A244740 - OEIS

%I #9 Apr 28 2016 12:40:41

%S 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 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 1,1,1,1,1,1,8,3,2,2,3,1,3,1,2,1,2,1

%N 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 (sum of numbers in row n) = prime(n+1)-2; (column 1) = A244796; (column 2) = A244797; (column 3) = A244798.

%H Clark Kimberling, <a href="/A244740/b244740.txt">Table of n, a(n) for n = 1..500</a>

%e First 12 rows:

%e 1

%e 2 1

%e 3 1 1

%e 3 2 2 1 1

%e 5 1 2 1 1 1

%e 4 3 2 1 2 1 1 1

%e 5 1 3 2 2 1 1 1 1

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

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

%e 7 1 4 2 2 1 3 1 2 1 1 1 1 1 1

%e 8 3 2 2 3 1 3 1 2 1 2 1 1 1 1 1 1 1

%e 7 3 2 1 5 2 2 2 2 1 2 1 2 1 1 1 1 1 1 1

%e For row 4, count these congruences:

%e 11 = (1 mod m) for m = 2, 5, 10;

%e 11 = (2 mod m) for m = 3, 9;

%e 11 = (3 mod m) for m = 4, 8;

%e 11 = (4 mod m) for m = 7;

%e 11 = (5 mod m) for m = 6, so that (row 4) = (3,2,2,1,1).

%t z = 60; f[n_, m_, k_] := f[n, m, k] = If[Mod[Prime[n], m] == k, 1, 0];

%t t[k_] := t[k] = Table[f[n, m, k], {n, 1, z}, {m, 1, -1 + Prime[n]}];

%t u = Table[Count[t[k][[i]], 1], {k, 1, 40}, {i, 1, z}];

%t v = Table[u[[n, k]], {k, 2, 20}, {n, 1, (-1 + Prime[k])/2}]

%t Flatten[v] (* A244740 *)

%Y Cf. A000040, A244796, A244797, A244798, A244799, A244800.

%K nonn,easy,tabf

%O 1,2

%A _Clark Kimberling_, Jul 06 2014