A329563 For all n >= 1, exactly five sums are prime among a(n+i) + a(n+j), 0 <= i < j < 5; lexicographically earliest such sequence of distinct positive numbers.
1, 2, 3, 4, 5, 8, 9, 14, 6, 23, 17, 7, 12, 24, 10, 13, 19, 16, 18, 25, 22, 15, 28, 21, 26, 32, 75, 20, 11, 27, 56, 30, 41, 53, 29, 38, 60, 44, 35, 113, 36, 31, 48, 61, 37, 42, 46, 33, 34, 55, 39, 40, 49, 58, 45, 43, 52, 51, 106, 57, 62, 50, 87, 47, 54, 59, 80, 66, 83, 68
Offset: 1
Keywords
Examples
For n = 1, we consider pairwise sums among the first 5 terms chosen as small as possible, a(1..5) = (1, 2, 3, 4, 5). We see that we have indeed 5 primes among the sums 1+2, 1+3, 1+4, 1+5, 2+3, 2+4, 2+5, 3+4, 3+5, 4+5. Then, to get a(6), consider first the pairwise sums among terms a(2..5), (2+3, 2+4, 2+5; 3+4, 3+5; 4+5), among which there are 3 primes, counted with multiplicity (i.e., the prime 7 is there two times). So the new term a(6) must give exactly two more prime sums with the terms a(2..5). We find that 6 or 7 would give just one more (5+6 resp. 4+7), but a(6) = 8 gives exactly two more, 3+8 and 5+8.
Links
- M. F. Hasler, Prime sums from neighboring terms, OEIS wiki, Nov. 23, 2019
Crossrefs
Programs
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PARI
{A329563(n,show=1,o=1,N=5,M=4,p=[],u=o,U)=for(n=o,n-1, show>0&& print1(o","); show<0&& listput(L,o); U+=1<<(o-u); U>>=-u+u+=valuation(U+1,2); p=concat(if(#p>=M,p[^1],p),o); my(c=N-sum(i=2,#p, sum(j=1,i-1, isprime(p[i]+p[j])))); if(#p
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