the 8 known terms I think the above conjecture should perhaps refer to A274264 rather than to the present sequence.
From the link "A trillion triangles": "The calculation found of these most mysterious congruent numbers up to a trillion = 3148379694."
That number corresponds to a(10) = 108744287 + A274264(10).
With A274264(10) = 3039635407. Now
3/Pi^2 = 0.303963550927013314...
A274264(08) = 0030396356.
A274264(10) = 003039635407.
A274264(18) = 00303963550927001730.
The sequence A274264 tends to this limit. This sequence may not. (End)

Crossrefs

Programs

  • Mathematica
    CongruentQ[n_] := Module[{x, y, z, ok=False}, (Which[!SquareFreeQ[n], Null[], MemberQ[{5,6,7}, Mod[n, 8]], ok=True, OddQ@n&&Length@Solve[x^2 + 2 y^2 + 8 z^2 == n, {x, y, z}, Integers]==2Length@Solve[x^2+2y^2+32z^2==n, {x, y, z}, Integers], ok=True, EvenQ@n&&Length@Solve[x^2+4y^2+8z^2==n/2, {x, y, z}, Integers]==2Length@Solve[x^2+4y^2+32z^2==n/2, {x, y, z}, Integers], ok=True]; ok)]; Table[Length@Select[Range[10^n], CongruentQ], {n, 1, 5}]

Extensions

a(7) corrected by Frank M Jackson, Jul 25 2016
a(8) from Jose Aranda, Jul 04 2024
Showing 1-3 of 3 results.