A329415
Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any seven consecutive terms there are exactly two prime sums.
Original entry on oeis.org
1, 2, 3, 7, 13, 19, 23, 5, 20, 9, 15, 25, 29, 10, 35, 21, 27, 11, 17, 28, 14, 37, 38, 4, 8, 6, 12, 18, 31, 24, 26, 32, 16, 22, 34, 33, 30, 40, 36, 42, 44, 45, 52, 46, 48, 50, 54, 39, 41, 69, 67, 43, 47, 56, 49, 51, 55, 59, 61, 53, 60, 62, 58, 65, 57, 63, 64, 71, 70, 77, 83, 72, 73, 76, 82, 88, 68, 66, 74
Offset: 1
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 one prime sum (on the required two) with the septet {1,2,a(3),a(4),a(5),a(6),a(7)}.
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 septet {1,2,3,a(4),a(5),a(6),a(7)}.
a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, the septets {1,2,3,4,a(5),a(6),a(7)}, {1,2,3,5,a(5),a(6),a(7)} and {1,2,3,6,a(5),a(6),a(7)} will produce more than the two required prime sums. With a(4) = 7 we have no contradiction as the septet {1,2,3,7,a(5),a(6),a(7)} has two prime sums so far: 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 septet); in combination with any other term before it, a(5) = 13 will produce only composite sums.
a(6) = 19 as 19 is the smallest available integer not leading to a contradiction: indeed, the septet {1,2,3,7,13,19,a(7)} shows two prime sums so far: 1 + 2 = 3 and 2 + 3 = 5.
a(7) = 23 as 23 is the smallest available integer not leading to a contradiction; indeed, the septet {1,2,3,7,13,19,23} shows now exactly two prime sums, which are 1 + 2 = 3 and 2 + 3 = 5.
a(8) = 5 as 5 is the smallest available integer not leading to a contradiction and producing exactly two prime sums with the septet {2,3,7,13,19,23,5}, which are 2 + 3 = 5 and 2 + 5 = 7.
And so on.
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A329415(n, show=0, o=1, N=2, M=6, 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(#pM. 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
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.
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.
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{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
A329574
For every n >= 0, exactly 10 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, 8, 9, 10, 14, 33, 15, 20, 27, 26, 11, 32, 16, 41, 21, 57, 116, 22, 51, 38, 23, 50, 63, 86, 6, 17, 24, 77, 65, 18, 13, 114, 25, 36, 28, 35, 43, 12, 31, 61, 66, 40, 19, 47, 42, 90, 241, 7, 52, 37, 34, 45, 30, 55, 49, 394, 58, 73, 39, 48, 64, 109, 115
Offset: 0
At the beginning of the sequence, we must avoid the choice of 6 or 7 for a(6): both appear to be possible at first sight, giving exactly 10 prime sums with n = 0 in the definition, but then make it impossible to find a successor term a(7) for which the definition is satisfied with n = 1.
The same happens again for a(37) and a(58), where the apparently possible value 19 resp. 46 must be avoided.
-
{A329574(n, show=0, o=0, N=10, M=6, X=[[6,6],[6,7],[37,19],[58,46]], 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.
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
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.
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).
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{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
A329407
Among the pairwise sums of any five 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, 12, 14, 4, 20, 21, 6, 18, 15, 10, 3, 17, 5, 11, 16, 25, 9, 19, 23, 30, 26, 32, 22, 33, 24, 27, 28, 36, 29, 34, 35, 40, 31, 41, 37, 44, 38, 43, 39, 42, 45, 46, 47, 48, 49, 68, 51, 57, 54, 53, 61, 58, 62, 50, 52, 59, 56, 60, 55, 67, 63, 65, 66, 69, 75, 77, 64, 71, 70, 72, 73, 76, 74, 80
Offset: 1
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 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) = 12 and we have the single prime sum we need among the last 5 integers {2,7,8,13,12}, which is 19 = 12 + 7.
And so on.
Cf.
A329333 (3 consecutive terms, exactly 1 prime sum).
Cf.
A329405: no prime among the pairwise sums of 3 consecutive terms.
Cf.
A329406 ..
A329410: exactly 1 prime sum using 4, ..., 10 consecutive terms.
Cf.
A329411 ..
A329416: exactly 2 prime sums using 3, ..., 10 consecutive terms.
A329408
Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any six consecutive terms there is exactly one prime sum.
Original entry on oeis.org
1, 2, 7, 8, 13, 14, 12, 20, 4, 22, 35, 10, 6, 16, 28, 29, 5, 34, 21, 15, 3, 11, 17, 18, 9, 27, 31, 19, 33, 24, 25, 32, 30, 26, 36, 38, 39, 40, 42, 46, 48, 45, 23, 54, 69, 37, 43, 41, 50, 44, 47, 49, 55, 61, 53, 62, 51, 57, 59, 63, 60, 58, 52, 64, 56, 77, 67, 65, 68, 66, 75, 78, 70, 74, 72, 80, 73, 71, 81
Offset: 1
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 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) = 12 as 12 is the smallest available integer that produces the single prime sum we need among the last 6 integers {2,7,8,13,14,12}, which is 19 = 12 + 7.
And so on.
A329414
Lexicographically earliest sequence of distinct positive numbers such that among the pairwise sums of any six consecutive terms there are exactly two prime sums.
Original entry on oeis.org
1, 2, 3, 7, 13, 19, 5, 8, 9, 17, 16, 40, 4, 6, 11, 12, 10, 14, 22, 18, 15, 20, 24, 26, 25, 29, 28, 52, 30, 35, 21, 23, 33, 31, 32, 27, 39, 37, 38, 43, 36, 48, 44, 46, 34, 45, 42, 50, 41, 54, 49, 69, 51, 47, 57, 60, 53, 55, 59, 58, 61, 56, 66, 65, 63, 67, 62, 78, 68, 70, 64, 71, 72, 73, 75, 81, 82, 80
Offset: 1
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 one prime sum (on the required two) with the sextuplet {1,2,a(3),a(4),a(5),a(6)}.
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 sextuplet {1,2,3,a(4),a(5),a(6)}.
a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, the sextuplets {1,2,3,4,a(5),a(6)}, {1,2,3,5,a(5),a(6)} and {1,2,3,6,a(5),a(6)} will produce more than the two required prime sums. With a(4) = 7 we have no contradiction as the sextuplet {1,2,3,7,a(5),a(6)} 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 sextuplet); in combination with any other term before it, a(5) = 13 will produce only composite sums.
a(6) = 19 as 19 is the smallest available integer not leading to a contradiction: indeed, the sextuplet {1,2,3,7,13,19} shows exactly the two prime sums we are looking for: 1 + 2 = 3 and 2 + 3 = 5.
a(7) = 5 as 5 is the smallest available integer not leading to a contradiction; indeed, the sextuplet {2,3,7,13,19,5} shows exactly two prime sums, which are 2 + 3 = 5 and 2 + 5 = 7.
And so on.
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A329414(n,show=0,o=1,N=2,M=5,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(#pM. F. Hasler, Nov 15 2019
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
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.
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).
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{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
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).
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{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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