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T(n,k) = greatest e such that k^e divides n!, 2<=k<=n (triangle read by rows).
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%I #14 Sep 15 2021 03:29:36

%S 1,1,1,3,1,1,3,1,1,1,4,2,2,1,2,4,2,2,1,2,1,7,2,3,1,2,1,2,7,4,3,1,4,1,

%T 2,2,8,4,4,2,4,1,2,2,2,8,4,4,2,4,1,2,2,2,1,10,5,5,2,5,1,3,2,2,1,5,10,

%U 5,5,2,5,1,3,2,2,1,5,1,11,5,5,2,5,2,3,2,2,1,5,1,2,11,6,5,3,6,2,3,3,3,1,5,1,2,3

%N T(n,k) = greatest e such that k^e divides n!, 2<=k<=n (triangle read by rows).

%H <a href="/index/Fa#factorial">Index entries for sequences related to factorial numbers</a>.

%F T(n,2) = A011371(n); T(n,3) = A054861(n) for n>2; T(n,n) = A011776(n).

%e Array begins:

%e 1

%e 1 1

%e 3 1 1

%e 3 1 1 1

%e 4 2 2 1 2

%e ...

%t T[n_, k_] := IntegerExponent[n!, k];

%t Table[T[n, k], {n, 2, 15}, {k, 2, n}] // Flatten (* _Jean-François Alcover_, Sep 15 2021 *)

%o (PARI) T(n,k) = valuation(n!, k); \\ _Michel Marcus_, Sep 15 2021

%Y Cf. A011371, A054861, A011776.

%K nonn,tabl

%O 2,4

%A _Reinhard Zumkeller_, Sep 14 2004