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T(n,k) = highest power of prime(k) dividing n!, read by rows.
7

%I #14 Sep 19 2021 10:47:02

%S 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,

%T 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,

%U 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

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

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

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

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

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

%C T(n,k) = A115627(n,k) for n > 1 and k=1..A000720(n). - _Reinhard Zumkeller_, Nov 01 2013

%H Reinhard Zumkeller, <a href="/A085604/b085604.txt">Rows n = 1..125 of triangle, flattened</a>

%e 0;

%e 1,0;

%e 1,1,0;

%e 3,1,0,0;

%e 3,1,1,0,0;

%e 4,2,1,0,0,0;

%e 4,2,1,1,0,0,0;

%e 7,2,1,1,0,0,0,0;

%e 7,4,1,1,0,0,0,0,0;

%e 8,4,2,1,0,0,0,0,0,0;

%t T[n_, k_] := Module[{p = Prime[k], jm}, jm = Floor[Log[p, n]]; Sum[Quotient[n, p^j], {j, 1, jm}]];

%t Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Sep 19 2021 *)

%o (Haskell)

%o a085604 n k = a085604_tabl !! (n-2) !! (k-1)

%o a085604_row 1 = [0]

%o a085604_row n = a115627_row n ++ (take $ a062298 $ fromIntegral n) [0,0..]

%o a085604_tabl = map a085604_row [1..]

%o -- _Reinhard Zumkeller_, Nov 01 2013

%Y Cf. A141809, A115627, A000142.

%K nonn,tabl

%O 1,7

%A _Reinhard Zumkeller_, Jul 07 2003