1,1

A heuristic argument suggests that all primes except 11, 13 and 23 are included in this series (tested on first million primes). - Ryan Murphy (murphy(AT)minegoboom.com), Jan 24 2006

Except for 17 which uses all 4 of the previous terms, all the other terms so far use only two or three of the previous terms. This is a more restrictive application of the Goldbach conjecture. - Robert G. Wilson v, Apr 08 2007, May 05 2007

Up to 10^4, all a(n) requiring 4 terms are of the form a(n)=2+7+m+p with m=5 or m=19, i.e. of the form 14+p or 28+p; no a(n)<10^6 requires more than 4 terms. - M. F. Hasler (maximilian.hasler(AT)gmail.com), May 04 2007

5 = 2+3 follows 3, 7 = 5+2 follows 5, 17 = 2+3+5+7 follows 7.

(* first do *) Needs["DiscreteMath`Combinatorica`"] (* then *) f[lst_List] := Block[{k = Length@ lst, p = Infinity, q}, lmt = If[k > 5, Sum[Binomial[k, i], {i, 2, 4}], 2^k + 1]; k++; While[k < lmt, q = Plus @@ NthSubset[k, lst]; If[ ! MemberQ[lst, q] && PrimeQ@q && q < p, p = q]; k++ ]; Append[lst, p]]; Nest[f, {2, 3}, 58] (* Robert G. Wilson v *)

(PARI) prevprime(p)={ if( nextprime(p-1)<p | p<3, return((p-1)*(p>2))); p=bitor(p-3, 1); while( nextprime(p) > p, p-=2 ); p } \ decomp(n, p)={local(d); if(!p, if(n==2|n==3, return([n]), p=n), p=min(n, p)); while( p=prevprime(p), if( bittest(disallowed, p), next); if( (n<2*p & isprime(n-p) & !bittest(disallowed, n-p) & d=[n-p]) | d=decomp( n-p, p ), return(concat(d, p)) ))} \ disallowed=0; forprime(p=1, 10^4, if(decomp(p), print1(p", "), disallowed+=1<<p)) - M. F. Hasler (maximilian.hasler(AT)gmail.com), May 04 2007

Sequence in context: A090725 A089968 A164060 this_sequence A090432 A127042 A069802

Adjacent sequences: A113026 A113027 A113028 this_sequence A113030 A113031 A113032

easy,nonn

Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Jan 03 2006

More terms from Ryan Murphy (murphy(AT)minegoboom.com), Jan 24 2006

Edited by N. J. A. Sloane (njas(AT)research.att.com) at the suggestion of Andrew Plewe, May 14 2007