OFFSET
1,1
COMMENTS
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" with "noncomposite", 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 were 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_{i=1..n-1} a(i). As a consequence, the sequence grows like a(n) ~ 2^(n-0.256...) and thus is not a counterexample to Erdős's conjecture mentioned in the cited paper.
The sequence of partial sums, s(n) = Sum_{i=1..n} a(i) =
(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.
LINKS
S. J. Benkoski and P. Erdős, On weird and pseudoperfect numbers, Math. Comp., 28 (1974), pp. 617-623. Alternate link; 1975 corrigendum
FORMULA
a(n) > a(n-1) and a(n) <= nextprime((Sum_{i=1..n-1} a(i)) - (-1)^n); but in fact a(n) ~ Sum_{i=1..n-1} a(i) and thus a(n) ~ constant*2^n.
EXAMPLE
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 as 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), yet {0,2,3,5,7,9,10,12} and 11 + {0,2,3,5,7,9,10,12} are disjoint.
PROG
(PARI) s=p=1; for( n=1, 30, while( bitand(s, s>>p=nextprime(p+1)), ); s+=s<<p; print1(p", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Apr 08 2008
EXTENSIONS
a(23)-a(30) from Donovan Johnson, Feb 18 2009
a(31)-a(35) from Giovanni Resta, Feb 28 2020
a(31) corrected and a(36) added by Seth A. Troisi, May 13 2022
STATUS
approved