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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A303112 Primes p such that (r-q)/(q-p) = 2 or 1/2, and p < q < r are three consecutive primes.

Original entry on oeis.org

2, 5, 7, 11, 13, 17, 37, 41, 67, 89, 97, 101, 103, 107, 191, 193, 223, 227, 277, 307, 311, 347, 389, 397, 449, 457, 461, 479, 487, 491, 503, 613, 641, 739, 757, 761, 821, 823, 853, 857, 877, 881, 907, 929, 991, 1087, 1091, 1231, 1277, 1297, 1301, 1423, 1427, 1439, 1447, 1453
Offset: 1

Views

Author

Andres Cicuttin, Apr 18 2018

Keywords

Comments

Conjecture: The two most frequent ratios between consecutive prime gaps are 2 and 1/2, and both ratios occur with about the same frequency.

Examples

			The first three consecutive primes are 2, 3 and 5, and (5-3)/(3-2)=2, so the first term is a(1)=2, that is, the first prime of (2,3,5).
The next three consecutive primes are 3, 5 and 7, and (7-5)/(5-3)=1, so the first prime of (3,5,7) is not in the list.
The next three consecutive primes are 5, 7 and 11, and (11-7)/(7-5)=2, so the second term is a(2)=5, that is, the first prime of (5,7,11).
The prime 13 is also in the list because (19-17)/(17-13)=1/2.
		

Crossrefs

Cf. A257762 (indices of primes with above ratio = 2).

Programs

  • Mathematica
    b={};
    Do[If[Abs[Log[2,(Prime[j+2]-Prime[j+1])/(Prime[j+1]-Prime[j])]]==1,AppendTo[b,Prime[j]]],{j,1,200}];
    Print@b
    Select[Partition[Prime[Range[250]],3,1],(#[[3]]-#[[2]])/(#[[2]]-#[[1]]) == 2||(#[[3]]-#[[2]])/(#[[2]]-#[[1]])==1/2&][[All,1]] (* Harvey P. Dale, Mar 14 2022 *)
  • PARI
    isok(p) = my(q = nextprime(p+1), r = nextprime(q+1), f = (r-q)/(q-p)); (f == 2) || (f == 1/2);
    forprime(p=2, 1000, if (isok(p), print1(p, ", "))); \\ Michel Marcus, Apr 23 2018

Formula

Conjecture: lim_{n->inf} n/primepi(a(n)) > k > 0 for some k.

A373299 Numbers prime(k) such that prime(k) - prime(k-1) = prime(k+2) - prime(k+1).

Original entry on oeis.org

7, 11, 13, 17, 29, 41, 59, 79, 101, 103, 107, 113, 139, 163, 181, 193, 227, 257, 269, 311, 359, 379, 397, 419, 421, 439, 461, 487, 491, 547, 569, 577, 599, 691, 701, 709, 761, 811, 823, 857, 863, 881, 887, 919, 983, 1021, 1049, 1051, 1091, 1109, 1163
Offset: 1

Views

Author

Alexandre Herrera, May 31 2024

Keywords

Examples

			7 is in the list because the prime previous to 7 is 5 and the next primes after 7 are 11 and 13, so we have 7 - 5 = 13 - 11 = 2.
		

Crossrefs

Programs

  • Maple
    P:= select(isprime,[seq(i,i=3..10^4,2)]):
    G:= P[2..-1]-P[1..-2]: nG:= nops(G):
    J:= select(t -> G[t-1]=G[t+1],[$2..nG-1]):
    P[J]; # Robert Israel, May 31 2024
  • Mathematica
    Select[Partition[Prime[Range[200]], 4, 1], #[[2]] - #[[1]] == #[[4]] - #[[3]] &][[;; , 2]] (* Amiram Eldar, May 31 2024 *)
  • Python
    from sympy import prime
    def ok(k):
        return prime(k)-prime(k-1) == prime(k+2)-prime(k+1)
    print([prime(k) for k in range(2,200) if ok(k)])
    
  • Python
    from sympy import nextprime
    from itertools import islice
    def agen(): # generator of terms
        p, q, r, s = [2, 3, 5, 7]
        while True:
            if q-p == s-r: yield q
            p, q, r, s = q, r, s, nextprime(s)
    print(list(islice(agen(), 60))) # Michael S. Branicky, May 31 2024

Formula

a(n) = A151800(A022885(n)).

A064103 Primes p = p(k) such that p(k) + p(k+9) = p(k+1) + p(k+8) = p(k+2) + p(k+7) = p(k+3) + p(k+6) = p(k+4) + p(k+5).

Original entry on oeis.org

13, 139, 6091, 19843, 51787, 55793, 113143, 179029, 205157, 302551, 346361, 460949, 895799, 970447, 1150651, 1180847, 1697257, 1929553, 2334781, 2580631, 2797447, 3056561, 3086009, 3416717, 3598943, 4024667, 4026107, 4067123, 4077583, 4389503, 4541083, 4790503
Offset: 1

Views

Author

Robert G. Wilson v, Sep 17 2001

Keywords

Examples

			13 + 47 = 17 + 43 = 19 + 41 = 23 + 37 = 29 + 31.
		

Crossrefs

Programs

  • Mathematica
    a = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}; Do[ a = Delete[ a, 1 ]; a = Append[ a, Prime[ n ] ]; If[ a[ [ 1 ] ] + a[ [ 10 ] ] == a[ [ 2 ] ] + a[ [ 9 ] ] == a[ [ 3 ] ] + a[ [ 8 ] ] == a[ [ 4 ] ] + a[ [ 7 ] ] == a[ [ 5 ] ] + a[ [ 6 ] ], Print[ a[ [ 1 ] ] ] ], {n, 1, 3 10^5} ]

Formula

Primes p = prime(k) = A000040(k) such that A359440(k+4) >= 4. - Peter Munn, Jan 13 2023

Extensions

More terms from Sean A. Irvine, Jun 11 2023

A064104 Primes p = p(k) such that p(k) + p(k+11) = p(k+1) + p(k+10) = p(k+2) + p(k+9) = p(k+3) + p(k+8) = p(k+4) + p(k+7) = p(k+5) + p(k+6).

Original entry on oeis.org

137, 55787, 113131, 179021, 895789, 1150649, 3086003, 4026103, 4077559, 8021753, 8750857, 12577063, 14355559, 19136527, 19412863, 20065961, 21865339, 22633141, 25880177, 30404971, 33926159, 38202173, 41905891, 42925699
Offset: 1

Views

Author

Robert G. Wilson v, Sep 17 2001

Keywords

Examples

			137 + 193 = 139 + 191 = 149 + 181 = 151 + 179 = 157 + 173 = 163 + 167.
		

Crossrefs

Programs

  • Mathematica
    a = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}; Do[ a = Delete[ a, 1 ]; a = Append[ a, Prime[ n ] ]; If[ a[ [ 1 ] ] + a[ [ 12 ] ] == a[ [ 2 ] ] + a[ [ 11 ] ] == a[ [ 3 ] ] + a[ [ 10 ] ] == a[ [ 4 ] ] + a[ [ 9 ] ] == a[ [ 5 ] ] + a[ [ 8 ] ] == a[ [ 6 ] ] + a[ [ 7 ] ], Print[ a[ [ 1 ] ] ] ], {n, 1, 10^6} ]
    okQ[n_]:=Length[Union[Take[n,6]+Reverse[Take[n,-6]]]]==1; Transpose[ Select[Partition[Prime[Range[2700000]],12,1],okQ]][[1]] (* Harvey P. Dale, Apr 25 2011 *)

Formula

Primes p = prime(k) = A000040(k) such that A359440(k+5) >= 5. - Peter Munn, Jan 13 2023

A266882 Primes p(n) such that p(n) + p(n+3) = p(n+1) + p(n+2) and p(n) + p(n+4) = p(n+2) + p(n+3).

Original entry on oeis.org

13, 37, 223, 1087, 1423, 1483, 2683, 4783, 6079, 7331, 7547, 11057, 12269, 12401, 12641, 17333, 19471, 20743, 21799, 23027, 27733, 28097, 29017, 29389, 30631, 30859, 33191, 33343, 33587, 33613, 35527, 36551, 42457, 44263, 45817, 48857, 49459, 54499, 55813, 57329, 58151, 59207
Offset: 1

Views

Author

Keywords

Comments

All terms = {1, 5} mod 6. - Muniru A Asiru, Aug 19 2017

Examples

			Starting from 13, the five consecutive primes are 13, 17, 19, 23, 29; and they satisfy 13 + 23 = 17 + 19 and 13 + 29 = 23 + 19. So 13 is in the sequence.
		

Crossrefs

Subsequence of A022885.

Programs

  • GAP
    K:=10^7+1;; # to get all terms <= K.
    P:=Filtered([1,3..K],IsPrime);;
    A:=[];; for n in [1..Length(P)-4] do if P[n]+P[n+3]=P[n+1]+P[n+2] and  P[n]+P[n+4]=P[n+2]+P[n+3] then Add(A,P[n]); fi; od; A; # Muniru A Asiru, Aug 19 2017
  • Maple
    for i from 1 to 10^5 do if ithprime(i)+ithprime(i+3) = ithprime(i+1)+ithprime(i+2) and ithprime(i)+ithprime(i+4) = ithprime(i+2)+ithprime(i+3) then print(ithprime(i)); fi; od; # Muniru A Asiru, Aug 19 2017
  • Mathematica
    Prime@ Select[Range@ 6000, And[Prime@ # + Prime[# + 3] == Prime[# + 1] + Prime[# + 2], Prime@ # + Prime[# + 4] == Prime[# + 2] + Prime[# + 3]] &] (* Michael De Vlieger, Jan 05 2016 *)
  • PARI
    lista(nn) = {for (n=1, nn, if ((prime(n) + prime(n+3) == prime(n+1) + prime(n+2)) && (prime(n) + prime(n+4) == prime(n+2) + prime(n+3)), print1(prime(n), ", ")););} \\ Michel Marcus, Jan 05 2016
    
  • Python
    from sympy import primerange
    b, c, d, e = 2, 3, 5, 7
    for p in primerange(11, 10**9):
        a, b, c, d, e = b, c, d, e, p
        if a + d == b + c and a + e == c + d:
            print(a, end=', ')
    

Extensions

More terms from Michel Marcus, Jan 05 2016
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