A168253 -id:A168253 - cp's OEIS frontend

A179234 a(n) is the smallest prime q such that, for the previous prime p and the following prime r, the fraction (r-q)/(q-p) has denominator n in lowest terms.

3, 11, 29, 367, 149, 521, 127, 1847, 1087, 1657, 1151, 4201, 2503, 2999, 5779, 10831, 1361, 9587, 30631, 19373, 16183, 36433, 81509, 28277, 31957, 25523, 40343, 82129, 44351, 102761, 34123, 89753, 282559, 134581, 173429, 705389, 404671, 212777, 371027, 1060861, 265703, 461801, 156007, 544367, 576881, 927961, 1101071, 1904407, 604171, 396833

Offset: 1

Vladimir Shevelev, Jan 05 2011

nonn

The conjecture that a(n) exists for every n is a weaker conjecture than a related one in the comment to A179210.

			For q=3 we have (r-q)/(q-p)=2/1. Therefore, a(1)=3.
For q=5: (r-q)/(q-p) = 1/1; for q = 7: (r-q)/(q-p) = 2/1; for q = 11: (r-q)/(q-p) = 1/2. Therefore, a(2)=11.

Robert G. Wilson v, Table of n, a(n) for n = 1..123

Cf. A001223, A168253, A179210, A279067.

f[n_] := Block[{p = 2, q = 3, r = 5}, While[ Denominator[(r - q)/(q - p)] != n, p = q; q = r; r = NextPrime@ r]; q]; Array[f, 50]
p = 2; q = 3; r = 5; t[] = 0; While[q < 100000000, If[ t[ Denominator[(r - q)/(q - p)]] == 0, t[ Denominator[(r - q)/(q - p)]] = q]; p = q; q = r; r = NextPrime@ r]; t@# & /@ Range@100 (* _Robert G. Wilson v, Dec 11 2016 *)

a(n)=my(p=2,q=3);forprime(r=5,default(primelimit),if(denominator((r-q)/(q-p))==n,return(q));p=q;q=r)

Revised definition, new program, and terms past a(5) from Charles R Greathouse IV, Jan 12 2011

A179256 a(n) is the smallest prime q such that (q-p)/(r-q) = n, where p

Original entry on oeis.org

5, 11, 29, 6421, 149, 521, 84913, 1949, 1277, 43391, 1151, 4547, 933151, 2999, 6947, 1568867, 10007, 32297, 4131223, 25301, 78779, 12809491, 91079, 28277, 13626407, 35729, 117497, 37305881, 399851, 102761, 217795433, 288647, 296909, 240485461, 173429, 1026029, 213158501, 1053179, 371027, 1163010421, 1885151, 461801, 1661688551, 1155821, 576881, 3403741987, 4876607, 4252679, 10394432611, 838349, 1775171

Offset: 1

Vladimir Shevelev, Jan 05 2011

nonn

Conjecture: a(n)>0 for n >= 1.
It would appear that a(3n+1) is greater than either a(3n) or a(3n+2).
Records appear for a(n) for n's: 1, 2, 3, 4, 7, 13, 16, 19, 22, 25, 28, 31, 34, 40, 43, 46, 49, ..., .

Robert G. Wilson v, Table of n, a(n) for n = 1..57

Cf. A179210, A179234, A168253, A001223

f[n_] := f[n] = Block[{p = 3, q = 5, r = 7}, While[p + n*r != (n + 1) q, p = q; q = r; r = NextPrime@ r]; q]; Array[f, 33]
p = 2; q = 3; r = 5; t[] = 0; While[p < 10^9, If[ Mod[q - p, r - q] == 0 && t[(q - p)/(r - q)] == 0, t[(q - p)/(r - q)] = q; Print[{(q - p)/(r - q), q}]]; p = q; q = r; r = NextPrime@ r]; t@# & /@ Range@45 (* _Robert G. Wilson v, Dec 10 2016 *)

A275785 Primes such that the ratio between the distance to the next prime and from the previous prime appears for the first time.

Original entry on oeis.org

3, 5, 11, 23, 29, 31, 37, 89, 113, 127, 139, 149, 199, 251, 293, 331, 337, 367, 409, 521, 523, 631, 701, 787, 797, 953, 1087, 1129, 1151, 1259, 1277, 1327, 1361, 1381, 1399, 1657, 1669, 1847, 1933, 1949, 1951, 1973, 2477, 2503, 2579, 2633, 2861, 2879, 2971, 2999, 3089, 3137, 3163, 3229, 3407

Offset: 1

Andres Cicuttin, Nov 14 2016

nonn

Number of terms less than 10^n: 2, 8, 26, 85, 224, 511, 1035, 1905, 3338, ..., . - Robert G. Wilson v, Nov 30 2016

			a(1) = 3 because this is the first prime for which it is possible to determine the ratio between the distance to the next prime (5) and from the previous prime (2). This first ratio is 2.
a(2) = 5 because the ratio between the distance to the next prime (7) and from the previous prime (3) is 1 and this ratio has not appeared before.
The third element a(3) is not 7 because (11-7)/(7-5) = 2, a ratio that appeared before with a(1), so a(3) = 11 because (13-11)/(11-7) = 1/2, a ratio that did not appear before.

Robert G. Wilson v, Table of n, a(n) for n = 1..7589

Cf. A168253, A179210, A179234, A179256, A274263, A276309, A276812.

nmax = 720;
a = Prime[Range[nmax]];
gaps = Rest[a] - Most[a];
gapsratio = Rest[gaps]/Most[gaps];
newpindex = {}; newgratios = {}; i = 1;
While[i < Length[gapsratio] + 1,
If[Cases[newgratios, gapsratio[[i]]] == {},
  AppendTo[newpindex, i + 1];
  AppendTo[newgratios, gapsratio[[i]]] ];
  i++];
Prime[newpindex]
p = 2; q = 3; r = 5; rtlst = qlst = {}; While[q < 10000, rt = (r - q)/(q - p); If[ !MemberQ[rtlst, rt], AppendTo[rtlst, rt]; AppendTo[qlst, q]]; p = q; q = r; r = NextPrime@ r]; qlst (* Robert G. Wilson v, Nov 30 2016 *)

A179240 a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator equal to A006843(n) (or 0, if such a prime does not exist).

Original entry on oeis.org

5, 11, 17, 19, 29, 41, 47, 67, 73, 97, 101, 359, 367, 379, 383, 389, 397, 419, 421, 449, 467, 547, 613, 631, 647, 683, 691, 733, 769, 797, 811, 929, 941, 1021, 1087, 1153, 1181, 1193, 1249, 1709, 1721, 1747, 1847, 1889, 2017, 2153, 2357

Offset: 1

Vladimir Shevelev, Jan 06 2011

nonn

Conjecture: a(n) > 0 for all n.

			For n = 1..3, A006843(n) = 1, and p,q,r have to obey the condition
r-q | q-p. Thus a(1) = 5, a(2) = 11, a(3) = 17.

Cf. A006843, A168253, A179210, A179234, A179256, A001223

More terms from Alois P. Heinz, Jan 06 2011

A179328 a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator prime(n) (or 0, if such a prime does not exist).

Original entry on oeis.org

3, 23, 139, 293, 1129, 2477, 8467, 30593, 81463, 85933, 190409, 404597, 535399, 840353, 1100977, 2127163, 4640599, 6613631, 6958667, 10343761, 24120233, 49269581, 83751121, 101649649, 166726367, 273469741, 310845683, 568951459

Offset: 1

Vladimir Shevelev, Jan 06 2011

nonn

Conjecture: a(n) > 0 for all n.

Cf. A168253, A179210, A179234, A179240, A179256, A001223

with(numtheory):
a:= proc(n) option remember; local k, p, q, r, pn;
      pn:= ithprime(n);
      for k from `if`(n=1, 1, pi(a(n-1))) do
        p:= ithprime(k);
        q:= ithprime(k+1);
        r:= ithprime(k+2);
        if denom((q-p)/(r-q)) = pn then break fi
      od; q
    end:
seq(a(n), n=1..10);  # Alois P. Heinz, Jan 06 2011

a[n_] := a[n] = Module[{k, p, q, r, pn},
     pn = Prime[n];
     For[k = If[n == 1, 1, PrimePi[a[n - 1]]], True, k++,
     p = Prime[k];
     q = Prime[k + 1];
     r = Prime[k + 2];
     If [Denominator[(q - p)/(r - q)] == pn, Break[]]]; q];
Table[a[n], {n, 1, 10}] (* Jean-François Alcover, Mar 18 2022, after Alois P. Heinz *)

More terms from Alois P. Heinz, Jan 06 2011

A178942 a(1) = 3; for n >= 2, a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator equal to A001223(n)/2 (or 0, if no such prime exists).

Original entry on oeis.org

3, 5, 11, 13, 17, 19, 29, 37, 47, 53, 61, 67, 71, 79, 83, 131, 137, 151, 163, 173, 233, 277, 331, 359, 379, 397, 401, 419, 439, 773, 823, 941, 947, 1021, 1031, 1033, 1063, 1087, 1097, 1117, 1123, 1153, 1187, 1237, 1277, 1709, 1789, 1823

Offset: 1

Vladimir Shevelev, Jan 06 2011

nonn

Conjecture: a(n) > 0 for all n.

The smallest prime(k) > a(n-1) such that the denominator of A001223(k-1)/A001223(k) equals A001223(n)/2. - R. J. Mathar, Jan 07 2011

Cf. A001223, A168253, A179210, A179234, A179240, A179328.

A001223 := proc(n) ithprime(n+1)-ithprime(n) ; end proc:
A178942 := proc(n) option remember; local p,q,r ; if n = 1 then 3; else for q from procname(n-1)+1 do if isprime(q) then p := prevprime(q) ; r := nextprime(q) ; denom((q-p)/(r-q)) ; if % = A001223(n)/2 then return q; end if; end if; end do: end if; end proc: # R. J. Mathar, Jan 07 2011

A001223[n_] := Prime[n + 1] - Prime[n];
a[n_] := a[n] = Module[{p, q, r, d}, If[n == 1, 3, For[q = a[n - 1] + 1, True, q++, If [PrimeQ[q], p = NextPrime[q, -1]; r = NextPrime[q]; d = Denominator[(q - p)/(r - q)]; If[d == A001223[n]/2, Return[q]]]]]];
Array[a, 48] (* Jean-François Alcover, May 21 2020, after Maple *)

More terms from Alois P. Heinz, Jan 06 2011

A179479 a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator 2, for odd n and 1 for even n (or 0, if such a prime does not exist).

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 19, 29, 43, 53, 67, 71, 79, 97

Offset: 1

Vladimir Shevelev, Jan 08 2011

nonn

Conjecture: a(n) > 0 for all n.

Since, as it is accepted in the OEIS, we consider the uncancelled fractions, then, by the condition, for even n, we have (r-q)|(q-p).

			If n=1, then denominator should be 2. Thus a(1)=3, since (3-2)/(5-3)=1/2. If n=2, then denominator should be 1. Thus a(2)=5, since (5-3)/(7-5)=1/1, etc.

Cf. A168253, A179210, A179234, A179240, A179256, A179328.

A179921 a(n) = prime(n) if n<=3; for n>3, a(n) is the smallest prime >a(n-1), such that the denominator of fraction (a(n-1)-a(n-2))/(a(n)-a(n-1)) did not appear earlier.

Original entry on oeis.org

2, 3, 5, 7, 13, 23, 31, 53, 67, 79, 113, 131, 151, 193, 233, 271, 307, 353, 379, 409, 457, 557, 613, 691, 761, 809, 883, 907, 1013, 1069, 1123, 1181, 1213, 1279, 1361, 1423, 1483, 1571, 1657, 1709, 1811, 1933, 1997, 2087, 2179, 2273, 2341, 2459

Offset: 1

Vladimir Shevelev, Jan 12 2011

nonn

Using Dirichlet's theorem on arithmetic progressions, it is easy to prove that the sequence is infinite. The sequence of the corresponding denominators begins with 2,1,3,5,4,11,7,6,17, ...

			The first four terms 2,3,5,13 give three denominators: 2,1,3. Then a(5) is not in {17, 19}, since (13-5)/(17-13) = 2/1, (13-5)/(19-13) = 4/3 and denominators 1 and 3 already appeared earlier. Since (13-5)/(23-13) = 4/5 and 5 is not yet in the denominator sequence, a(5) = 23.

Cf. A168253, A178942, A179210, A179234, A179256, A179328.

Edited by Alois P. Heinz, Jan 12 2011

cp's OEIS Frontend

A179234 a(n) is the smallest prime q such that, for the previous prime p and the following prime r, the fraction (r-q)/(q-p) has denominator n in lowest terms.

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A179256 a(n) is the smallest prime q such that (q-p)/(r-q) = n, where p

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A275785 Primes such that the ratio between the distance to the next prime and from the previous prime appears for the first time.

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A179240 a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator equal to A006843(n) (or 0, if such a prime does not exist).

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A179328 a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator prime(n) (or 0, if such a prime does not exist).

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A178942 a(1) = 3; for n >= 2, a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator equal to A001223(n)/2 (or 0, if no such prime exists).

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A179479 a(n) is the smallest prime q > a(n-1) such that, for the previous prime p and the following prime r, the fraction (q-p)/(r-q) has denominator 2, for odd n and 1 for even n (or 0, if such a prime does not exist).

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A179921 a(n) = prime(n) if n<=3; for n>3, a(n) is the smallest prime >a(n-1), such that the denominator of fraction (a(n-1)-a(n-2))/(a(n)-a(n-1)) did not appear earlier.

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