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A244740 Irregular triangular array read by rows: T(n,k) = number of positive integers m such that (prime(n) mod m) = k, for k=1..(-1 + prime(k))/2.

%I A244740 #9 Apr 28 2016 12:40:41
%S A244740 1,2,1,3,1,1,3,2,2,1,1,5,1,2,1,1,1,4,3,2,1,2,1,1,1,5,1,3,2,2,1,1,1,1,
%T A244740 3,3,4,1,3,1,2,1,1,1,1,5,3,2,2,4,1,2,1,2,1,1,1,1,1,7,1,4,2,2,1,3,1,2,
%U A244740 1,1,1,1,1,1,8,3,2,2,3,1,3,1,2,1,2,1
%N A244740 Irregular triangular array read by rows:  T(n,k) = number of positive integers m such that (prime(n) mod m) = k, for k=1..(-1 + prime(k))/2.
%C A244740 (sum of numbers in row n) = prime(n+1)-2; (column 1) = A244796; (column 2) = A244797; (column 3) = A244798.
%H A244740 Clark Kimberling, <a href="/A244740/b244740.txt">Table of n, a(n) for n = 1..500</a>
%e A244740 First 12 rows:
%e A244740 1
%e A244740 2 1
%e A244740 3 1 1
%e A244740 3 2 2 1 1
%e A244740 5 1 2 1 1 1
%e A244740 4 3 2 1 2 1 1 1
%e A244740 5 1 3 2 2 1 1 1 1
%e A244740 3 3 4 1 3 1 2 1 1 1
%e A244740 5 3 2 2 4 1 2 1 2 1 1 1 1 1
%e A244740 7 1 4 2 2 1 3 1 2 1 1 1 1 1 1
%e A244740 8 3 2 2 3 1 3 1 2 1 2 1 1 1 1 1 1 1
%e A244740 7 3 2 1 5 2 2 2 2 1 2 1 2 1 1 1 1 1 1 1
%e A244740 For row 4, count these congruences:
%e A244740 11 = (1 mod m) for m = 2, 5, 10;
%e A244740 11 = (2 mod m) for m = 3, 9;
%e A244740 11 = (3 mod m) for m = 4, 8;
%e A244740 11 = (4 mod m) for m = 7;
%e A244740 11 = (5 mod m) for m = 6, so that (row 4) = (3,2,2,1,1).
%t A244740 z = 60; f[n_, m_, k_] := f[n, m, k] = If[Mod[Prime[n], m] == k, 1, 0];
%t A244740 t[k_] := t[k] = Table[f[n, m, k], {n, 1, z}, {m, 1, -1 + Prime[n]}];
%t A244740 u = Table[Count[t[k][[i]], 1], {k, 1, 40}, {i, 1, z}];
%t A244740 v = Table[u[[n, k]], {k, 2, 20}, {n, 1, (-1 + Prime[k])/2}]
%t A244740 Flatten[v] (* A244740 *)
%Y A244740 Cf. A000040, A244796, A244797, A244798, A244799, A244800.
%K A244740 nonn,easy,tabf
%O A244740 1,2
%A A244740 _Clark Kimberling_, Jul 06 2014