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.

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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

A329406 Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any four consecutive terms there is exactly one prime sum.

Original entry on oeis.org

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

Views

Author

Eric Angelini and Jean-Marc Falcoz, Nov 13 2019

Keywords

Comments

For all n >= 1, there is exactly one prime in {a(n+i) + a(n+j), 0 <= i < j <= 3}. See A329450, A329452 onwards for variants for nonnegative integers. - M. F. Hasler, Nov 14 2019

Examples

			a(1) = 1 by minimality.
a(2) = 2 as 2 is the smallest available integer not leading to a contradiction. Note that as 1 + 2 = 3 we already have our prime sum.
a(3) = 7 as a(3) = 3, 4, 5 or 6 would produce one prime sum too many.
a(4) = 8 as a(4) = 3, 4, 5 or 6 would again produce one prime sum too many.
a(5) = 4 as a(5) = 3 would produce two primes instead of one (3 + 2 = 5 and 3 + 8 = 11); with a(5) = 4 we have the single prime sum we need among the last 4 integers {2,7,8,4}: 11 = 4 + 7.
And so on.

Links

Crossrefs

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

A329566 For all n >= 0, exactly six 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, 24, 5, 7, 6, 8, 9, 10, 11, 13, 18, 19, 16, 12, 28, 31, 17, 15, 14, 22, 26, 20, 21, 27, 23, 30, 32, 80, 41, 38, 51, 39, 62, 29, 35, 44, 34, 45, 54, 25, 49, 33, 64, 36, 37, 40, 46, 61, 47, 42, 43, 55, 66, 58, 65, 48, 72, 79, 52, 53, 59, 78, 50, 57, 60, 89, 71, 56, 68, 63, 74, 75, 76, 69, 82, 81, 67, 91, 88, 70, 100
Offset: 0

Views

Author

M. F. Hasler, Nov 17 2019

Keywords

Comments

That is, there are 6 primes, counted with multiplicity, among the 15 pairwise sums of any 6 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, 7, 6, 8, 9, 10, 11, 13, 18, 19, 16, 12, 24, ...).

Examples

			For n = 0, we consider pairwise sums of the first 6 terms a(0..5) = (0, 1, 2, 3, 4, 24): We have (a(i) + a(j), 0 <= i < j < 6) = (1; 2, 3; 3, 4, 5; 4, 5, 6, 7; 24, 25, 26, 27, 28) among which there are 6 primes, counted with repetition. This justifies taking a(0..4) = (0, ..., 4), the smallest possible choices for these first 5 terms. Since no smaller a(5) between 5 and 23 has this property, this is the start of the lexicographically earliest nonnegative sequence with this property and no duplicate terms.
Then we find that a(6) = 5 is possible, also giving 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 we find that a(7) = 6 is not possible, it would give only 5 prime sums using the 6 consecutive terms (2, 3, 4, 24, 5, 6). However, a(7) = 7 is a valid continuation, and so on.

Links

Crossrefs

Cf. A329425 (6 primes using 5 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
    A329566(n,show=0,o=0,N=6,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

A329410 Among the pairwise sums of any ten consecutive terms there is exactly one prime sum: lexicographically earliest such sequence of distinct positive numbers.

Original entry on oeis.org

1, 2, 7, 8, 13, 14, 19, 20, 25, 26, 108, 32, 37, 38, 44, 50, 10, 40, 12, 18, 28, 48, 105, 6, 4, 16, 24, 30, 36, 42, 54, 56, 9, 46, 22, 60, 66, 68, 72, 76, 78, 82, 93, 34, 52, 62, 43, 83, 92, 102, 23, 53, 3, 29, 31, 27, 33, 41, 15, 88, 5, 11, 17, 35, 45, 47, 55, 57, 21, 64, 51, 59, 61, 65, 39, 69, 71, 77, 79, 136, 49, 67, 63
Offset: 1

Views

Author

Eric Angelini and Jean-Marc Falcoz, Nov 13 2019

Keywords

Examples

			a(1) = 1 by minimality.
a(2) = 2 as 2 is the smallest available integer not leading to a contradiction. Note that as 1 + 2 = 3 we have already the prime sum we need.
a(3) = 7 as a(3) = 3, 4, 5 or 6 would produce at least one prime sum too many.
a(4) = 8 as a(4) = 3, 4, 5 or 6 would again produce at least one prime sum too many.
a(5) = 13 as a(5) = 3, 4, 5, 6, 9, 10, 11 or 12 would also produce at least one prime sum too many.
a(6) = 14 as a(6) = 14 doesn't produce an extra prime sum - only composite sums.
a(7) = 19 as a(7) = 3, 4, 5, 6, 9, 10, 11, 12, 15, 16, 17 or 18 would produce at least a prime sum too many.
a(8) = 20 as a(8) = 20 doesn't produce an extra prime sum - only composite sums.
a(9) = 25 as a(9) = 3, 4, 5, 6, 9, 10, 11, 12, 15, 16, 17, 18, 21, 22, 23 or 24 would produce at least a prime sum too many.
a(10) = 26 as(10) = 26 doesn't produce an extra prime sum - only composite sums.
a(11) = 108 is the smallest available integer that produces the single prime sum we need among the last 10 integers {2,7,8,13,14,19,20,25,26,108}, which is 127 = 108 + 19.
And so on.

Links

Crossrefs

Cf. A329333 (3 consecutive terms, exactly 1 prime sum).
Cf. A329405 (no prime among the pairwise sums of 3 consecutive terms).
Cf. A329406 .. A329409 (exactly 1 prime sum using 4, ..., 7 consecutive terms).
Cf. A329411 .. A329416 (exactly 2 prime sums using 3, ..., 10 consecutive terms).
See also A329450, A329452 onwards for "nonnegative" variants.

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

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

Views

Author

Eric Angelini and Jean-Marc Falcoz, Nov 14 2019

Keywords

Comments

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

Examples

			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.

Links

Crossrefs

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

Programs

  • PARI
    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

A329409 Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any seven consecutive terms there is exactly one prime sum.

Original entry on oeis.org

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

Views

Author

Eric Angelini and Jean-Marc Falcoz, Nov 13 2019

Keywords

Examples

			a(1) = 1 by minimality.
a(2) = 2 as 2 is the smallest available integer not leading to a contradiction. Note that as 1 + 2 = 3 we have already the prime sum we need.
a(3) = 7 as a(3) = 3, 4, 5 or 6 would produce at least a prime sum too many.
a(4) = 8 as a(4) = 3, 4, 5 or 6 would again produce at least a prime sum too many.
a(5) = 13 as a(5) = 3, 4, 5, 6, 9, 10, 11 or 12 would also produce at least one prime sum too many.
a(6) = 14 as a(6) = 14 doesn't produce an extra prime sum - only composite sums.
a(7) = 19 as a(7) = 15, 16, 17 or 18 would produce at least a prime sum too many.
a(8) = 36 is the smallest available integer that produces the single prime sum we need among the last 7 integers {2, 7, 8, 13, 14, 19, 36}, which is 43 = 36 + 7.
And so on.

Links

Crossrefs

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