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
KEYWORD
nonn
AUTHOR
Zak Seidov, Jul 01 2005
STATUS
approved