A109411 Partition the sequence of positive integers into minimal groups so that sum of terms in each group is a semiprime; sequence gives sizes of the groups.

3, 1, 4, 1, 1, 5, 2, 3, 1, 1, 13, 3, 1, 3, 2, 2, 2, 1, 4, 6, 2, 1, 6, 1, 2, 2, 1, 14, 4, 1, 1, 1, 3, 5, 2, 1, 2, 2, 1, 3, 1, 10, 2, 7, 5, 4, 2, 1, 2, 2, 2, 6, 1, 2, 3, 5, 2, 3, 4, 5, 6, 2, 3, 2, 2, 4, 1, 14, 1, 1, 4, 7, 5, 2, 3, 6, 1, 2, 2, 2, 1, 2, 2, 1, 4, 2, 2, 2, 3, 17, 2, 3, 1, 10, 3, 1, 3, 6, 1, 4, 2, 1

Offset

1,1

Comments

Is the sequence finite? If a group begins with a and ends with b then sum of terms is s=(a+b)(b-a+1)/2 and it is not evident that a) there are a's such that it is impossible to find b>=a such that s is semiprime, b) such a's will appear in A109411.

The question is equivalent to the following: Given an odd integer n (=2a-1), can it be represented as p-2q or 2q-p where p,q are prime? I believe the answer is "yes" but the problem may have the same complexity as the Goldbach conjecture. - Max Alekseyev, Jul 01 2005

Links

Alois P. Heinz, Table of n, a(n) for n = 1..20000

Example

The partition begins {1-3},{4},{5-8},{9},{10},{11-15},{16-17},{18-20},{21},{22},{23-35}, {36-38},{39},{40-42},{43-44},{45-46},{47-48},{49},{50-53}, {54-59},{60-61},{62},{63-68},{69},{70-71},{72-73},{74},{75-88}, {89-92},{93},{94},{95},{96-98},{99-103},{104-105}...

Maple

s:= proc(n) option remember; `if`(n<1, 0, a(n)+s(n-1)) end:
a:= proc(n) option remember; local i, k, t; k:=0; t:=s(n-1);
      for i from 1+t do k:=k+i;
        if numtheory[bigomega](k)=2 then return i-t fi
      od
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Nov 26 2015

Mathematica

s={{1, 2, 3}}; a=4; Do[Do[If[Plus@@Last/@FactorInteger[(a+x)(x-a+1)/2]==2, AppendTo[s, Range[a, x]]; (*Print[Range[a, x]]; *)a=x+1; Break[]], {x, a, 20000}], {k, 1, 1000}]; s

Crossrefs

Cf. A133837.

Sequence in context: A055187 A217780 A329316 * A302240 A130307 A130314

Adjacent sequences: A109408 A109409 A109410 * A109412 A109413 A109414

Keyword

nonn

Author

Zak Seidov, Jul 01 2005

Status

approved