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.

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, 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, 1, 1, 1, 1, 1, 1, 8, 3, 2, 2, 3, 1, 3, 1, 2, 1, 2, 1

Offset

1,2

Comments

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

Links

Clark Kimberling, Table of n, a(n) for n = 1..500

Example

First 12 rows:
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
3 3 4 1 3 1 2 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 1 1 1 1 1 1
8 3 2 2 3 1 3 1 2 1 2 1 1 1 1 1 1 1
7 3 2 1 5 2 2 2 2 1 2 1 2 1 1 1 1 1 1 1
For row 4, count these congruences:
11 = (1 mod m) for m = 2, 5, 10;
11 = (2 mod m) for m = 3, 9;
11 = (3 mod m) for m = 4, 8;
11 = (4 mod m) for m = 7;
11 = (5 mod m) for m = 6, so that (row 4) = (3,2,2,1,1).

Mathematica

z = 60; f[n_, m_, k_] := f[n, m, k] = If[Mod[Prime[n], m] == k, 1, 0];
t[k_] := t[k] = Table[f[n, m, k], {n, 1, z}, {m, 1, -1 + Prime[n]}];
u = Table[Count[t[k][[i]], 1], {k, 1, 40}, {i, 1, z}];
v = Table[u[[n, k]], {k, 2, 20}, {n, 1, (-1 + Prime[k])/2}]
Flatten[v] (* A244740 *)

Crossrefs

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

Sequence in context: A349685 A218578 A006346 * A088742 A256435 A367412

Adjacent sequences: A244737 A244738 A244739 * A244741 A244742 A244743

Keyword

nonn,easy,tabf

Author

Clark Kimberling, Jul 06 2014

Status

approved