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Triangle, read by rows, of exponents of primes in canonical prime factorization of n: T(n,k) = greatest number such that prime(k)^T(n,k) divides n, 1 <= k <= n.
3

%I #56 Feb 14 2019 12:00:54

%S 0,1,0,0,1,0,2,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,0,1,0,0,0,3,0,0,0,0,0,

%T 0,0,0,2,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,2,1,

%U 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0

%N Triangle, read by rows, of exponents of primes in canonical prime factorization of n: T(n,k) = greatest number such that prime(k)^T(n,k) divides n, 1 <= k <= n.

%C n = Product_{k=1..n} prime(k)^T(n,k);

%C T(n, A055396(n)) > 0 and T(n,k) = 0 for 1 <= k < A055396(n);

%C T(n, A061395(n)) > 0 and T(n,k) = 0 for A061395(n) < k <= n;

%C Sum_{k=1..n} T(n,k) = A001222(n);

%C Sum_{k=1..n} A057427(T(n,k)) = A001221(n);

%C Sum_{k=1..n} T(n,k)*prime(k) = A001414(n);

%C Sum_{k=1..n} A057427(T(n,k))*prime(k) = A008472(n);

%C Min(T(n,k): 1<=k<=n) = A051904(n);

%C Max(T(n,k): 1<=k<=n) = A051903(n);

%C T(n,1) = A007814(n); T(n,2) = A007949(n), n>1.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeFactorization.html">Prime Factorization</a>

%e Triangle begins:

%e 0,

%e 1, 0,

%e 0, 1, 0,

%e 2, 0, 0, 0,

%e 0, 0, 1, 0, 0,

%e 1, 1, 0, 0, 0, 0,

%e 0, 0, 0, 1, 0, 0, 0,

%e 3, 0, 0, 0, 0, 0, 0, 0,

%e ...

%t Table[IntegerExponent[n, Prime[k]], {n,1,15}, {k,1,n}] // Flatten (* _Amiram Eldar_, Dec 14 2018 *)

%o (PARI) row(n) = vector(n, k, valuation(n, prime(k)));

%o tabl(nn) = for (n=1, nn, print(row(n))); \\ _Michel Marcus_, Dec 14 2018

%Y Cf. A000040, A049084.

%Y Cf. A067255 (same as irregular triangle).

%K nonn,tabl

%O 1,7

%A _Reinhard Zumkeller_, May 22 2003