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a(n) is the smallest number that belongs simultaneously to the two arithmetic progressions prime(n) + m*prime(n+1) and prime(n+1) + m*prime(n+2), m >= 1, n >= 1.
2

%I #187 Dec 23 2024 14:53:45

%S 8,33,40,128,115,302,226,226,835,401,734,1718,1030,842,3121,3475,1401,

%T 2339,5108,1969,3233,2486,6491,9692,10298,5560,11552,6211,4177,7987,

%U 6022,18763,16678,21893,8001,25585,13523,9682,30961,32035,7057,36089,19105,39002,7162,47041,50163,51752

%N a(n) is the smallest number that belongs simultaneously to the two arithmetic progressions prime(n) + m*prime(n+1) and prime(n+1) + m*prime(n+2), m >= 1, n >= 1.

%C Construct a table T in which T(m,n) = prime(n) + m*prime(n+1) as shown below. Then a(n) is defined as the smallest number appearing both in column n and column n+1, so a(1)=8, a(2)=33, a(3)=40, etc.

%C .

%C m\n| 1 2 3 4 5 6 7 8 ...

%C ----+--------------------------------------------------

%C 1 | 5 --8 12 18 24 30 36 42 ...

%C |

%C 2 | 8-- 13 19 29 37 47 55 65 ...

%C |

%C 3 | 11 18 26 40 50 64 74 88 ...

%C | /

%C 4 | 14 23 33 / 51 63 81 93 111 ...

%C | / /

%C 5 | 17 28 / 40- 62 76 98 112 134 ...

%C | /

%C 6 | 20 33- 47 73 89 115 131 157 ...

%C | /

%C 7 | 23 38 54 84 102 / 132 150 180 ...

%C | /

%C 8 | 26 43 61 95 115 149 169 203 ...

%C |

%C 9 | 29 48 68 106 128 166 188 226 ...

%C | / /

%C 10 | 32 53 75 117 / 141 183 207 / 249 ...

%C | / /

%C 11 | 35 58 82 128 154 200 226 272 ...

%C |

%C 12 | 38 63 89 139 167 217 245 295 ...

%C |

%C 13 | 41 68 96 150 180 234 264 318 ...

%C |

%C 14 | 44 73 103 161 193 251 283 341 ...

%C |

%C 15 | 47 78 110 172 206 268 302 364 ...

%C | /

%C 16 | 50 83 117 183 219 285 / 321 387 ...

%C | /

%C 17 | 53 88 124 194 232 302 340 410 ...

%C |

%C ... |... ... ... ... ... ... ... ... ...

%C Conjectures:

%C 1. There are infinitely many pairs of consecutive equal terms. (Note that the first pair is (a(7), a(8)).)

%C 2. There exists no N such that the sequence is monotonic for n > N.

%C From _Amiram Eldar_, Sep 22 2018: (Start)

%C Theorem 1: The intersection of the two mentioned arithmetic progressions is always nonempty.

%C Corollary: The sequence is infinite. (End)

%C Sequences that derive from this:

%C 1. Positions in {s(n)} at which a(n) occurs: (2,6,5,11,8,17,19,...).

%C 2. Positions in {s(n+1)} at which a(n) occurs: (1,4,3,9,6,15,15,...).

%C 3. Differences between these two sequences: (1,2,2,2,2,4,...).

%H Alois P. Heinz, <a href="/A319524/b319524.txt">Table of n, a(n) for n = 1..20000</a> (first 600 terms from Muniru A Asiru)

%H Fourth International contest of logical problems, <a href="http://users.skynet.be/albert.frank/fourth_international_contest3.html">Problem 7</a>, the Ludomind Society.

%H Fifth International contest of logical problems, <a href="http://www.sigmasociety.com/Fifth_international_contest.doc">Problem 6</a>, the Ludomind Society, 2009.

%H Olivier Gérard, in reply to Zak Seidov, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2016-April/016273.html">11 related sequences</a>, SeqFan list, Apr 14 2016.

%t a[n_]:=ChineseRemainder[{Prime[n],Prime[n+1]},{Prime[n+1],Prime[n+2]} ];Array[a,44] (* _Amiram Eldar_, Sep 22 2018 *)

%o (GAP) P:=Filtered([1..10000],IsPrime);;

%o T:=List([1..Length(P)-1],n->List([1..Length(P)-1],m->P[n]+m*P[n+1]));;

%o a:=List([1..50],k->Minimum(List([1..Length(T)-1],i->Intersection(T[i],T[i+1]))[k])); # _Muniru A Asiru_, Sep 26 2018

%Y Cf. A001043, A016789, A016885, A017041, A017473, A269100.

%K nonn,look,changed

%O 1,1

%A _Alexandra Hercilia Pereira Silva_, Sep 22 2018

%E Table from _Jon E. Schoenfield_, Sep 23 2018

%E More terms from _Amiram Eldar_, Sep 22 2018