cp's OEIS Frontend

Consider integers 1 <= r <= n where all the prime divisors p of r also divide n. Call such r of n "regulars" of n, or "r regular to n".

Consider rho = smallest power e of n such that r | n^e. a(n) is the largest value of rho found among regular 1 <= r <= n of n.

For n=1, r=1 | n^0; since it is the only integer in the range 1 <= k <= n and since 1 is the empty product, a(1) = 0.

a(p) = 1 for prime p, since all 1 <= k <= n must either divide or be coprime to p; if k is coprime to p it is nonregular and does not qualify. Thus only k=1 and k=p divide some power of n; 1 | p^0 and p | p^1 by definition. Thus the largest value of rho among r={1,p} is 1.

a(p^x) = 1 for prime powers p^x with x >= 1, since the only regular 1 <= r <= p^x are divisors that are powers {1, p^1, p^2, ... p^(x-1), p^x}; since these r are all divisors d, d | n^1 by definition, thus the largest rho among these r is 1. Thus, a(n) = 1 for n with omega(n) = 1.

a(4) = 1 since all regular 1 <= r <= 4 divide 4; all divisors d divide n^1 by definition, thus the largest rho among these r is 1.

a(n) > 1 for composite n > 4, since there is at least one nondivisor regular ("semidivisor") 1 <= r <= n. (See A243822.) If r does not divide n, then it divides some power n^e with e > 1, since all the prime divisors p of r divide n. Another way to look at this is that the set of prime divisors p of semidivisor r is a subset of the set of prime divisors p of n, yet r does not itself divide n. The multiplicity of at least one prime divisor p of r exceeds that of the corresponding prime divisor p of n. The maximum possible multiplicity of any number m < n pertains to the largest power of 2 < n, thus the greatest possible value of rho = floor(log_2(n)).

a(n) for n = 2 (mod 4) = floor(log_2(n)), since such n is an odd multiple of 2. Thus the largest power of 2^x less than n has multiplicity x for 2 while that in n is 1. Any other prime divisor q of n has multiplicity less than that of 2. Thus 2^x | n^x, and the largest rho among 1 <= r <= n = x = floor(log_2(n)).

a(n) for squarefree n = floor(log_p(n)) = A280363(n), where p is the smallest distinct prime divisor of n.

Consider the standard form prime decomposition of n = p_1^e_1 * p_2^3_2 * ... p_i^e_i. a(n) for all other n is the maximum value of ceiling(floor(log_p_i(n))/e_i), i.e., ceiling(A280363(n)/e_i).

a(n) < n.

a(n) is the longest terminating base-n expansion of 1/r for 1 <= r <= n. Example: in base 10, the longest terminating expansion of 1/r is 3 for r = 8. 1/8 = .125. a(10) = 3.

Also maximum value in row n of A280269.

T(n,k) > 0 for k in row n of A027750. - Michael De Vlieger, May 20 2017

Compare rows to those of triangle A279907, smallest exponent e of n divisible by k. The values of k > -1 in row n of A279907 pertain to k in row n of A162306 rather than k in row n of A027750. - Michael De Vlieger, May 21 2017

This table eliminates the negative values in row n of A279907.

Let k = A162306(n,m), i.e., the value in column m of row n.

T(n,1) = 0 since 1 | n^0.

T(n,p) = 1 for prime divisors p of n since p | n^1.

T(n,d) = 1 for divisors d > 1 of n since d | n^1.

Row n for prime p have two terms, {0,1}, the maximum value 1, since all k < p are coprime to p, and k | p^1 only when k = p.

Row n for prime power p^i have (i+1) terms, one zero and i ones, since all k that appear in corresponding row n of A162306 are divisors d of p^i.

Values greater than 1 pertain only to composite k of composite n > 4, but not in all cases. T(n,k) = 1 for squarefree kernels k of composite n.

Numbers k > 1 coprime to n and numbers that are products of at least one prime q coprime to n and one prime p | n do not appear in A162306; these do not divide n^e evenly.

T(n,k) is nonnegative for all numbers k for which n^k (mod k) = 0, i.e., all the prime divisors p of k also divide n.

The largest possible value s in row n of T = floor(log_2(n)), since the largest possible multiplicity of any number m <= n pertains to perfect powers of 2, as 2 is the smallest prime. This number s first appears at T(2^s + 2, 2^s) for s > 1.

1/k terminates T(n,k) digits after the radix point in base n for values of k that appear in row n of A162306.

Originally from Robert Israel at A279907: (Start)

T(a*b,c*d) = max(T(a,c),T(b,d)) if GCD(a,b)=1, GCD(b,d)=1,T(a,c)>=0 and T(b,d)>=0.

T(n,a*b) = max(T(n,a),T(n,b)) if GCD(a,b)=1 and T(n,a)>=0 and T(n,b)>=0.

(End)