A329411 Among the pairwise sums of any three consecutive terms there are exactly two prime sums: lexicographically earliest such sequence of distinct positive numbers.
1, 2, 3, 4, 7, 6, 5, 8, 9, 10, 13, 16, 15, 14, 17, 12, 11, 18, 19, 22, 21, 20, 23, 24, 29, 30, 31, 28, 25, 33, 34, 26, 27, 32, 35, 36, 37, 42, 41, 38, 45, 44, 39, 40, 43, 46, 51, 50, 47, 53, 54, 48, 49, 52, 55, 57, 82, 56, 75, 62, 64, 87, 63, 76, 61, 66, 65, 71, 86, 60, 77, 67, 72, 59, 68, 69, 58, 70
Offset: 1
Keywords
Examples
a(1) = 1 is the smallest possible choice; there's no restriction on the first term. a(2) = 2 as 2 is the smallest available integer not leading to a contradiction. Note that as 1 + 2 = 3 we already have one prime sum (out of the required two) with the pair {1, 2}. a(3) = 3 as 3 is the smallest available integer not leading to a contradiction. Since 2 + 3 = 5 we now have our two prime sums with the triplet {1, 2, 3}. a(4) = 4 as 4 is the smallest available integer not leading to a contradiction. Since 3 + 4 = 7 we now have our two prime sums with the triplet {2, 3, 4}: they are 2 + 3 = 5 and 3 + 4 = 7. a(5) = 7 because 5 or 6 would lead to a contradiction: indeed, both the triplets {3, 4, 5} and {3, 4, 6} will produce only one prime sum (instead of two). With a(5) = 7 we have the triplet {3, 4, 7} and the two prime sums we were looking for: 3 + 4 = 7 and 4 + 7 = 11. And so on.
Links
- Jean-Marc Falcoz, Table of n, a(n) for n = 1..10000
- Eric Angelini, Prime sums from neighbouring terms [Cached copy of html file, with permission]
- Eric Angelini, Prime sums from neighbouring terms [Cached copy of pdf file, with permission]
- M. F. Hasler, Prime sums from neighboring terms, OEIS wiki, Nov. 23, 2019
Crossrefs
Programs
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Mathematica
a[1]=1;a[2]=2;a[n_]:=a[n]=(k=1;While[Length@Select[Plus@@@Subsets[{a[n-1],a[n-2],++k},{2}],PrimeQ]!=2||MemberQ[Array[a,n-1],k]];k);Array[a,100] (* Giorgos Kalogeropoulos, May 09 2021 *)
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PARI
A329411(n,show=0,o=1,N=2,M=2,p=[],U,u=o)={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])))); for(k=u,oo, bittest(U,k-u)|| min(c-#[0|p<-p, isprime(p+k)], #p>=M) ||[o=k,break]));show&&print([u]);o} \\ Optional args: show=1: print a(o..n-1), show=-1: append a(o..n-1) to the (global) list L, in both cases print [least unused number] at the end; o=0: start with a(o)=o; N, M: find N primes using M+1 consecutive terms. - M. F. Hasler, Nov 16 2019
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