%I #16 Feb 07 2020 16:41:40

%S 5,7,11,31,0,29,0,23,37,0,139,401,53,97,149,199,0,89,367,0,521,0,467,

%T 337,0,251,223,0,1933,113,509,409,701,1543,127,1949,523,0,953,1201,0,

%U 479,331,0,1277,0,2861,3643,0,797,631,0,3407,1087,0,1951,887,1069,1831,293,211,787,2609,541,907,1151

%N Table read by antidiagonals: T(n,k) is the smallest prime that differs from its predecessor and successor by 2n and 2k, respectively.

%C If two consecutive primes p and q appear in the table, then the column number in which p appears is the row number in which q appears. E.g., 23 is in column 3 and 29 is in row 3, 29 is in column 1 and 31 is in row 1, 113 is in column 7 and 127 is in row 7, 3643 is in column 8 and 3659 is in row 8.

%C Nonzero terms on the main diagonal are the terms of A054342.

%H Jon E. Schoenfield, <a href="/A308764/b308764.txt">Table of n, a(n) for n = 1..5050</a> (first 100 antidiagonals, flattened)

%F T(n,k)=0 if n == k (mod 3) !== 0 (mod 3), with the exception of T(1,1)=5.

%e T(1,1)=5 because 5 is the only prime p whose predecessor and successor primes are p-2 and p+2, respectively (i.e., 3 and 7).

%e T(7,2)=127 because 127 is the smallest prime p whose predecessor and successor primes are p-14 and p+4, respectively (i.e., 113 and 131).

%e T(2,2)=0: the only set of three numbers {p-4, p, p+4} that are all prime is the set {3, 7, 11}, but these are not consecutive primes. (For every set of three integers {m-4, m, m+4}, exactly one of the three is divisible by 3.)

%e Table begins:

%e 5, 7, 31, 0, 139, 199, 0, 1933, ...

%e 11, 0, 23, 401, 0, 467, 113, 0, ...

%e 29, 37, 53, 89, 337, 509, 953, 3643, ...

%e 0, 97, 367, 0, 409, 1201, 0, 1831, ...

%e 149, 0, 251, 701, 0, 797, 293, 0, ...

%e 521, 223, 1543, 479, 631, 211, 2633, 4111, ...

%e 0, 127, 331, 0, 787, 7057, 0, 13381, ...

%e 1949, 0, 3407, 2609, 0, 3659, 1847, 0, ...

%e ... ... ... ... ... ... ... ... ...

%Y Cf. A000040 (primes), A001223 (prime gaps), A054342 (first occurrence of distances of equidistant lonely primes).

%K nonn,tabl

%O 1,1

%A _Jon E. Schoenfield_, Jun 23 2019