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Rectangular array, read by antidiagonals: T(n,k) = greatest m such that 2^m divides prime(n+k+1)-prime(n+1).
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%I #5 Feb 26 2024 10:23:03

%S 1,2,1,3,1,2,1,3,1,1,1,2,1,1,2,4,1,2,3,1,1,2,1,4,2,1,1,2,1,3,1,1,4,2,

%T 1,1,2,1,3,2,1,1,2,3,1,1,5,1,1,3,2,1,1,3,1,1,2,1,1,2,3,1,1,2,1,2,3,1,

%U 2,5,1,1,3,2,1,2,1,1,2,1,3,2,1,1,2,3

%N Rectangular array, read by antidiagonals: T(n,k) = greatest m such that 2^m divides prime(n+k+1)-prime(n+1).

%e Corner:

%e 1 2 3 1 1 4 2 1 2 1 1 3 2 1 3 1

%e 1 1 3 2 1 1 3 1 5 2 1 1 4 1 3 1

%e 2 1 1 2 4 1 3 1 1 2 3 1 2 1 2 6

%e 1 1 3 2 1 2 1 1 5 2 1 4 1 3 2 1

%e 2 1 1 4 1 3 2 1 1 3 1 4 1 1 2 1

%e 1 1 2 1 2 3 1 1 2 1 2 1 1 3 1 1

%e 2 1 2 1 1 3 2 1 3 1 4 2 1 2 6 1

%e 1 3 1 1 2 3 1 2 1 2 4 1 3 2 1 1

%e 1 3 2 1 1 3 1 5 1 1 2 1 1 2 2 3

%e 1 1 2 4 1 2 1 2 3 1 4 2 1 1 1 3

%e Row 1 gives the greatest exponent m such that 2^m divides these differences of primes: 5-3, 7-3, 11-3, 13-3, 17-3, 19-3, 23-3,...

%t p[n_, k_] := p[n, k] = Prime[n + k + 1] - Prime[n + 1];

%t w[n_, k_] := Last[Select[Range[15], IntegerQ[p[n, k]/2^#] &]];

%t Table[w[n, k], {n, 1, 20}, {k, 1, 30}] (* array *)

%t Table[w[n - k + 1, k], {n, 13}, {k, n, 1, -1}] // Flatten (* sequence *)

%Y Cf. A000040, A023572 (row 1).

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Feb 17 2024