A329405
Among the pairwise sums of any three consecutive terms there is no prime: lexicographically earliest such sequence of distinct positive integers.
Original entry on oeis.org
1, 3, 5, 7, 9, 11, 13, 14, 2, 4, 6, 8, 10, 12, 15, 18, 17, 16, 19, 20, 25, 24, 21, 27, 23, 22, 26, 28, 29, 34, 31, 32, 33, 30, 35, 39, 37, 38, 40, 36, 41, 44, 43, 42, 45, 46, 47, 48, 51, 54, 57, 58, 53, 52, 59, 56, 49, 50, 55, 60, 61, 62, 63, 66, 67, 68, 65, 64, 69, 71, 72, 70, 73, 74, 79, 76, 77, 78
Offset: 1
a(1) = 1 from minimality.
a(2) = 3 since 2 would produce 3 (a prime) by making 1 + 2.
a(3) = 5 since 2 or 4 would produce a prime (e.g., 3 + 4 = 7).
a(4) = 7 since 2, 4 or 6 would produce a prime (e.g., 5 + 6 = 11).
...
a(8) = 14 as 2, 4, 6, 8, 10 or 12 would produce a prime together with a(7) = 13 or a(6) = 11.
a(9) = 2 as neither 2 + 13 = 15 nor 2 + 14 = 16 is prime.
And so on.
Cf.
A329333 (3 consecutive terms, exactly 1 prime sum).
Cf.
A329406 ..
A329410 (exactly 1 prime sum using 4, ..., 10 consecutive terms).
Cf.
A329411 ..
A329416 (exactly 2 prime sums using 3, ..., 10 consecutive terms).
-
a[1]=1;a[2]=3;a[n_]:=a[n]=(k=1;While[Or@@PrimeQ[Plus@@@Subsets[{a[n-1],a[n-2],++k},{2}]]||MemberQ[Array[a,n-1],k]];k);Array[a,100] (* Giorgos Kalogeropoulos, May 09 2021 *)
-
A329405(n, show=1, o=1, p=o, U=[])={for(n=o, n-1, show&&print1(p", "); U=setunion(U, [p]); while(#U>1&&U[1]==U[2]-1, U=U[^1]); for(k=U[1]+1, oo, setsearch(U, k) || isprime(o+k) || isprime(p+k) || [o=p, p=k, break])); p} \\ Optional args: show=0: don't print the list; o=0: start with a(0) = 0, i.e., compute A329450. See the wiki page for more general code returning a vector: S(n,0,3,1) = a(1..n).
A329449
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.
Original entry on oeis.org
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
Offset: 0
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.
- Eric Angelini, Prime sums from neighbouring terms, personal blog "Cinquante signes" (and post to the SeqFan list), Nov. 11, 2019.
- Eric Angelini, Prime sums from neighbouring terms [Cached copy of html file, with permission]
- Eric Angelini, Prime sums from neighbouring terms [Cached copy of pdf file, with permission]
- M. F. Hasler, Prime sums from neighboring terms, OEIS Wiki, Nov. 23, 2019.
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).
-
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.
A329416
Among the pairwise sums of any ten consecutive terms there are exactly two prime sums: lexicographically earliest such sequence of distinct positive numbers.
Original entry on oeis.org
1, 2, 3, 7, 13, 19, 23, 25, 31, 32, 17, 8, 26, 37, 43, 49, 14, 38, 55, 61, 11, 20, 35, 67, 73, 79, 57, 9, 5, 15, 21, 42, 27, 12, 33, 30, 39, 45, 47, 18, 48, 6, 51, 24, 63, 69, 72, 75, 16, 36, 54, 60, 22, 66, 10, 4, 40, 29, 28, 34, 44, 41, 46, 50, 52, 58, 64, 53, 70, 71, 59, 62, 76, 56, 82, 88, 94, 65, 100
Offset: 1
a(1) = 1 is the smallest possible choice, there's no restriction on 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 10-set {1,2,a(3),a(4),a(5),a(6),a(7),a(8),a(9),a(10)}.
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 10-set {1,2,a(3),a(4),a(5),a(6),a(7),a(8),a(9),a(10)}.
a(4) = 7 as a(4) = 4, 5 or 6 would lead to a contradiction: indeed, the 10-sets {1,2,3,4,a(5),a(6),a(7),a(8),a(9),a(10)}, {1,2,3,5,a(5),a(6),a(7),a(8),a(9),a(10)} and {1,2,3,6,a(5),a(6),a(7),a(8),a(9),a(10)} will produce more than the two required prime sums. With a(4) = 7 we have no contradiction as the 10-set {1,2,3,7,a(5),a(6),a(7),a(8),a(9),a(10)} has now 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 10-set); 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 10-set {1,2,3,7,13,19,a(7),a(8),a(9),a(10)} 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 10-set {1,2,3,7,13,19,23,a(8),a(9),a(10)} shows only two prime sums so far, which are 1 + 2 = 3 and 2 + 3 = 5.
a(8) = 25 as 25 is the smallest available integer not leading to a contradiction and producing two prime sums so far with the 10-set {1,2,3,7,13,19,23,25,a(9),a(10)}; etc.
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 ..
A329415 (exactly 2 prime sums using 3, ..., 7 consecutive terms).
See also "nonnegative" variants:
A329450 (0 primes using 3 terms),
A329452 (2 primes using 4 terms),
A329453 (2 primes using 5 terms),
A329454 (3 primes using 4 terms),
A329449 (4 primes using 4 terms),
A329455 (3 primes using 5 terms),
A329456 (4 primes using 5 terms).
-
A329416(n, show=0, o=1, N=2, M=9, 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
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
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.
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).
-
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
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.
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.
-
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
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
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.
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).
-
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
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.
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).
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
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.
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).
-
{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
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.
-
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
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
Showing 1-10 of 16 results.
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