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.

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A343118 Length of the longest sequence of equidistant primes among the first n primes.

Original entry on oeis.org

2, 2, 3, 3, 3, 3, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6
Offset: 2

Views

Author

Andres Cicuttin, Apr 05 2021

Keywords

Comments

This sequence is unbounded as stated by the Green-Tao theorem.

Examples

			For the first 2 primes {2,3}, the sequence is itself a list of two equidistant primes, so a(2) = 2.
For the first 3 primes {2,3,5}, there is at most two equidistant primes, so a(3) = 2.
For the first 4 primes {2,3,5,7}, the subsequence {3,5,7} is the longest subsequence with 3 equidistant primes, so a(4) = 3.
For the first 10 primes {2,3,5,7,11,13,17,19,23,29}, the subsequence {5,11,17,23,29} is the longest subsequence with 5 equidistant primes, so a(10) = 5.

Links

Crossrefs

Cf. A000720, A005115, A338869, A343122.

Programs

  • Mathematica
    nmax = 128; (* Last n *)
maxlen = 11 ; (* Maximum exploratory length of sequences of equidistant primes. "maxlen" must be larger than the maximum term obtained with "nmax" *)
(* a[n,p,s] returns the sequence of "s" equidistant primes with period "p" and last prime prime(n) if it exists, otherwise it returns {} *)
a[n_, period_, seqlen_] := Module[{tab, test},
(* Building sequences of equidistant numbers ending with prime(n) *)
  tab = Table[Prime[n] - k*period, {k, 0, seqlen - 1}];
(* Checking if all elements are primes and greater than 2 *)
  test = (And @@ PrimeQ@tab) && (And @@ Map[(# > 2 &), tab]);
  Return[If[test, tab, {}]]];
atab = {}; aterms = {};
(* For every n, exploring all sequences of equidistant primes among the first n primes with n > 2 *)
Do[
  Do[Do[
    If[a[n, period, seqlen] != {}, AppendTo[atab, seqlen]]
    , {period, 2, Ceiling[Prime[n]/(seqlen - 1)], 2}]
   , {seqlen, 2, maxlen}];
(* Saving the pairs {n, corresponding maximum lengths} *)
  AppendTo[aterms, {n, Max[atab]}]
  , {n, 3, nmax}];
(* Prepending the first term corresponding to the trivial case of first two primes {2,3} *)
Join[{2}, (Transpose[aterms][[2]])]

Formula

a(A000720(A005115(n))) = n. - Rémy Sigrist, Apr 15 2021

A122763 A093364 with duplicates removed.

Original entry on oeis.org

0, 1, 2, 6, 30, 150, 210, 13860, 60060, 420420, 4144140, 9699690, 87297210, 717777060, 4180566390, 18846497670, 26004868890
Offset: 1

Views

Author

Roger L. Bagula, Sep 22 2006

Keywords

Examples

			Succeedingly longer arithmetic progressions of primes where the end value is smallest (A005115) start 2,2+j,3+2j,5+6j,5+6j,7+30j,7+150j... This sequence collects the unique values of the common differences.

Crossrefs

Cf. A093364, A005115.

Extensions

Edited by Ralf Stephan, May 22 2014

A216539 Largest prime p such that there is no n-term arithmetic progression of primes ending with p.

Original entry on oeis.org

2, 37, 557, 100291, 2521081, 198702899, 5082095279
Offset: 2

Views

Author

Alex Ratushnyak, Sep 12 2012

Keywords

Comments

For n>=3, the value given is only a conjecture.

Examples

			It is conjectured (A216495) that 37 is the largest prime p with no 3-term arithmetic progression ending with p. This would imply a(3)=37. (7,19,31, for example, is a progression ending with 31.)

Crossrefs

A216495 conjectures a(3).
A094383 conjectures a(4).
A216497 conjectures a(5).
A216498 conjectures a(6).
A216468 conjectures a(7).
A216590 conjectures a(8).
Cf. A005115.

A360883 Smallest powerful (1) number which is at the end of an arithmetic progression of n terms.

Original entry on oeis.org

1, 4, 49, 144, 4500, 5400, 308700, 352800, 396900, 441000, 58697100, 64033200, 11723411700, 12625212600, 13527013500, 14428814400
Offset: 1

Views

Author

Charles R Greathouse IV, Feb 24 2023

Keywords

Comments

Makowski proves that this sequence is infinite.

Examples

			Corresponding arithmetic progressions:
1
1, 4
1, 25, 49
36, 72, 108, 144
900, 1800, 2700, 3600, 4500
900, 1800, 2700, 3600, 4500, 5400
44100, 88200, 132300, 176400, 220500, 264600, 308700
44100, 88200, 132300, 176400, 220500, 264600, 308700, 352800
44100, 88200, 132300, 176400, 220500, 264600, 308700, 352800, 396900
44100, 88200, 132300, 176400, 220500, 264600, 308700, 352800, 396900, 441000
5336100, 10672200, 16008300, 21344400, 26680500, 32016600, 37352700, 42688800, 48024900, 53361000, 58697100
5336100, 10672200, 16008300, 21344400, 26680500, 32016600, 37352700, 42688800, 48024900, 53361000, 58697100, 64033200
From _Bert Dobbelaere_, Feb 25 2023: (Start)
a(13): k*901800900 (k=1..13)
a(14): k*901800900 (k=1..14)
a(15): k*901800900 (k=1..15)
a(16): k*901800900 (k=1..16)
(End)

References

  • A. Makowski, Remarks on some problems in the elementary theory of numbers, Acta Math. Univ. Comenian. 50/51 (1987), 277-281.

Links

Crossrefs

Subsequence of A001694.
Smallest number at the end of an arithmetic progression of n terms of various types of numbers: A005115 (primes).

Programs

  • PARI
    pow(lim, mn=2)=my(v=List(), t); for(m=1, sqrtnint(lim\1, 3), t=m^3; for(n=1, sqrtint(lim\t), listput(v, t*n^2))); Set(v)
v=pow(1e11); \\ Set as needed
a(n)=my(k=n-1); for(i=n,#v, for(j=1,i-k, my(g=v[i]-v[j],m); if(g%k, next); m=g/k; forstep(a=v[j]+m,v[i]-m,m, if(!setsearch(v,a), next(2))); return(v[i]))); warning("More powerful numbers needed in v")

Extensions

a(13)-a(16) from Bert Dobbelaere, Feb 25 2023
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