cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-9 of 9 results.

A329456 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.

Original entry on oeis.org

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
Offset: 0

Views

Author

M. F. Hasler, based on an idea from Eric Angelini, Nov 15 2019

Keywords

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.

Examples

			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.

Links

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).

Programs

  • 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

A329581 For every n >= 0, exactly 11 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, 20, 9, 8, 11, 23, 7, 10, 21, 50, 30, 36, 17, 31, 37, 16, 12, 14, 25, 42, 22, 67, 15, 19, 28, 13, 34, 18, 40, 24, 41, 139, 27, 49, 43, 60, 124, 52, 26, 57, 75, 87, 32, 48, 35, 44, 92, 39, 29, 38, 45, 33, 59, 98, 64, 51, 46, 218, 53, 93, 58, 56, 47, 135, 54, 134, 55, 95, 72, 62, 65, 85
Offset: 0

Views

Author

M. F. Hasler, Nov 17 2019

Keywords

Comments

That is, there are 11 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, 8, 9, 10, 13, 24, 23, 30, 29, 14, ...).

Examples

			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 produce exactly one more prime when added to elements of P(7). We find that a(7) = 20 is the smallest possible term (with 20 + 3 = 23).
Then in P(8) = {1, 2, 3, 4, 5, 6, 20} there are S(8) = 8 primes among the pairwise sums, so a(8) must produce exactly 3 more primes when added to elements of P(8). We find a(8) = 9 is the smallest possibility (with 2+9, 4+9 and 20+9).
And so on.

Links

Crossrefs

Cf. A329580 (10 primes using 8 consecutive terms), 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.

Programs

  • PARI
    A329581(n,show=0,o=0,N=11,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

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.

Original entry on oeis.org

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

Views

Author

M. F. Hasler, Feb 09 2020

Keywords

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 positive integers.
This sequence is quite different from the restriction of the "nonnegative" variant A329564 to positive indices: it seems that the two 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.

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

Crossrefs

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.

Programs

  • 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

A329577 For every n >= 0, exactly seven sums are prime among a(n+i) + a(n+j), 0 <= i < j < 7; lexicographically earliest such sequence of distinct nonnegative numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 24, 9, 5, 7, 11, 10, 8, 14, 12, 29, 15, 17, 13, 16, 30, 18, 23, 19, 20, 41, 45, 22, 38, 26, 25, 27, 28, 75, 21, 33, 34, 39, 31, 40, 36, 32, 35, 37, 42, 47, 49, 54, 48, 52, 53, 43, 44, 55, 84, 46, 50, 57, 51, 59, 56, 60, 71, 92, 68, 63, 83, 66, 61, 131, 62, 96, 58, 65, 102, 69, 77, 164
Offset: 0

Views

Author

M. F. Hasler, Nov 17 2019

Keywords

Comments

That is, there are 7 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 not the lexicographically earliest one with this property, which starts (1, 2, 3, 4, 5, 6, 89, 8, 7, 9, 10, 11, 14, 12, 17, 19, 18, 13, ...).

Crossrefs

Cf. A329425 (6 primes using 5 consecutive terms), A329566 (6 primes using 6 consecutive terms).
Cf. A329449 (4 primes using 4 consecutive terms), A329455 (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), A329450 (0 primes using 3 terms).
Cf. A329405 ff: other variants defined for positive integers.

Programs

  • PARI
    A329577(n,show=0,o=0,N=7,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

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.

Original entry on oeis.org

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

Views

Author

M. F. Hasler, Nov 17 2019

Keywords

Comments

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.

Crossrefs

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.

Programs

  • PARI
    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

Views

Author

M. F. Hasler, Nov 17 2019

Keywords

Comments

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.

Examples

			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.

Crossrefs

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.

Programs

  • PARI
    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

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

Views

Author

M. F. Hasler, Feb 09 2020

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

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.

Programs

  • PARI
    {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

Views

Author

M. F. Hasler, Feb 09 2020

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

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.

Programs

  • PARI
    {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

Views

Author

M. F. Hasler, Feb 09 2020

Keywords

Comments

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.)

Examples

			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.

Links

Crossrefs

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).

Programs

  • PARI
    {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
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