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%I #57 Nov 28 2021 02:40:41
%S 3,7,3,23,5,89,23,3,139,19,7,3,199,47,17,5,113,83,23,17,3,1831,211,43,
%T 13,7,3,523,109,79,19,11,5,887,317,107,47,17,11,3,1129,619,109,79,19,
%U 7,1669,199,113,73,43,13,5,2477,1373,197,113,71,41,11,3,2971,1123,199,109
%N Special prime numbers arranged in a triangle: n-th row contains m primes p (where m = pi(2n + A020483(n)) - pi(A020483(n))) with following properties.
%C 1: q = p + 2n is also a prime, although not necessarily the next after p;
%C 2: the k-th position of the n-th row gives is a prime p such that the number of further primes between p and q = p + 2n (not counting p and q) is k-1;
%C 3: the primes p are the smallest with these properties.
%C Thus each row only contains primes. The first term in the n-th row is A000230(n). The last one in the same row is A020483(n). The length of the n-th row is pi(2n + A020483(n)) - pi(A020483(n)).
%C From _Martin Raab_, Aug 29 2021: (Start)
%C T(n,k) is zero if there is no admissible pattern with k+1 primes for the interval of length 2n under the given properties.
%C T(38,16) > 2^48. It requires a pattern of 17 primes with a difference of 76 between the first and the last prime. Admissible patterns of this kind exist, but solutions with 17 primes are rather hard to find. (End)
%H Martin Raab, <a href="/A086153/b086153.txt">Rows n = 1..37, flattened</a>
%H Martin Raab, <a href="/A086153/a086153.txt">The first 370 rows of the table arranged as a triangle, including unknown terms.</a>
%e The table begins as follows:
%e 3;
%e 7, 3;
%e 23, 5;
%e 89, 23, 3;
%e 139, 19, 7, 3;
%e 199, 47, 17, 5;
%e 113, 83, 23, 17, 3;
%e ...
%e For example, suppose n = 50: d = 2n = 100; the 50th row consists of 25 terms as follows: {396733, 58789, 142993, 38461, 37699, 7351, 5881, 1327, 2557, 1879, 1621, 1117, 463, 457, 283, 331, 211, 127, 73, 67, 31, ?, ?, 7, 3};
%e A000230(50)=396733, A020483(50)=3; between 143093 and 142993 two primes {143053,143063} occur because 142993 is the 3rd (from 2+1) entry in the 50th row.
%e The length of 50th row is pi(100+3) - pi(3) = pi(103) - pi(3) = 27 - 2 = 25, number of primes between 103 and 3 is 24 (not counting 103 and 3).
%t (* Program to generate the 19th row *) cp[x_, y_] := Count[Table[PrimeQ[i], {i, x, y}], True] {d=38, k=0, mxc=Ceiling[d/3]; vg=PrimePi[30593]} t=Table[0, {mxc}]; t1=Table[0, {mxc}]; Do[s=cp[1+Prime[n], Prime[n]+d-1]; np=d+Prime[n]; If[PrimeQ[np]&&s<(1+mxc)&&t[[s+1]]==0, t[[s+1]]=n; t1[[s+1]]=Prime[n]], {n, 1, 5000}]; {t, t1}
%o (PARI) {z=concat(vector(13),binary(8683781)); for(n=1, 37, p1=3; while(!isprime(p1+2*n), p1=nextprime(p1+2)); p2=p1+2*n; k=primepi(p2)-primepi(p1); r=vector(k); r[k]=p1; i1=1; i2=0; s=vecsort(r); while(s[1+z[n]]==0, while(i1*i2==0, p1+=2; p2+=2; i2=isprime(p2); k=k-i1+i2; i1=isprime(p1)); if(!r[k], r[k]=p1; s=vecsort(r)); i2=0); print("row "n": "r))} \\ _Martin Raab_, Oct 21 2021
%Y Cf. A000720, A000230, A020483, A023190, A086155, A086138-A086149, A086155.
%K nonn,tabf
%O 1,1
%A _Labos Elemer_, Aug 08 2003
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