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A086153
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.
3
3, 7, 3, 23, 5, 89, 23, 3, 139, 19, 7, 3, 199, 47, 17, 5, 113, 83, 23, 17, 3, 1831, 211, 43, 13, 7, 3, 523, 109, 79, 19, 11, 5, 887, 317, 107, 47, 17, 11, 3, 1129, 619, 109, 79, 19, 7, 1669, 199, 113, 73, 43, 13, 5, 2477, 1373, 197, 113, 71, 41, 11, 3, 2971, 1123, 199, 109
OFFSET
1,1
COMMENTS
1: q = p + 2n is also a prime, although not necessarily the next after p;
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;
3: the primes p are the smallest with these properties.
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)).
From Martin Raab, Aug 29 2021: (Start)
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.
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)
The next unknown values are T(43,19) and T(44,19), which require intervals of 20 primes with a diameter of 86 and 88, respectively. - Brian Kehrig, Jun 25 2024
EXAMPLE
The table begins as follows:
3;
7, 3;
23, 5;
89, 23, 3;
139, 19, 7, 3;
199, 47, 17, 5;
113, 83, 23, 17, 3;
...
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};
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.
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).
MATHEMATICA
(* 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}
PROG
(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
KEYWORD
nonn,tabf
AUTHOR
Labos Elemer, Aug 08 2003
STATUS
approved