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A244740
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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.
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4
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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
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OFFSET
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1,2
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COMMENTS
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(sum of numbers in row n) = prime(n+1)-2; (column 1) = A244796; (column 2) = A244797; (column 3) = A244798.
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LINKS
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EXAMPLE
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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).
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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nonn,easy,tabf
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AUTHOR
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STATUS
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approved
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