

A113029


a(1) = 2, a(2) = 3; for n > 2, a(n) = least prime equal to the sum of two or more previous terms.


0



2, 3, 5, 7, 17, 19, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283
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OFFSET

1,1


COMMENTS

A heuristic argument suggests that all primes except 11, 13 and 23 are included in this sequence (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, May 04 2007
Ryan Murphy's heuristic is correct: a(n) = prime(n+3) for n > 6. It suffices to check that all n in 36..72 are the sum of one or more members of this sequence. Thus, all n > 72 are the sum of two or more distinct members of this sequence, since by Bertrand's postulate there is a prime n/2 < p < n.  Charles R Greathouse IV, Aug 22 2011


LINKS



EXAMPLE

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


MATHEMATICA

(* 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, Apr 08 2007 *)


PROG

(PARI) prevprime(p)={ if( nextprime(p1)<p  p<3, return((p1)*(p>2))); p=bitor(p3, 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(np) && !bittest(disallowed, np) && d=[np])  d=decomp( np, p ), return(concat(d, p)) ))} \ disallowed=0; forprime(p=1, 10^4, if(decomp(p), print1(p", "), disallowed+=1<<p)) \\ M. F. Hasler, May 04 2007
(Python)
from sympy import isprime; P = [2, 3]; B = {2, 3, 5}
while P[1] < 300:
S = set()
for i in B:
if isprime(i) and i > P[1]: P.append(i); break
for j in B: S.add(j+P[1])
B = BS


CROSSREFS



KEYWORD

easy,nonn


AUTHOR



EXTENSIONS

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


STATUS

approved



