A085604 T(n,k) = highest power of prime(k) dividing n!, read by rows.

0, 1, 0, 1, 1, 0, 3, 1, 0, 0, 3, 1, 1, 0, 0, 4, 2, 1, 0, 0, 0, 4, 2, 1, 1, 0, 0, 0, 7, 2, 1, 1, 0, 0, 0, 0, 7, 4, 1, 1, 0, 0, 0, 0, 0, 8, 4, 2, 1, 0, 0, 0, 0, 0, 0, 8, 4, 2, 1, 1, 0, 0, 0, 0, 0, 0, 10, 5, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 10, 5, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 11, 5, 2, 2, 1, 1, 0, 0, 0

Offset

1,7

Comments

T(n,1) = A011371(n); T(n,2) = A054861(n) for n>1;

T(n,k) = number of occurrences of prime(k) as factor in numbers <= n (with repetitions);

Sum{T(n,k): 1<=k<=n} = A022559(n);

T(n, A000720(n)) = 1; T(n,k) = 0, A000720(n)<k<n.

T(n,k) = A115627(n,k) for n > 1 and k=1..A000720(n). - Reinhard Zumkeller, Nov 01 2013

Links

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

Example

0;
1,0;
1,1,0;
3,1,0,0;
3,1,1,0,0;
4,2,1,0,0,0;
4,2,1,1,0,0,0;
7,2,1,1,0,0,0,0;
7,4,1,1,0,0,0,0,0;
8,4,2,1,0,0,0,0,0,0;

Mathematica

T[n_, k_] := Module[{p = Prime[k], jm}, jm = Floor[Log[p, n]]; Sum[Quotient[n, p^j], {j, 1, jm}]];
Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Sep 19 2021 *)

Prog

(Haskell)
a085604 n k = a085604_tabl !! (n-2) !! (k-1)
a085604_row 1 = [0]
a085604_row n = a115627_row n ++ (take $ a062298 $ fromIntegral n) [0, 0..]
a085604_tabl = map a085604_row [1..]
-- Reinhard Zumkeller, Nov 01 2013

Crossrefs

Cf. A141809, A115627, A000142.

Sequence in context: A027200 A035654 A170846 * A345371 A306268 A354490

Adjacent sequences: A085601 A085602 A085603 * A085605 A085606 A085607

Keyword

nonn,tabl

Author

Reinhard Zumkeller, Jul 07 2003

Status

approved