A084299
Smallest primes such that the subsequent terms of consecutive prime differences (A001223) modulo 6 (A054763) displays repeatedly n times a {0,2,4} pattern of remainders of differences.
Original entry on oeis.org
83, 2903, 5897, 319499, 346943, 7974179, 15262433, 33954251, 5521833683, 83993232497, 848099080883, 1293322433639
Offset: 1
For n=1: a(1) = 83 is followed by [6,8,4].
For n=2: a(2) = 2903 is followed by [6,2,4,18,2,4].
For n=3: a(3) = 5897 is followed by [6,20,4,12,14,28,6,20,4].
For n=4: a(4) = 319499 is followed by [12,8,22,6,20,10,12,2,10,6,32,34].
For n=5: a(5) = 346943 is followed by [18,2,40,....,30,2,10] differences corresponding to n "wavelet" of [0,2,4] remainders modulo 6.
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(* generates a(5) *) d[x_] := Prime[x+1]-Prime[x]; md[x_] := Mod[Prime[x+1]-Prime[x], 6]; h={k1=0, k2=2, k3=4}; k=0; Do[If[Equal[md[n], k1]&&Equal[md[n+1], k2]&& Equal[md[n+2], k3]&&Equal[md[n+3], k1]&&Equal[md[n+4], k2]&&Equal[md[n+5], k3] &&Equal[md[n+6], k1]&&Equal[md[n+7], k2]&&Equal[md[n+8], k3] &&Equal[md[n+9], k1]&&Equal[md[n+10], k2]&&Equal[md[n+11], k3]&& Equal[md[n+12], k1]&&Equal[md[n+13], k2]&&Equal[md[n+14], k3], k=k+1; Print[{k, n, Prime[n], Table[md[n+j], {j, -1, 15}], Table[d[n+j], {j, -1, 15}]}]], {n, 2, 10000000}]
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lista(pmax) = {my(rec = 0, m = 0, c = 0, prv = 2, p0 = 0, d); forprime(p = 3, pmax, d = (p-prv)%6; if(d == 0 && m == 0, p0 = prv); if(d == c, m++; c = (c+2)%6; if(!(m%3) && m/3 > rec, print1(p0, ", "); rec++; m = 0), if(d == 0, p0 = prv; c = 2; m = 1, c = 0; m = 0)); prv = p);} \\ Amiram Eldar, Nov 04 2024
A263049
a(n) = smallest prime p(k) such that the gaps between the primes p(k), p(k+1), p(k+2), ..., p(k+n) are 2n, 2n-2, ... 6, 4, 2.
Original entry on oeis.org
3, 7, 31, 1979, 41203, 752251, 5647457, 32465047, 245333233, 245333213, 27797667517, 196559847120517, 3040284075731561, 253253149671986983
Offset: 1
Consider the consecutive primes 1979, 1987, 1993, 1997, 1999. The gaps are 8, 6, 4, 2, and this does not occur for any prime smaller than 1979, so a(4)=1979.
A094749
Triangle read by rows in which the n-th row contains the least set of n successive primes whose successive difference forms an arithmetic progression with common difference 2, (successive even numbers).
Original entry on oeis.org
2, 3, 5, 5, 7, 11, 17, 19, 23, 29, 347, 349, 353, 359, 367, 2903, 2909, 2917, 2927, 2939, 2953, 15373, 15377, 15383, 15391, 15401, 15413, 15427, 128981, 128983, 128987, 128993, 129001, 129011, 129023, 129037, 95285633, 95285639, 95285647, 95285657
Offset: 1
2
3 5
5 7 11
17 19 23 29
...
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Module[{prs=Prime[Range[551*10^4]],nn=9},Join[{2,3,5},Table[ SelectFirst[ Partition[ prs,n,1],Union[Differences[#,2]]=={2}&],{n,3,nn}]]// Flatten] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Aug 09 2020 *)
A349121
a(n) is the smallest prime p, such that p + k + k^2 are consecutive primes for 0 <= k <= n, but not for k>n.
Original entry on oeis.org
3, 5, 17, 347, 13901, 665111, 128981, 113575727, 2426256797, 137168442221, 4656625081181, 101951758179851, 484511389338941, 221860944705726407
Offset: 1
a(4)=347, because it is the smallest prime such that 347 + 2 = 349, 349 + 4 = 353, 353 + 6 = 359, 359 + 8 = 367 are 5 consecutive primes.
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Table[k=2;While[Flatten[k+(s={Range[0,n]})+s^2]!=NextPrime[k,Range[0,n]]||NextPrime[k,n+1]==k+n+1+(n+1)^2,k=NextPrime@k];k,{n,7}] (* Giorgos Kalogeropoulos, Nov 10 2021 *)
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isok(p, n) = my(q=p); for (k=1, n, my(r = p+k+k^2); if (nextprime(q+1) != r, return (0)); q=r); return(1);
a(n) = my(p=2); while (!isok(p, n), p=nextprime(p+1)); p; \\ Michel Marcus, Nov 09 2021
A090870
a(n) is the smallest m such that d(m+k-1) = 2k for k = 1, ..., n where d(t)= prime(t+1) - prime(t) (differences of consecutive primes in arithmetic progression).
Original entry on oeis.org
2, 3, 7, 69, 1642, 12073, 12073, 6496152, 118033638, 5575956036, 165534366186, 3265469041280, 14779996741980, 5701362336480884
Offset: 1
a(8) = 6496152 because prime(6496152) = 113575727 and 113575727, 113575729, 113575733, 113575739, 113575747, 113575757, 113575769, 113575783, and 113575799 are nine consecutive primes with differences respectively 2, 4, 6, 8, 10, 12, 14, 16.
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a[n_] := (For[m=1, !Sum[(d[m+k-1]-2k)^2, {k, n}]==0, m++ ];m); Do[Print[a[n]], {n, 8}]
Extended and edited by
T. D. Noe, May 23 2011
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