Displaying 1-10 of 12 results found.
a(n) is the smallest positive integer not already in the sequence such that a(n)+a(n-1) is prime, starting with a(1)=1.
+0
40
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, 34, 27, 26, 33, 38, 35, 32, 39, 40, 43, 36, 37, 42, 41, 48, 49, 52, 45, 44, 53, 50, 47, 54, 55, 46, 51, 56, 57, 70, 61, 66, 65, 62, 69, 58, 73, 64, 63, 68, 59, 72, 67, 60
COMMENTS
The sequence is well-defined (the terms must alternate in parity, and by Dirichlet's theorem a(n+1) always exists). - N. J. A. Sloane, Mar 07 2017 Does every positive integer eventually occur? - Dmitry Kamenetsky, May 27 2009. Reply from Robert G. Wilson v, May 27 2009: The answer is almost certainly yes, on probabilistic grounds. It appears that this is the limit of the rows of A051237. That those rows do approach a limit seems certain, and given that that limit exists, that this sequence is the limit seems even more likely, but no proof is known for either conjecture. - Robert G. Wilson v, Mar 11 2011, edited by Franklin T. Adams-Watters, Mar 17 2011 The sequence is also a particular case of "among the pairwise sums of any M consecutive terms, N are prime", with M = 2, N = 1. For other M, N see A055266 & A253074 (M = 2, N = 0), A329333, A329405 - A329416, A329449 - A329456, A329563 - A329581, and the OEIS Wiki page. - M. F. Hasler, Feb 11 2020
EXAMPLE
a(5) = 7 because 1, 2, 3 and 4 have already been used and neither 4 + 5 = 9 nor 4 + 6 = 10 are prime while 4 + 7 = 11 is prime.
MAPLE
local a, i, known ;
option remember;
if n =1 then
1;
else
for a from 1 do
known := false;
for i from 1 to n-1 do
if procname(i) = a then
known := true;
break;
end if;
end do:
if not known and isprime(procname(n-1)+a) then
return a;
end if;
end do:
end if;
end proc:
MATHEMATICA
f[s_List] := Block[{k = 1, a = s[[ -1]]}, While[ MemberQ[s, k] || ! PrimeQ[a + k], k++ ]; Append[s, k]]; Nest[f, {1}, 71] (* Robert G. Wilson v, May 27 2009 *) q=2000; a={1}; z=Range[2, 2*q]; While[Length[z]>q-1, k=1; While[!PrimeQ[z[[k]]+Last[a]], k++]; AppendTo[a, z[[k]]]; z=Delete[z, k]]; Print[a] (*200 times faster*) (* Vladimir Joseph Stephan Orlovsky, May 03 2011 *)
PROG
(HP 50G Calculator) << DUPDUP + 2 -> N M L << { 1 } 1 N 1 - FOR i L M FOR j DUP j POS NOT IF THEN j DUP 'L' STO M 'j' STO END NEXT OVER i GET SWAP WHILE DUP2 + DUP ISPRIME? NOT REPEAT DROP DO 1 + 3 PICK OVER POS NOT UNTIL END END ROT DROP2 + NEXT >> >> Gerald Hillier, Oct 28 2008 (Haskell)
import Data.List (delete)
a055265 n = a055265_list !! (n-1)
a055265_list = 1 : f 1 [2..] where
f x vs = g vs where
g (w:ws) = if a010051 (x + w) == 1
then w : f w (delete w vs) else g ws
(PARI) v=[1]; n=1; while(n<50, if(isprime(v[#v]+n)&&!vecsearch(vecsort(v), n), v=concat(v, n); n=0); n++); v \\ Derek Orr, Jun 01 2015 (PARI) U=-a=1; vector(100, k, k=valuation(1+U+=1<<a, 2); while(bittest(U, k)|| !isprime(a+k), k++); a=k) \\ M. F. Hasler, Feb 11 2020
CROSSREFS
Cf. A086527 (the primes a(n)+a(n-1)). Cf. A070942 (n's such that a(1..n) is a permutation of (1..n)). - Zak Seidov, Oct 19 2011 See A282695 for deviation from identity sequence. A073659 is a version where the partial sums must be primes.
a(n) is the least number not occurring earlier such that a(n)+a(n-1) is prime, a(0) = 0.
+0
19
0, 2, 1, 4, 3, 8, 5, 6, 7, 10, 9, 14, 15, 16, 13, 18, 11, 12, 17, 20, 21, 22, 19, 24, 23, 30, 29, 32, 27, 26, 33, 28, 25, 34, 37, 36, 31, 40, 39, 44, 35, 38, 41, 42, 47, 50, 51, 46, 43, 54, 49, 48, 53, 56, 45, 52, 55, 58, 69, 62, 65, 66, 61, 70, 57, 74, 63, 64, 67, 60, 71, 68, 59
COMMENTS
Original definition: start with a(1) = 2. See A055265 for start with a(1) = 1. The sequence may well be a rearrangement of natural numbers. Interestingly, subsets of first n terms are permutations of 1..n for n = {2, 4, 8, 10, 18, 22, 24, 56, ...}. E.g., first 56 terms: {2, 1, 4, 3, 8, 5, 6, 7, 10, 9, 14, 15, 16, 13, 18, 11, 12, 17, 20, 21, 22, 19, 24, 23, 30, 29, 32, 27, 26, 33, 28, 25, 34, 37, 36, 31, 40, 39, 44, 35, 38, 41, 42, 47, 50, 51, 46, 43, 54, 49, 48, 53, 56, 45, 52, 55} are a permutation of 1..56.
Without altering the definition nor the existing values, one can as well start with a(0) = 0 and get (conjecturally) a permutation of the nonnegative integers. This sequence is in some sense the "arithmetic" analog of the "digital" variant A231433: Here we add subsequent terms, there the digits are concatenated. - M. F. Hasler, Nov 09 2013 The sequence is also a particular case of "among the pairwise sums of any M consecutive terms, N are prime", with M = 2, N = 1. For other M, N see A329333, A329405 ff, A329449 ff and the OEIS Wiki page. - M. F. Hasler, Nov 24 2019
PROG
(PARI) {a=0; u=0; for(n=1, 99, u+=1<<a; print1(a", "); for(k=1, 9e9, bittest(u, k)&&next; isprime(a+k)&&(a=k)&&next(2)))}
For all n >= 0, six among (a(n+i) + a(n+j), 0 <= i < j < 5) are prime: lexicographically first such sequence of distinct nonnegative integers.
+0
18
0, 1, 2, 3, 4, 9, 8, 10, 33, 14, 93, 20, 17, 23, 44, 6, 24, 35, 65, 5, 18, 32, 11, 12, 29, 30, 7, 31, 72, 16, 22, 25, 37, 15, 46, 64, 43, 28, 85, 19, 54, 13, 88, 34, 49, 39, 40, 27, 100, 57, 26, 52, 111, 21, 38, 45, 62, 41, 51, 56, 47, 116, 50, 81, 63, 68, 59, 170, 69, 71
COMMENTS
The restriction to [1, oo) is the lexicographically first such sequence of positive integers. (This is rather exceptional, cf. A128280 vs A055265, A329405 vs A329450, ..., see the wiki page for more.) Conjectured to be a permutation, i.e., all n >= 0 appear. The restriction to [1, oo) is then the lexicographically first such permutation of the positive integers.
Among pairwise sums of 5 consecutive terms, there cannot be more than 2 x 3 = 6 primes: see the wiki page for this and further considerations and variants.
MAPLE
R:= 0, 1, 2, 3, 4:
S:= {R}:
for i from 1 to 100 do
for x from 5 do
if member(x, S) then next fi;
n1:= nops(select(isprime, [seq(seq(R[i+j]+R[i+k], j=1..k-1), k=1..4)]));
if nops(select(isprime, [seq(R[i+j]+x, j=1..4)]))+n1 = 6 then
R:= R, x; S:= S union {x}; break
fi
od od:
PROG
(PARI) A329425_upto(N) = S(N, 6, 5, 0) \\ see the wiki page for the function S().
For any n >= 0, exactly four sums a(n+i) + a(n+j) are prime, for 0 <= i < j <= 3: lexicographically earliest such sequence of distinct nonnegative integers.
+0
22
0, 1, 2, 3, 4, 9, 8, 15, 14, 5, 26, 17, 6, 11, 12, 7, 30, 29, 24, 13, 18, 19, 10, 43, 28, 31, 16, 25, 22, 21, 46, 37, 52, 27, 34, 45, 44, 39, 58, 69, 20, 51, 32, 41, 38, 35, 48, 23, 36, 53, 50, 47, 54, 59, 42, 55, 72, 65, 84, 67, 114, 79, 60, 49, 78, 71, 102, 61, 66, 91, 40, 73, 76, 33, 64, 63, 68
COMMENTS
That is, there are exactly four primes (counted with multiplicity) among the 6 pairwise sums of any four consecutive terms. This is the theoretical maximum: there can't be a sequence with more than 4 prime sums in any 4 consecutive terms, see the wiki page for details.
This map is defined with offset 0 as to have a permutation of the nonnegative integers in case each of these eventually appears, which is so far only conjectured, see below. The restriction to positive indices would then be a permutation of the positive integers, and as it happens, also the smallest one with the given property. (This is in contrast to most other cases where that one is not the restriction of the other one: see crossrefs).
Concerning the existence of the sequence with infinite length: If the sequence is to be computed in a greedy manner, this means that for given P(n) := {a(n-1), a(n-2), a(n-3)} and thus 0 <= N(n) := #{ primes x + y with x, y in P(n), x < y} <= 4, we have to find a(n) such that we have exactly 4 - N(n) primes in a(n) + N(n). It is easy to prove that this is always possible when 4 - N(n) = 0 or 1. Otherwise, similar to A329452, ..., A329456, we see that P(n) is an "admissible constellation" in the sense that a(n-4) + P(n) already gave the number of primes required now. So a weaker variant of the k-tuple conjecture would ensure we can find this a(n). But the sequence need not be computable in greedy manner! That is, if ever for given P(n) no a(n) would exist such that a(n) + P(n) contains 4 - N(n) primes, this simply means that the considered value of a(n-1) (and possibly a(n-2)) was incorrect, and the next larger choice has to be made. Given this freedom, there is no doubt that this sequence is well defined up to infinity. Concerning surjectivity: If a number m would never appear, this means that m + P(n) will never have the required number of 4 - N(n) primes for all n with a(n) > m, in spite of having found for each of these n at least two other solutions, a(n-4) + P(n) and a(n) + P(n) which both gave 4 - N(n) primes. This appears extremely unlikely and thus as strong evidence in favor of surjectivity.
See examples for further computational evidence.
EXAMPLE
We start with a(0) = 0, a(1) = 1, a(2) = 2, a(3) = 3, the smallest possibilities which do not lead to a contradiction. Indeed, the four sums 0 + 2, 0 + 3, 1 + 2 and 2 + 3 are prime.
Now we have 2 prime sums using {1, 2, 3}, so the next term must give two more prime when added to these. We find that a(4) = 4 is the smallest possible choice, with 1 + 4 = 5 and 3 + 4 = 7.
Then there are again 2 primes among the pairwise sums using {2, 3, 4}, so the next term must again produce two more prime sums. We find that a(5) = 9 is the smallest possibility, with 2 + 9 = 11 and 4 + 9 = 13.
a(10^4) = 9834 and all numbers up to 9834 occurred by then.
a(10^5) = 99840 and all numbers below 99777 occurred by then.
a(10^6) = 1000144 and all numbers below 999402 occurred by then.
PROG
(PARI) A329449(n, show=0, o=0, N=4, M=3, p=[], U, u=o)={for(n=o, n-1, if(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=1: start with a(1)=1; N, M: get N primes using M+1 consecutive terms.
CROSSREFS
Other sequences with N primes among pairwise sums of M consecutive terms, starting with a(o) = o, sorted by decreasing N and lowest possible M: A329581 (N=11, M=8, o=0), A329580 (N=10, M=8, o=0), A329569 (N=9, M=6, o=0), A329568 (N=9, M=6, o=1), A329425 (N=6, M=5, o=0), A329449 (N=4, M=4, o=0), A329411 (N=2, M=3, o=0 or 1), A128280 (N=1, M=2, o=0), A055265 (N=1, M=2, o=1), A055266 (N=0, M=2; o=1), A253074 (N=0, M=2; o=0).
For any n >= 0, exactly four sums a(n+i) + a(n+j) are prime, for 0 <= i < j <= 4: lexicographically earliest such sequence of distinct nonnegative integers.
+0
13
0, 1, 2, 3, 24, 4, 5, 7, 8, 6, 9, 10, 11, 13, 18, 12, 16, 19, 29, 25, 42, 14, 15, 17, 20, 21, 22, 23, 26, 38, 45, 27, 28, 33, 40, 32, 31, 39, 30, 41, 48, 49, 36, 35, 34, 37, 43, 66, 47, 50, 46, 51, 52, 53, 55, 54, 44, 56, 83, 63, 59, 68, 64, 67, 72, 85, 57, 70, 79, 78, 58, 60, 61, 121, 76, 71, 90, 73
COMMENTS
That is, there are exactly four primes (counted with multiplicity) among the 10 pairwise sums of any five consecutive terms. (It is possible to have 4 primes among the pairwise sums of any 4 consecutive elements, see A329449.) This map is defined with offset 0 so as to have a permutation of the nonnegative integers in case each of these eventually appears, which is not yet proved (cf. below). The restriction to positive indices would then be a permutation of the positive integers with the same property, but not the lexicographically earliest such, which starts (1, 2, 3, 4, 23, 8, 5, 6, 10, 7, 9, 11, 12, ...).
Concerning the existence of the sequence with infinite length: If the sequence is to be computed in a greedy manner, this means that for given P(n) := {a(n-1), a(n-2), a(n-3), a(n-4)} and thus N(n) := #{ primes x + y with x, y in P(n), x < y} in {0, ..., 4}, we have to find a(n) such that we have exactly 4 - N(n) primes in a(n) + N(n). It is easy to prove that this is always possible when 4 - N(n) = 0 or 1. Otherwise, similar to A329452, ..., A329455, we see that P(n) is an "admissible constellation" in the sense that a(n-5) + P(n) already gave the number of primes required now. So a (weaker) variant of the k-tuple conjecture ensures we can find this a(n). But the sequence need not be computable in greedy manner! That is, if ever for given P(n) no convenient a(n) would exist, this just means that the considered value of a(n-1) (and possibly a(n-2)) was incorrect, and the next larger choice has to be made. Given this freedom, there is no doubt that this sequence is well defined up to infinity. Concerning surjectivity: If a number m would never appear, this means that m + P(n) will never have the required number of 4 - N(n) primes for all n with a(n) > m, in spite of having found for each of these n at least two other solutions, a(n-4) + P(n) and a(n) + P(n) which both gave 4 - N(n) primes. This appears extremely unlikely and thus as strong evidence in favor of surjectivity.
See examples for further computational evidence.
EXAMPLE
We start with a(0) = 0, a(1) = 1, a(2) = 2, a(3) = 3, the smallest possibilities which do not lead to a contradiction. Indeed, the four sums 0 + 2, 0 + 3, 1 + 2 and 2 + 3 are prime.
Now the next term must not give an additional prime when added to any of {0, 1, 2, 3}. We find that a(4) = 24 is the smallest possible choice.
Then there are 2 primes (1+2, 2+3) among the pairwise sums using {1, 2, 3, 24}, so the next term must produce two more prime sums. We find that a(5) = 4 is correct, with 1+4 and 3+4.
a(10^5) = 99948.
a(10^6) = 999923 and all numbers below 999904 occurred by then.
PROG
(PARI) A329455(n, show=0, o=0, N=4, M=4, p=[], U, u=o)={for(n=o, n-1, show>0&& print1(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<M && sum(i=1, #p, isprime(p[i]+u))<=c, o=u)|| for(k=u, oo, bittest(U, k-u) || sum(i=1, #p, isprime(p[i]+k))!=c || [o=k, break])); show&&print([u]); o} \\ Optional args: show=1: print a(o..n-1), show=-1: print only [least unused number] at the end; o=1: start with a(1)=1; N, M: get N primes using M+1 consecutive terms.
CROSSREFS
Other sequences with N primes among pairwise sums of M consecutive terms, starting with a(o) = o, sorted by decreasing N: A329581 (N=11, M=8, o=0), A329580 (N=10, M=8, o=0), A329579 (N=9, M=7, o=0), A329577 (N=7, M=7, o=0), A329566 (N=6, M=6, o=0), A329449 (N=4, M=4, o=0), this A329456 (N=4, M=5, o=0), A329454 (3, 4, 0), A329455 (3, 5, 0), A329411 (2, 3, o=1 and 0), A329452 (2, 4, 0), A329412 (2, 4, 1), A329453 (2, 5, 0), A329413 (2, 5, 1), A329333 (N=1, M=3, o=0 and 1), A329450 (0, 3, 0), A329405 (0, 3, 1).
For all n >= 0, exactly five sums are prime among a(n+i) + a(n+j), 0 <= i < j < 5; lexicographically earliest such sequence of distinct nonnegative numbers.
+0
2
0, 1, 2, 3, 6, 5, 8, 11, 7, 12, 29, 18, 19, 4, 13, 9, 22, 10, 21, 14, 57, 16, 15, 17, 26, 27, 20, 23, 33, 34, 38, 45, 25, 28, 51, 46, 31, 43, 58, 30, 24, 37, 49, 35, 36, 102, 47, 42, 55, 32, 41, 48, 65, 39, 62, 44, 40, 63, 69, 50, 68, 59, 80, 71, 54, 77, 60, 53, 56, 74, 75
COMMENTS
That is, there are 5 primes, counted with multiplicity, among the 10 pairwise sums of any 5 consecutive terms.
Conjectured to be a permutation of the nonnegative integers.
If so, then the restriction to [1..oo) is a permutation of the positive integers, but not the smallest such, which is given in A329563. It seems that the two sequences have no common terms beyond a(6) = 8, except for the accidental a(22) = 15 and maybe some later coincidences of this type. There also appears to be no other simple relation between the terms of these sequences, in contrast to, e.g., A055265 vs. A128280. - M. F. Hasler, Feb 12 2020
EXAMPLE
For n = 0, we consider pairwise sums among the first 5 terms a(0..4), among which we must have 5 primes. To get a(4), consider first a(0..3) = (0, 1, 2, 3) and the pairwise sums (a(i) + a(j), 0 <= i < j <= 3) = (1; 2, 3; 3, 4, 5) among which there are 4 primes, counted with multiplicity (i.e., the prime 3 is there two times). So the additional term a(4) must give exactly one more prime sum with all of a(0..3). We find that 4 or 5 would give two more primes, but a(4) = 6 gives exactly one more, 1 + 6 = 7.
Now, for n = 1 we forget the initial 0 and consider the pairwise sums of the remaining terms {1, 2, 3, 6}. There are 3 prime sums, so the next term must give two more. The term 4 would give two more (1+4 and 3+4) primes, but thereafter we would have {2, 3, 6, 4} with only 2 prime sums and impossibility to add one term to get three more prime sums: 2+x, 6+x and 4+x can't be all prime for x > 1.
Therefore 4 isn't the next term, and we try a(5) = 5 which indeed gives the required number of primes, and also allows us to continue.
PROG
(PARI) { A329564(n, show=1, o=0, N=5, M=4, X=[[4, 4]], p=[], u, 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<M&&sum(i=1, #p, isprime(p[i]+u))<=c, o=u)|| for(k=u, oo, bittest(U, k-u)|| sum(i=1, #p, isprime(p[i]+k))!=c|| setsearch(X, [n, k])|| [o=k, break])); show&&print([u]); o} \\ optional args: show=1: print a(o..n-1), show=-1: append them on global list L, in both cases print [least unused number] at the end. See the wiki page for a function S() which returns a vector: a(0..n-1) = S(5, 5; 0).
CROSSREFS
Cf. A329425 (6 primes using 5 consecutive terms). Cf. A055266 & A253074 (0 primes using 2 terms), A329405 & A329450 (0 primes using 3 terms), A055265 & A128280 (1 prime using 2 terms), A329333, A329406 - A329410 (1 prime using 3, ..., 10 terms), A329411 - A329416 and A329452, A329453 (2 primes using 3, ..., 10 terms), A329454 & A329455 (3 primes using 4 resp. 5 terms), A329449 & A329456 (4 primes using 4 resp. 5 terms), A329568 & A329569 (9 primes using 6 terms), A329572 & A329573 (12 primes using 7 terms), A329563 - A329581: other variants.
For all n >= 1, exactly 9 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6: lexicographically earliest such sequence of distinct positive numbers.
+0
4
1, 2, 3, 9, 4, 10, 27, 14, 33, 57, 26, 40, 87, 50, 21, 63, 16, 20, 51, 8, 81, 93, 46, 56, 15, 58, 135, 183, 28, 44, 39, 88, 69, 123, 34, 68, 105, 128, 45, 129, 22, 52, 141, 38, 75, 159, 32, 82, 99, 64, 117, 147, 80, 94, 177, 116, 237, 273, 74, 100, 387, 76, 207, 357, 62, 104, 165, 86, 77, 95
COMMENTS
That is, there are nine primes, counted with multiplicity, among the 15 pairwise sums of any six consecutive terms. This is the maximum number of possible prime sums for any set of 6 numbers > 1, see wiki page for details.
Conjectured to be a permutation of the positive integers. See A329569 = (0, 1, 2, 5, 6, 11, 12, 17, ...) for the quite different variant for nonnegative integers. For n > 6, a(n) is the smallest number not used earlier such that the set a(n) + {a(n-5}, ..., a(n-1)} has the same number of primes as a(n-6) + {a(n-5), ..., a(n-1)}. Such a number always exists, by definition of the sequence. (If it would not exist for a given n, the term a(n-1) (or earlier) "was wrong and must be corrected", so to say.) See the wiki page for further considerations about existence and surjectivity.
For a(4), one must exclude the values {4, ..., 8} to get an infinite sequence, but for all other (at least several hundred) terms, the greedy choice gives the correct solution.
PROG
(PARI) { A329568(n, show=0, o=1, N=9, M=5, X=[[4, x]|x<-[4..8]], p=[], u=o, U)=for(n=o+1, n, 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|x<-p, isprime(x+k)], #p>=M)|| setsearch(X, [n, k])|| [o=k, break])); show&&print([u]); o} \\ optional args: show=1: print a(o..n-1), show=-1: append them on global list L, in both cases print [least unused number] at the end. Parameters N, M, o, ... allow getting other variants, see the wiki page for more.
For all n >= 0, exactly 9 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6: lexicographically earliest such sequence of distinct nonnegative numbers.
+0
4
0, 1, 2, 5, 6, 11, 12, 17, 26, 35, 36, 47, 24, 77, 32, 65, 62, 149, 74, 9, 8, 39, 14, 15, 4, 3, 28, 33, 38, 69, 10, 51, 20, 21, 58, 93, 16, 81, 46, 13, 70, 27, 76, 37, 34, 97, 52, 7, 30, 49, 40, 31, 22, 67, 82, 19, 42, 25, 64, 85, 18, 109, 54, 43, 88, 139, 84, 145, 94, 79, 112, 55, 48, 289, 144
COMMENTS
That is, there are nine primes, counted with multiplicity, among the 15 pairwise sums of any six consecutive terms. This is the maximum: there can't be more than 9 primes among the pairwise sums of any 6 numbers > 1, cf. wiki page in LINKS.
Conjectured to be a permutation of the nonnegative integers. The restriction to [1,oo) is then a permutation of the positive integers with similar properties, but different from the lexico-smallest one, A329568 = (1, 2, 3, 9, 4, 10, 27, ...). For n > 5, a(n) is the smallest number not used earlier such that the set a(n) + {a(n-5}, ..., a(n-1)} has the same number of primes as a(n-6) + {a(n-5), ..., a(n-1)}. Such a number always exists, by definition of the sequence. (If it would not exist for a given n, the term a(n-1) (or earlier) "is wrong and must be corrected", so to say.) See the wiki page for further considerations about existence and surjectivity.
For a(3) and a(4), one must exclude values 3 & 4 to be able to continue the sequence indefinitely, but in all other cases (at least for several hundred terms), the greedy choice gives the correct solution.
The values 3, 4 and 7 appear quite late at indices 25, 24 resp. 47.
PROG
(PARI) { A329569(n, show=0, o=0, N=9, M=5, X=[[3, 3], [3, 4], [4, 3], [4, 4]], p=[], u=o, U)=for(n=o+1, n, 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|x<-p, isprime(x+k)], #p>=M)|| setsearch(X, [n, k])|| [o=k, break])); show&&print([u]); o} \\ optional args: show=1: print a(o..n-1), show=-1: append them on global list L, in both cases print [least unused number] at the end. Parameters N, M, o, ... allow getting other variants, see the wiki page for more.
For all n >= 0, exactly 12 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 7; lexicographically earliest such sequence of distinct nonnegative numbers.
+0
3
0, 1, 2, 5, 6, 11, 12, 17, 26, 35, 36, 47, 24, 54, 77, 7, 43, 60, 13, 30, 96, 4, 67, 97, 16, 133, 34, 3, 40, 27, 63, 100, 10, 20, 171, 9, 8, 51, 21, 22, 52, 15, 32, 38, 75, 141, 56, 41, 71, 122, 152, 45, 68, 29, 59, 14, 39, 44, 50, 23, 53, 57, 74, 107, 170, 176, 93, 134, 137, 86, 177, 65, 476, 62, 87, 92, 101
COMMENTS
That is, there are 12 primes, counted with multiplicity, among the 21 pairwise sums of any 7 consecutive terms.
This is the theoretical maximum: there can't be more than 12 primes in pairwise sums of 7 distinct numbers > 1. See the wiki page for more details.
Conjectured to be a permutation of the nonnegative integers. See A329573 for the "positive" variant: same definition but with offset 1 and positive terms, leading to a quite different sequence. For a(3) and a(4) resp. a(5) one must forbid the values < 5 resp. < 11 which would be the greedy choices, in order to get a solution for a(7), but from then on, the greedy choice gives the correct solution, at least for several hundred terms.
PROG
(PARI) { A329572(n, show=0, o=0, N=12, M=6, D=[3, 5, 4, 6, 5, 11], p=[], u=o, U)=for(n=o+1, n, 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); D&& D[1]==n&& [o=D[2], D=D[3..-1]]&& next; 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 them on global list L, in both cases print [least unused number] at the end. See the wiki page for more.
For all n >= 1, exactly 12 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 7; lexicographically earliest such sequence of distinct positive numbers.
+0
3
1, 2, 3, 4, 9, 10, 27, 14, 20, 33, 34, 69, 39, 28, 40, 13, 19, 70, 31, 43, 180, 220, 61, 36, 66, 91, 127, 7, 12, 5, 102, 186, 11, 6, 25, 18, 55, 41, 42, 48, 65, 72, 59, 38, 125, 24, 29, 35, 54, 32, 47, 77, 164, 26, 407, 15, 116, 63, 75, 404, 416, 8, 215, 45, 56, 183, 23, 134, 206, 17, 44, 50
COMMENTS
That is, there are 12 primes, counted with multiplicity, among the 21 pairwise sums of any 7 consecutive terms.
This is the theoretical maximum: there can't be more than 12 primes in pairwise sums of 7 distinct numbers > 1. See the wiki page for more details.
Conjectured to be a permutation of the positive integers. See A329572 for the nonnegative variant (same definition but with n >= 0 and terms >= 0), leading to a quite different sequence. For a(5) and a(6) one must forbid values up to 8 in order to be able to find a solution for a(7), but from then on, the greedy choice gives the correct solution, at least for several hundred terms. Small values appearing late are a(30) = 5, a(34) = 6, a(28) = 7, a(62) = 8.
EXAMPLE
Up to and including the 6th term, there is no constraint other than not using a term more than once, since it is impossible to have more than 12 primes as pairwise sums of 6 numbers. So one would first try to use the lexicographically smallest possible choice a(1..6) =?= (1, 2, ..., 6). But then one would have only 7 pairs (i,j) such that a(i) + a(j) is prime, 1 <= i < j <= 6. So one would need 12 - 7 = 5 more primes in {1, 2, ..., 6} + a(7), which is impossible. One can check that even a(1..5) =?= (1,...,5) does not allow one to find a(6) and a(7) in order to have 12 prime sums a(i) + a(j), 1 <= i < j <= 7. Nor is it possible to find a solution with a(5) equal to 6 or 7 or 8. One finds that a(5) = 9, and a(6) = 10, are the smallest possible choices for which a(7) can be found as to satisfy the requirement. In that case, a(7) = 27 is the smallest possible solution, which yields the 12 prime sums 1+2, 2+3, 1+4, 3+4, 2+9, 4+9, 1+10, 3+10, 9+10, 2+27, 4+27, 10+27.
Now, to satisfy the definition of the sequence for n = 2, we drop the initial 1 from the set of consecutive terms, and search for a(8) producing the same number of additional primes together with {2, 3, 4, 9, 10, 27} as did a(1) = 1, namely 3. We see that a(8) = 14 is the smallest possibility. And so on.
It seems that once a(5) and a(6) are chosen, one may always take the smallest possible choice for the next term without ever again running into difficulty. This is in strong contrast to the (exceptional) case of the variant where we require 10 prime sums among 7 consecutive terms, cf. sequence A329574.
PROG
(PARI) { A329573(n, show=0, o=1, N=12, M=6, D=[5, 9, 6, 10], p=[], u=o, U)=for(n=o+1, n, 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); D&& D[1]==n&& [o=D[2], D=D[3..-1]]&& next; 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 them on global list L, in both cases print [least unused number] at the end. See the wiki page for more.
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