Search: id:A138000 Results 1-1 of 1 results found. %I A138000 %S A138000 2,3,7,11,29,53,107,211,431,853,1709,3433,6857,13709,27427,54851,109717, %T A138000 219409,438827,877651,1755319,3510623,7021249,14042491,28084997, %U A138000 56169977,112339957,224679913,449359829,898719707 %N A138000 Least prime such that the subsets of { a(1),...,a(n) } sum up to 2^n different values. %C A138000 Obviously one must exclude previously used primes. Here %C A138000 this is done by requiring 2^n %C A138000 subsets, equivalently one could require a(n) to be different from %C A138000 preceding terms, or larger than a(n-1). %C A138000 (If a value smaller than a(n-1) were possible, then a(n-1) would not %C A138000 have been the minimal choice.) %C A138000 If we replace "prime" by "non-composite", the sequence %C A138000 starts 1, 2, 5, 11, 23, 43, 89, 179, 359, 719, 1433, 2879,... and %C A138000 seems to coincide with A064934, having a different definition, though. %C A138000 The present sequence clearly (cf. a(4)) would not be the same if the %C A138000 definition be changed to "least prime larger than the sum of preceding %C A138000 terms" (as in A064934). %C A138000 It can be seen that a(n) is always very close to sum(a(i),i=1..n-1). %C A138000 As a consequence, the sequence grows like a(n) ~ 2^(n-0.256...) and %C A138000 thus is not a counter-example to Erdos' conjecture mentioned in the %C A138000 cited paper. %C A138000 The sequence of partial sums, s(n) = sum(a(i),i=1..n) = %C A138000 (2,5,12,23,52,105,...) is of alternating parity. If s(n)-1 is prime, %C A138000 this is an upper bound for a(n+1), since the smallest element of the %C A138000 sequence is 2; e.g., a(4)=s(3)-1. Thus if s(n) is even, a(n+1)<=nextprime(s(n)-1). %C A138000 If s(n) is odd, then a(n+1) may be nextprime(s(n)+2) (since the value %C A138000 of s(n) itself is never admissible), %C A138000 as in the case of a(3)=5+2>s(2)=5 which is prime. %D A138000 S. J. Benkoski and P. Erdos, On weird and pseudoperfect numbers, Math. Comp. 28 (1974) 617-623. %F A138000 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 %e A138000 a(1)=2, the smallest prime, since subsets of {2} are {},{2} %e A138000 summing to 0 resp. 2. %e A138000 a(2)=3, the second smallest prime, since subsets %e A138000 {},{2},{3},{2,3} have sums 0, 2, 3, 5 which are all different. %e A138000 Then, 5 is not allowed for a(3), since for {2,3,5}, the sum %e A138000 of the subset {2,3} would be the same than that of {5}. %e A138000 For a(3)=7, however, the set of the previously possible sums, %e A138000 {0,2,3,5} and the set of possible sums using the new element, %e A138000 7+{0,2,3,5} = {7,9,10,12} are disjoint. %e A138000 Obviously this is always true for a(n) larger than the sum of all %e A138000 preceding terms. %e A138000 However, a(4)=11 is smaller than this sum (7+3+2=12); %e A138000 but yet {0,2,3,5,7,9,10,12} and 11+{0,2,3,5,7,9,10,12} are disjoint. %o A138000 (PARI) s=p=1; for( n=1,30, while( bitand(s,s>>p=nextprime(p+1)),); s+=s<