A329453 -id:A329453 - cp's OEIS frontend

A329579 For every n >= 0, exactly nine sums are prime among a(n+i) + a(n+j), 0 <= i < j < 7; lexicographically earliest such sequence of distinct nonnegative numbers.

0, 1, 2, 3, 4, 5, 20, 9, 10, 8, 33, 11, 6, 50, 21, 17, 56, 12, 47, 14, 26, 7, 125, 15, 24, 83, 54, 66, 13, 35, 22, 18, 19, 48, 23, 31, 28, 30, 25, 16, 36, 42, 121, 29, 43, 37, 46, 70, 72, 60, 27, 79, 67, 40, 34, 39, 32, 69, 38, 41, 44, 45, 51, 58, 62, 86, 52, 53, 105, 171, 65, 74, 146, 68, 63, 123, 76

Offset: 0

M. F. Hasler, Nov 17 2019

nonn

That is, there are 9 primes, counted with multiplicity, among the 21 pairwise sums of any 7 consecutive terms.
Is this a permutation of the nonnegative integers?
If so, then the restriction to [1..oo) is a permutation of the positive integers, but maybe not the lexicographically earliest one with this property.

Cf. A329577 (7 primes using 7 consecutive terms), A329566 (6 primes using 6 consecutive terms), A329449 (4 primes using 4 consecutive terms).
Cf. A329425 (6 primes using 5 consecutive terms), A329455 (4 primes using 5 consecutive terms), A329455 (3 primes using 5 consecutive terms), A329453 (2 primes using 5 consecutive terms), A329452 (2 primes using 4 consecutive terms).
Cf. A329454 (3 primes using 4 consecutive terms), A329411 (2 primes using 3 consecutive terms), A329333 (1 odd prime using 3 terms), A329450 (0 primes using 3 terms).
Cf. A329405 ff: other variants defined for positive integers.

A329579(n,show=0,o=0,N=9,M=6,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]))));if(#p

A329580 For every n >= 0, exactly 10 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 8: lexicographically earliest such sequence of distinct nonnegative numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 90, 7, 11, 8, 9, 10, 12, 13, 30, 29, 31, 14, 16, 15, 17, 22, 42, 19, 25, 18, 24, 20, 23, 28, 33, 43, 35, 36, 38, 26, 21, 32, 27, 34, 71, 37, 39, 40, 44, 63, 64, 68, 41, 46, 183, 50, 45, 333, 51, 98, 47, 58, 62, 69, 65, 48, 101, 66, 49, 61, 78, 57, 53, 180, 52, 55, 96, 631, 54, 56, 83, 75, 95, 74, 116, 60

Offset: 0

M. F. Hasler, Nov 17 2019

nonn

That is, there are 10 primes, counted with multiplicity, among the 28 pairwise sums of any 8 consecutive terms.
Is this a permutation of the nonnegative integers?
If so, then the restriction to [1..oo) is a permutation of the positive integers, but not the lexicographically earliest one with this property, which starts (1, 2, 3, 4, 5, 6, 7, 19, 10, 8, 9, 12, 11, 18, 13, 29, ...).
We remark the surprisingly large numbers 333 and 631 among the first terms.

			In P(7) := {0, 1, 2, 3, 4, 5, 6} there are already S(7) := 10 primes 0+2, 0+3, 0+5, 1+2, 1+4, 1+6, 2+3, 2+5, 3+4, 5+6 among the pairwise sums, so the next term a(7) must not produce any more primes when added to elements of P(7). We find that a(7) = 90 is the smallest possible term.
Then in P(8) = {1, 2, 3, 4, 5, 6, 90} there are S(8) = 7 primes among the pairwise sums, so a(8) must produce 3 more primes when added to elements of P(8). We find a(8) = 7 is the smallest possibility (with 4+7, 6+7 and 90+7).
And so on.

Cf. A329579 (9 primes using 7 consecutive terms), A329425 (6 primes using 5 consecutive terms).
Cf. A329455 (4 primes using 5 consecutive terms), A329455 (3 primes using 5 consecutive terms), A329453 (2 primes using 5 consecutive terms), A329452 (2 primes using 4 consecutive terms).
Cf. A329577 (7 primes using 7 consecutive terms), A329566 (6 primes using 6 consecutive terms), A329449 (4 primes using 4 consecutive terms).
Cf. A329454 (3 primes using 4 consecutive terms), A329411 (2 primes using 3 consecutive terms), A329333 (1 odd prime using 3 terms), A329450 (0 primes using 3 terms).
Cf. A329405 ff: other variants defined for positive integers.

A329580(n,show=0,o=0,N=10,M=7,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]))));if(#p

A329413 Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any five consecutive terms there are exactly two prime sums.

Original entry on oeis.org

1, 2, 3, 7, 13, 5, 8, 9, 17, 16, 4, 6, 11, 12, 10, 14, 15, 21, 18, 19, 20, 30, 22, 24, 29, 26, 23, 25, 36, 32, 33, 27, 28, 37, 31, 39, 35, 34, 38, 42, 44, 41, 40, 43, 45, 46, 47, 50, 52, 49, 65, 53, 51, 54, 55, 57, 48, 60, 56, 59, 61, 71, 70, 67, 58, 64, 62, 63, 68, 66, 73, 72, 69, 76, 75, 74, 78, 80

Offset: 1

Eric Angelini and Jean-Marc Falcoz, Nov 14 2019

nonn

Conjectured to be a permutation of the positive integers: a(10^6) = 10^6 + 9 and all numbers below 10^6 - 7 are used at that point. - M. F. Hasler, Nov 15 2019

			a(1) = 1 is the smallest possible choice for 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 (on the required two) with the 5-tuple {1,2,a(3),a(4),a(5)}.
a(3) = 3 as 3 is the smallest available integer not leading to a contradiction. Note that as 2 + 3 = 5 we now have the two prime sums required with the 5-tuple {1,2,3,a(4),a(5)}.
a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, the quintuplets {1,2,3,4,a(5)}, {1,2,3,5,a(5)} and {1,2,3,6,a(5)} will produce more than the two required prime sums. With a(4) = 7 we have no contradiction as the 5-tuple {1,2,3,7,a(5)} has now exactly two prime sums: 1 + 2 = 3 and 2 + 3 = 5.
a(5) = 13 as a(5) = 4, 5, 6, 8, 9, 10, 11 or 12 would again lead to a contradiction (more than 2 prime sums with the 5-tuple); in combination with any other term before it, a(5) = 13 will produce only composite sums.
a(6) = 5 as 5 is the smallest available integer not leading to a contradiction: indeed, the 5-tuple {2,3,7,13,5} shows exactly the two prime sums we are looking for: 2 + 3 = 5 and 2 + 5 = 7.
And so on.

Jean-Marc Falcoz, Table of n, a(n) for n = 1..10000

Cf. A329333 (3 consecutive terms, exactly 1 prime sum). See also A329450, A329452 onwards.

A329413(n,show=0,o=1,N=2,M=4,p=[],U,u=o)={for(n=o,n-1, show&&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(#pA329453), N, M: produce N primes using M+1 consecutive terms. - M. F. Hasler, Nov 15 2019

A329564 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.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Feb 09 2020

nonn

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

			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.

M. F. Hasler, Prime sums from neighboring terms, OEIS wiki, Nov. 23, 2019

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.

{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

A329576 For all n >= 1, exactly seven sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6; lexicographically earliest such sequence of distinct positive numbers.

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 11, 26, 15, 9, 14, 32, 17, 20, 21, 27, 10, 16, 19, 7, 12, 13, 24, 6, 23, 35, 25, 37, 18, 36, 22, 31, 61, 28, 30, 39, 40, 43, 33, 64, 38, 45, 34, 29, 63, 50, 44, 53, 42, 59, 47, 54, 48, 41, 90, 49, 55, 52, 108, 58, 46, 51, 121, 73, 78, 76, 100, 79, 81, 151, 60, 67, 112, 70, 69

Offset: 1

M. F. Hasler, Feb 09 2020

nonn

That is, there are 7 primes, counted with multiplicity, among the 15 pairwise sums of any 6 consecutive terms.

Conjectured to be a permutation of the positive integers.

			For n = 1, we must forbid the greedy choice for a(6) which would be 6, which leads to a dead end: there is no possibility to find a subsequent term that would give 7 prime sums together with {2, 3, 4, 5, 6}. If we take the next larger possibility, a(6) = 8, then it works for the next and all subsequent terms.

M. F. Hasler, Prime sums from neighboring terms, OEIS wiki, Nov. 23, 2019

Cf. A329425 (6 primes using 5 consecutive terms), A329566 (6 primes using 6 consecutive terms).
Cf. A329449 (4 primes using 4 consecutive terms), A329456 (4 primes using 5 consecutive terms).
Cf. A329454 (3 primes using 4 consecutive terms), A329455 (3 primes using 5 consecutive terms).
Cf. A329411 (2 primes using 3 consecutive terms), A329452 (2 primes using 4 consecutive terms), A329453 (2 primes using 5 consecutive terms).
Cf. A329333 (1 (odd) prime using 3 terms), A128280 & A055265 (1 prime using 2 terms); A055266 & A253074 (0 primes using 2 terms), A329405 & A329450 (0 primes using 3 terms), A329406 - A329416, A329563 - A329581: other variants.

{A329576(n,show=1,o=1,N=7,M=5,X=[[6,6]],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])))); if(#p

A329565 For all n >= 0, exactly five sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6; lexicographically earliest such sequence of distinct nonnegative numbers.

Original entry on oeis.org

0, 1, 2, 3, 6, 24, 4, 5, 8, 9, 10, 11, 7, 13, 12, 17, 16, 14, 15, 19, 22, 18, 21, 20, 26, 23, 25, 27, 33, 34, 28, 29, 32, 38, 39, 30, 31, 41, 40, 36, 35, 42, 61, 44, 43, 66, 37, 52, 45, 47, 46, 51, 50, 57, 48, 49, 53, 55, 56, 59, 54, 58, 72, 95, 62, 65, 67, 63, 84, 64, 60, 68, 89, 71, 69, 73, 80, 78, 70, 79, 87, 76, 75, 74, 88, 77, 81, 82, 189, 85

Offset: 0

M. F. Hasler, Feb 09 2020

nonn

That is, there are 5 primes, counted with multiplicity, among the 15 pairwise sums of any 6 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.

			For n = 0, we consider pairwise sums of the first 6 terms a(0..5) = (0, 1, 2, 3, 6, 24): We have (a(i) + a(j), 0 <= i < j < 6) = (1; 2, 3; 3, 4, 5; 6, 7, 8, 9; 24, 25, 26, 27, 30) among which there are 5 primes, counted with repetition. If one tries to take a(4) equal to 4 or 5, this yields already 6 primes among the pairwise sums of the first 5 terms, so the smallest possible choice is a(4) = 6, and thereafter any a(5) less than 24 would again yield too many prime sums. So (0, 1, 2, 3, 6, 24) is indeed the start of the lexicographically earliest nonnegative sequence with the required properties.
Then one finds that a(6) = 4 is possible, giving also 6 prime sums for n = 1, so this is the correct continuation (modulo later confirmation that the sequence can be continued without contradiction given this choice).
Next one finds that a(7) = 5 is also possible, and so on.

M. F. Hasler, Prime sums from neighboring terms, OEIS wiki, Nov. 23, 2019

Cf. A329425 (6 primes using 5 consecutive terms), A329566 (6 primes using 6 consecutive terms).
Cf. A329449 (4 primes using 4 consecutive terms), A329456 (4 primes using 5 consecutive terms).
Cf. A329454 (3 primes using 4 consecutive terms), A329455 (3 primes using 5 consecutive terms).
Cf. A329411 (2 primes using 3 consecutive terms), A329452 (2 primes using 4 consecutive terms), A329453 (2 primes using 5 consecutive terms).
Cf. A329333 (1 (odd) prime using 3 terms), A128280 & A055265 (1 prime using 2 terms); A055266 & A253074 (0 primes using 2 terms), A329405 & A329450 (0 primes using 3 terms), A329406 ff: other variants.

{A329565(n,show=0,o=0,N=5/*#primes*/,M=5,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]))));if(#p

A329567 For all n >= 0, exactly seven sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6: lexicographically earliest such sequence of distinct nonnegative numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 8, 5, 11, 26, 15, 9, 32, 14, 17, 20, 21, 27, 10, 16, 19, 7, 12, 13, 24, 6, 23, 35, 25, 37, 18, 36, 22, 31, 61, 28, 30, 39, 40, 43, 33, 64, 38, 45, 34, 29, 63, 50, 44, 53, 42, 59, 47, 54, 48, 41, 90, 49, 55, 52, 108, 58, 46, 51, 121, 73, 78, 76, 100, 79, 81, 151, 60, 67, 112, 70, 69, 82, 62, 87, 57, 80, 111, 56, 71, 66, 68, 86, 83, 65

Offset: 0

M. F. Hasler, Feb 09 2020

nonn

That is, there are 7 primes, counted with multiplicity, among the 15 pairwise sums of any 6 consecutive terms.
Conjectured to be a permutation of the nonnegative integers. The restriction to [1,oo) is then a permutation of the positive integers with the same property, but not the smallest one which is A329576 = (1, 2, 3, 4, 5, 8, 11, ...).
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, this means the term a(n-1) (or earlier) "was wrong and must be corrected", so to say. Of course this only refers to an incorrect computation.)

			Using the smallest possible 5 initial terms a(0..4) = (0, 1, 2, 3, 4), we have a total of 6 primes among the pairwise sums, namely 0+2, 0+3, 1+2, 1+4, 2+3 and 3+4. To satisfy the definition for n = 0, the next term a(5) must give exactly one more prime when added to these 5 initial terms. The smallest number with this property is 6, but this choice of a(5) would make it impossible to find a suitable a(7): Indeed, for n = 1 we must consider the pairwise sums of (1, 2, 3, 4, a(5), a(6)). If we had (1, 2, 3, 4, 6, a(6)), the first 5 terms would give 5 prime sums 1+2, 1+4, 1+6, 2+3 and 3+4. Then a(6) should give two more prime sums, which is easily possible, either with even a(6) such that 1+a(6) and 3+a(6) are prime, or odd a(6) such that two among {2, 4, 6} + a(6) are prime. Thereafter, for n = 2, we drop the 1 and include a(7) instead, which must produce the same number of prime sums when added to {2, 3, 4, 6, a(6)} as was the case for 1. For even a(6) this was 4 (1+2, 1+4, 1+6 and 1+a(6)), which is impossible to achieve with a(7) > 1, since 2+x, 4+x and 6+x can't be all prime for x > 1. For odd a(6) it is 3 (1+2, 1+4 and 1+6), which is also impossible as well for odd a(7) (same reason as before) as for even a(7) (since only 3 and a(6) are odd and can give a prime sum).
This shows that we can't take a(5) equal to 6, and must consider the next larger possibility, which is a(5) = 8 (with prime sum 3+8 = 11, while 7 would give more than one, 0+7 and 4+7). Now we find that the smallest possible a(6) = 5 yields a solution and all subsequent terms can also be chosen greedily.

Éric Angelini, Prime sums from neighbouring terms, SeqFan list, Nov 11 2019.
Éric Angelini, Prime sums from neighbouring terms, SeqFan list, Nov 11 2019. [Cached copy, with permission]
Eric Angelini, Prime sums from neighbouring terms, personal blog "Cinquante signes" [Cached copy of html file, with permission]
Eric Angelini, Prime sums from neighbouring terms, personal blog "Cinquante signes" [Cached copy of pdf file, with permission]
M. F. Hasler, Prime sums from neighboring terms, OEIS wiki, Nov. 23, 2019.

Cf. related sequences with N prime sums using M consecutive terms, labeled (N,M): A329425 (6,5), A329566 (6,6), A329449 (4,4), A329456 (4,5), A329454 & A329416 (3,4), A329455 (3,5), A329411 (2,3), A329452 (2,4), A329453 (2,5), A329333 (1,3), A128280 & A055265 (1,2); A055266 & A253074 (0,2), A329405 & A329450 (0,3), A329406 - A329416: (1,4) ... (2,10).

{A329567(n,show=1,o=0,N=7,M=5,X=[[4,6]],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

cp's OEIS Frontend

A329579 For every n >= 0, exactly nine sums are prime among a(n+i) + a(n+j), 0 <= i < j < 7; lexicographically earliest such sequence of distinct nonnegative numbers.

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A329580 For every n >= 0, exactly 10 sums are prime among a(n+i) + a(n+j), 0 <= i < j < 8: lexicographically earliest such sequence of distinct nonnegative numbers.

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A329413 Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any five consecutive terms there are exactly two prime sums.

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A329564 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.

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A329576 For all n >= 1, exactly seven sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6; lexicographically earliest such sequence of distinct positive numbers.

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A329565 For all n >= 0, exactly five sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6; lexicographically earliest such sequence of distinct nonnegative numbers.

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A329567 For all n >= 0, exactly seven sums are prime among a(n+i) + a(n+j), 0 <= i < j < 6: lexicographically earliest such sequence of distinct nonnegative numbers.

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