A329333 There is exactly one odd prime among the pairwise sums of any three consecutive terms: Lexicographically earliest sequence of distinct nonnegative integers with this property.
0, 1, 2, 7, 3, 6, 4, 5, 8, 10, 11, 9, 12, 14, 15, 13, 18, 17, 19, 20, 21, 24, 16, 23, 25, 22, 26, 27, 28, 31, 29, 32, 33, 34, 30, 39, 37, 36, 38, 41, 40, 42, 43, 46, 35, 44, 47, 45, 50, 51, 48, 49, 56, 52, 53, 54, 57, 55, 58, 59, 68, 60, 63, 64, 61, 66, 62, 69, 67, 72, 71, 65, 74, 70, 75, 76, 77
Offset: 0
Keywords
Examples
For the first two terms there is no restriction regarding primality, so a(0) = 0, a(1) = 1. (If only positive values and indices are considered, then a(1) = 1 and a(2) = 2.) Then a(2) must be such that among { 0+1, 0+a(2), 1+a(2) } there is exactly one odd prime, and 2 works. Then a(3) must be such that among { 1+2, 1+a(3), 2+a(3) } there is only one (odd) prime. Since 1+2 = 3, the other two sums must both yield a composite. This excludes 3, 4, 5 and 6 and the smallest possibility is a(3) = 7. And so on.
Links
- Jean-Marc Falcoz, Table of n, a(n) for n = 0..20000.
- Eric Angelini, Prime sums from neighbouring terms, personal blog "Cinquante signes" (and post to the SeqFan list), Nov. 11, 2019.
- 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[0]=0;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]!=1||MemberQ[Array[a,n-1,0],k]];k);Array[a,100,0] (* Giorgos Kalogeropoulos, May 07 2021 *)
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PARI
A329333(n,show=0,o=0,p=0,U=[])={for(n=o,n-1, show&&print1(o","); U=setunion(U,[o]); while(#U>1&&U[1]==U[2]-1,U=U[^1]); for(k=U[1]+1,oo, setsearch(U,k)|| if(isprime(o+p), isprime(o+k)|| isprime(p+k), isprime(o+k)==isprime(p+k)&&p)||[o&&p=o, o=k, break]));o} \\ Optional args: show = 1: print all values up to a(n); o = 1: start with a(1) = 1; p = 1: compute the variant with a(2) = 3. See the wiki page for more general code which returns the whole vector: Use S(n_max,1,3,1) or S(n_max,1,3,2,[0,1]); S(n_max,1,3,0) gives the variant (0, 1, 3, ...)
Extensions
Entry revised by N. J. A. Sloane, Nov 14 2019 and M. F. Hasler, Nov 15 2019
Comments