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.

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.

Original entry on oeis.org

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

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Author

Zak Seidov, Jul 01 2005

Keywords

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

Examples

			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}...
		

Crossrefs

Cf. A133837.

Programs

  • 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