1,1

Obviously one must exclude previously used primes. Here

this is done by requiring 2^n

subsets, equivalently one could require a(n) to be different from

preceding terms, or larger than a(n-1).

(If a value smaller than a(n-1) were possible, then a(n-1) would not

have been the minimal choice.)

If we replace "prime" by "non-composite", the sequence

starts 1, 2, 5, 11, 23, 43, 89, 179, 359, 719, 1433, 2879,... and

seems to coincide with A064934, having a different definition, though.

The present sequence clearly (cf. a(4)) would not be the same if the

definition be changed to "least prime larger than the sum of preceding

terms" (as in A064934).

It can be seen that a(n) is always very close to sum(a(i),i=1..n-1).

As a consequence, the sequence grows like a(n) ~ 2^(n-0.256...) and

thus is not a counter-example to Erdos' conjecture mentioned in the

cited paper.

The sequence of partial sums, s(n) = sum(a(i),i=1..n) =

(2,5,12,23,52,105,...) is of alternating parity. If s(n)-1 is prime,

this is an upper bound for a(n+1), since the smallest element of the

sequence is 2; e.g., a(4)=s(3)-1. Thus if s(n) is even, a(n+1)<=nextprime(s(n)-1).

If s(n) is odd, then a(n+1) may be nextprime(s(n)+2) (since the value

of s(n) itself is never admissible),

as in the case of a(3)=5+2>s(2)=5 which is prime.

S. J. Benkoski and P. Erdos, On weird and pseudoperfect numbers, Math. Comp. 28 (1974) 617-623.

a(n) > a(n-1) and a(n) <= nextprime(sum(a(i),i=1..n-1)-(-1)^n) ; but in fact a(n) ~ sum(a(i),i=1..n-1) and thus a(n) ~ constant*2^n

a(1)=2, the smallest prime, since subsets of {2} are {},{2}

summing to 0 resp. 2.

a(2)=3, the second smallest prime, since subsets

{},{2},{3},{2,3} have sums 0, 2, 3, 5 which are all different.

Then, 5 is not allowed for a(3), since for {2,3,5}, the sum

of the subset {2,3} would be the same than that of {5}.

For a(3)=7, however, the set of the previously possible sums,

{0,2,3,5} and the set of possible sums using the new element,

7+{0,2,3,5} = {7,9,10,12} are disjoint.

Obviously this is always true for a(n) larger than the sum of all

preceding terms.

However, a(4)=11 is smaller than this sum (7+3+2=12);

but yet {0,2,3,5,7,9,10,12} and 11+{0,2,3,5,7,9,10,12} are disjoint.

(PARI) s=p=1; for( n=1, 30, while( bitand(s, s>>p=nextprime(p+1)), ); s+=s<<p; print1(p", "))

Cf. A064934.

Sequence in context: A096362 A005479 A120856 this_sequence A034295 A140108 A056354

Adjacent sequences: A137997 A137998 A137999 this_sequence A138001 A138002 A138003

nonn,more

M. F. Hasler (www.univ-ag.fr/~mhasler), Apr 08 2008

a(23)-a(30) from Donovan Johnson (donovan.johnson(AT)yahoo.com), Feb 18 2009