A329452
There are exactly two primes in {a(n+i) + a(n+j), 0 <= i < j <= 3} for any n: lexicographically earliest such sequence of distinct nonnegative integers.
Original entry on oeis.org
0, 1, 2, 8, 4, 5, 6, 3, 7, 11, 10, 9, 12, 13, 28, 15, 17, 16, 20, 14, 21, 22, 19, 23, 25, 24, 29, 30, 26, 18, 35, 31, 32, 27, 34, 36, 33, 38, 37, 40, 63, 39, 41, 44, 42, 45, 47, 50, 51, 43, 52, 49, 46, 48, 53, 54, 57, 55, 56, 58, 69, 62, 59, 65, 66, 61, 60, 67, 64, 68, 70, 81, 72, 76, 73, 75, 71
Offset: 0
We start with a(0) = 0, a(1) = 1, a(2) = 2, the smallest possibilities which do not lead to a contradiction.
Now there are already 2 primes, 0 + 2 and 1 + 2, among the pairwise sums, so the next term must not generate any further prime. Given 0 and 1, primes and (primes - 1) are excluded, and a(3) = 8 is the smallest possible choice.
Now there is only one prime, 1 + 2 = 3, among the pairwise sums using {1, 2, 8}; the next term must produce exactly one additional prime as sum with these. We see that 3 is not possible (2 + 3 = 5 and 8 + 3 = 11), but a(4) = 4 is possible.
Now using {2, 8, 4} we have no prime as a pairwise sum, so the next term must produce two primes among the sums with these terms. Again, 3 would give three primes, but 5 yields exactly two primes, 2 + 5 = 7 and 8 + 5 = 13.
- 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
Cf.
A329412 (analog for positive integers),
A329453 (2 primes in a(n+i)+a(n+j), i < j < 5).
Cf.
A329333 (one odd prime among a(n+i)+a(n+j), 0 <= i < j < 3),
A329450 (no prime in a(n+i)+a(n+j), i < j < 3).
-
A329452(n,show=0,o=0,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(2<#p,p[^1],p),o); my(c=2-sum(i=2,#p,sum(j=1,i-1,isprime(p[i]+p[j])))); if(#p<3, o=u;next); for(k=u,oo, bittest(U,k-u) || sum(i=1,#p,isprime(p[i]+k))!=c || [o=k, break]));print([u]);o} \\ Optional args: show=1: print a(o..n-1); o=1: use indices & terms >= 1, i.e., compute A329412. See the wiki page for more general code returning a vector: S(n,2,4) = a(0..n-1).
Edited (deleted comments now found on the wiki) by
M. F. Hasler, Nov 24 2019
A329411
Among the pairwise sums of any three consecutive terms there are exactly two prime sums: lexicographically earliest such sequence of distinct positive numbers.
Original entry on oeis.org
1, 2, 3, 4, 7, 6, 5, 8, 9, 10, 13, 16, 15, 14, 17, 12, 11, 18, 19, 22, 21, 20, 23, 24, 29, 30, 31, 28, 25, 33, 34, 26, 27, 32, 35, 36, 37, 42, 41, 38, 45, 44, 39, 40, 43, 46, 51, 50, 47, 53, 54, 48, 49, 52, 55, 57, 82, 56, 75, 62, 64, 87, 63, 76, 61, 66, 65, 71, 86, 60, 77, 67, 72, 59, 68, 69, 58, 70
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 (out of the required two) with the pair {1, 2}.
a(3) = 3 as 3 is the smallest available integer not leading to a contradiction. Since 2 + 3 = 5 we now have our two prime sums with the triplet {1, 2, 3}.
a(4) = 4 as 4 is the smallest available integer not leading to a contradiction. Since 3 + 4 = 7 we now have our two prime sums with the triplet {2, 3, 4}: they are 2 + 3 = 5 and 3 + 4 = 7.
a(5) = 7 because 5 or 6 would lead to a contradiction: indeed, both the triplets {3, 4, 5} and {3, 4, 6} will produce only one prime sum (instead of two). With a(5) = 7 we have the triplet {3, 4, 7} and the two prime sums we were looking for: 3 + 4 = 7 and 4 + 7 = 11.
And so on.
- Jean-Marc Falcoz, Table of n, a(n) for n = 1..10000
- 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
Cf.
A055265 (sum of two consecutive terms is always prime: differs from a(30) on).
Cf.
A329412 ..
A329416 (exactly 2 prime sums using 4, ..., 10 consecutive terms).
Cf.
A055266 (no prime sum among 2 consecutive terms),
A329405 (no prime among the pairwise sums of 3 consecutive terms).
-
a[1]=1;a[2]=2;a[n_]:=a[n]=(k=1;While[Length@Select[Plus@@@Subsets[{a[n-1],a[n-2],++k},{2}],PrimeQ]!=2||MemberQ[Array[a,n-1],k]];k);Array[a,100] (* Giorgos Kalogeropoulos, May 09 2021 *)
-
A329411(n,show=0,o=1,N=2,M=2,p=[],U,u=o)={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])))); 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=0: start with a(o)=o; N, M: find N primes using M+1 consecutive terms. - M. F. Hasler, Nov 16 2019
A329454
There are exactly three primes among a(n+i) + a(n+j), 0 <= i < j <= 3, for any n >= 0: lexicographically earliest such sequence of distinct nonnegative integers.
Original entry on oeis.org
0, 1, 2, 4, 5, 3, 8, 6, 11, 7, 10, 12, 9, 19, 22, 14, 15, 16, 13, 18, 21, 40, 43, 20, 27, 46, 17, 26, 33, 24, 35, 38, 32, 23, 29, 30, 31, 28, 25, 34, 36, 39, 37, 64, 42, 41, 67, 47, 60, 49, 48, 52, 45, 55, 44, 58, 69, 51, 50, 62, 53, 77, 54, 56, 83, 57, 66, 74, 65, 61, 102, 70, 71, 79, 78, 59, 68, 63, 72, 95, 86, 81, 76, 73, 75, 82, 106
Offset: 0
We start with a(0) = 0, a(1) = 1, a(2) = 2, the smallest possibilities which do not lead to a contradiction.
Now there are already 2 primes, 0 + 2 and 1 + 2, among the pairwise sums, so the next term must generate exactly one further prime. It appears that a(3) = 4 is the smallest possible choice.
Then there are again two primes among the pairwise sums using {1, 2, 4}, and the next term must again produce one additional prime as sum with these. We find that a(4) = 5 is the smallest possibility.
Cf.
A329452 (2 primes among a(n+i)+a(n+j), 0 <= i < j < 4),
A329453 (2 primes among a(n+i)+a(n+j), 0 <= i < j < 5).
Cf.
A329333 (1 odd prime among a(n+i)+a(n+j), 0 <= i < j < 3),
A329450 (no primes among a(n+i)+a(n+j), 0 <= i < j < 3).
Cf.
A329405 ff: variants defined for positive integers.
-
Nest[Block[{k = 3}, While[Nand[FreeQ[#, k], Count[Subsets[Append[Take[#, -3], k], {2}], ?(PrimeQ@ Total@ # &)] == 3], k++]; Append[#, k]] &, {0, 1, 2}, 84] (* _Michael De Vlieger, Nov 15 2019 *)
-
A329454(n, show=0, o=0, N=3, M=3, 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(#p
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
A329455
There are exactly three primes in {a(n+i) + a(n+j), 0 <= i < j <= 4} for any n >= 0: lexicographically earliest such sequence of distinct nonnegative integers.
Original entry on oeis.org
0, 1, 2, 4, 8, 6, 3, 10, 14, 11, 5, 9, 15, 26, 12, 17, 13, 7, 18, 16, 20, 21, 19, 23, 27, 40, 22, 31, 24, 25, 29, 28, 30, 32, 33, 39, 34, 36, 35, 38, 41, 46, 37, 43, 48, 42, 55, 47, 44, 45, 52, 49, 50, 53, 56, 58, 54, 57, 51, 73, 76, 61, 59, 63, 64, 68, 60, 69, 67, 62, 65, 66, 70, 71, 72, 79, 77, 74, 81, 86, 78, 89, 82, 85, 80, 99, 84, 83, 75, 92, 87, 88, 90, 91, 93, 94, 100
Offset: 0
We start with a(0) = 0, a(1) = 1, a(2) = 2, the smallest possibilities which do not lead to a contradiction.
Now there are already 2 primes, 0 + 2 and 1 + 2, among the pairwise sums, so the next term must generate exactly one further prime. It appears that a(3) = 4 is the smallest possible choice.
Then there are 3 primes among the pairwise sums using {0, 1, 2, 4}, and the next term must not produce an additional prime as sum with these. The terms 0 and 1 exclude primes and (primes - 1). We find that a(4) = 8 is the smallest possibility.
Then there are 2 primes (1+2 and 1+4) among the pairwise sums using {1, 2, 4, 8}, and the next term must produce exactly one additional prime as sum with these terms. We find that a(5) = 6 is the smallest possibility (since 5+2 and 5+8 would give 2 primes).
Cf.
A329454 (3 primes among a(n+i)+a(n+j), 0 <= i < j <= 3).
Cf.
A329452 (2 primes among a(n+i)+a(n+j), 0 <= i < j <= 3),
A329453 (2 primes among a(n+i)+a(n+j), 0 <= i < j <= 4).
Cf.
A329333 (1 odd prime among a(n+i)+a(n+j), 0 <= i < j <= 2),
A329450 (0 primes among a(n+i)+a(n+j), 0 <= i < j <= 2).
Cf.
A329405 ff: variants defined for positive integers.
-
A329455(n, show=0, o=0, N=3, 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
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
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.
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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
Showing 1-10 of 17 results.
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