cp's OEIS Frontend

The k are the "semitotatives" of n as counted by A243823(n).

All nonzero terms k are composite and pertain to composite rows n. This is because prime k must either divide or be coprime to n, and k = 1 is both a divisor of and coprime to n. Further, the terms k must have at least two distinct prime divisors p and q.

Row n for prime p contains zero, since numbers 1 <= k < p must either divide or be coprime to prime p.

Row n for prime powers p^e contains all the numbers k in the corresponding row of A133995. There is only one prime divisor p of p^e and every power 1 <= m <= e of p divides p^e, thus none of the terms of the corresponding row of A133995 are in A272618(n).

Rows n = 4 and 6 are special cases of composite n that contains zero. 4 is the smallest composite number; there are no composites k < n. 6 has the prime divisors 2 and 3, thus 5 is the smallest prime coprime to 6; the product of the minimum prime divisor and minimum prime coprime to 6 is 10, which exceeds 6 and falls outside the considered range. The situation is not so for composite n > 6. Thus rows n for composite n > 6 contain at least 1 nonzero value.

The smallest k of row n = A096014(n) < n, i.e., those values of A096014(n) pertaining to composite n > 6, a product of the smallest prime divisor p of n and the smallest prime q coprime to n. The smallest k of n are even squarefree semiprimes since 2 either divides n or is coprime to n and k is by definition a number with at least two distinct primes. The smallest k = 2p for p^2 sets record values for A096014(n) when we ignore values pertaining to prime n, n = 4, and n = 6.

In base n, 1/a(n) has a mixed recurrent expansion.

Semitotatives m < n have a regular factor that is the product of prime divisors of n, and a coprime factor that is the product of primes q that are coprime to n.

The unit fractions of semitotatives have a mixed recurrent expansion in base n (See Hardy & Wright).

a(n) = 0 for prime powers and squarefree numbers.

Numbers k that are neither prime powers nor squarefree, such that rad(k) * A053669(k) < k and k/rad(k) >= A119288(k), where rad(k) = A007947(k).

Numbers k such that A360480(k), A360543(k), A361235(k), and A355432(k) are positive.

Subset of A126706. All terms are neither prime powers nor squarefree.

From Michael De Vlieger, Aug 03 2023: (Start)

Superset of A286708 = A001694 \ {{1} U A246547}, which in turn is a superset of A303606. We may write k in A286708 as m*rad(k)^2, m >= 1. Since omega(k) > 1, it is clear both k/rad(k) > A053669(k) and k/rad(k) >= A119288(k). Also superset of A359280 = A286708 \ A303606.

This sequence contains {A002182 \ A168263}. (End)

a(n) = A000203(n) when n is prime or a perfect prime power (A000961). This is because all products of the prime divisor p in such numbers produce divisors.

a(n) > A000203(n) when n is composite and not a perfect prime power.

Product matrix of tensors T = 1,p,p^2,...,p^e that include the powers 1 <= e of prime divisors p such that p^e <= n.

This sequence is analogous to A275055 but differs from it in that the tensors T include not only powers p^e that divide n but all powers p^e <= n.

The matrix a(n) is bounded by n, thus all products m <= n.

Let omega(n) = A001221(n). The matrix a(n) has omega(n) dimensions and is an omega(n)-dimensional simplex with (omega(n)-1) "right-angle: sides and 1 irregular surface that is bounded by n.

A027750(n) is a subset of A162306(n) and in a(n), the terms of A275055(n) appear in an contiguous omega(n)-dimensional parallelepiped (parallelotope) with 1 at the origin and n at the opposite corner. Thus the omega(n)-dimensional array described by A275055(n) is fully contained in the simplex-like matrix described by a(n). Divisors appear within the parallelepiped while nondivisors appear in the field outside the parallelepiped (see examples). Terms within the parallelepiped appear in A027750(n) while those outside appear in A272618(n).

For a(2^x + 2) there is a term m = (n-2); m != (n-1) except for n=2, since GCD(n, n-1)=1.

a(p^e) = A027750(p^e) = A162306(p^e) = A275055(p^e) for e >= 1.

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 maximum value 1, since all k < p are coprime to p, and k | p^1 only when k = p.

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.

T(n,k) = -1 for numbers k > 1 coprime to n and for numbers that are products of at least one prime q coprime to n and one prime p | n.

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.

If T(n,k) is positive, 1/k terminates T(n,k) digits after the radix point in base n. If T(n,k) is negative, 1/k is recurrent in base n.

From Robert Israel, Dec 28 2016: (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)

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)

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.