cp's OEIS Frontend

Replace "5" with "x" and extend the definition of a to positive rationals and a becomes an isomorphism between positive rationals under multiplication and polynomials over Z under addition. This remark generalizes A001222, A048675 and A054841: evaluate said polynomial at x=1, x=2 and x=10, respectively.

In the definition, replace "e_i*n^(i-1)" with "e_i*x^(i-1)" for all i to define a function P:N+ -> N[x]. If we extend this definition to positive rationals by allowing negative e_i, P(.) becomes an isomorphism between positive rationals under multiplication and polynomials over Z under addition. We can use P to generalize A001222, A048675 and A054841: evaluate each term of the sequence of polynomials P(1), P(2), ... at x=1, x=2 and x=10, respectively. [Edited and corrected by Peter Munn, Aug 12 2022]

Previous name: There exists an isomorphism from the positive rationals under multiplication to Z[x] under addition, defined by f(q) = e1 + (e2)x + (e3)(x^2) +...+ (ek)(x^(k-1)) + ... (where e_i is the exponent of the i-th prime in q's prime factorization) The a(n) above are calculated by a(n) = f^(-1)[d/dx f(n)] (In other words: differentiate n's image in Z[x] and return to Q).

With primes noted p_0 = 2, p_1 = 3, etc., let f be the function that maps n = Product_{i=0..d} p_i^e_i to P = Sum_{i=0..d} e_i*X^i; and let g be the inverse function of f. a(n) is by definition g(P') = g((f(n))'). - Luc Rousseau, Aug 06 2018

Replace "4" with "x" and extend the definition of a to positive rationals and a becomes an isomorphism between positive rationals under multiplication and polynomials over Z under addition. This remark generalizes A001222, A048675 and A054841: evaluate said polynomial at x=1, x=2 and x=10, respectively.

For each n>=2 the indices i of primes p(i), i>=1, in the prime number decomposition of n are ordered from right to left.

The mapping n->a(n) is from {2,3,...} onto {1,2,3,...}=N but not injective; hence not invertible.

There are at most pa(k):=A000041(k) (partition numbers) different numbers which map to any a(n) with k digits. 10 and 12 are the smallest numbers for which this is not equality; 10 because 1,0 is not a partition, and 12 because 1,2 lists partition parts in the wrong order.

For the invertible map onto lists of prime number indices see the W. Lang link; also A112798.

Encode n with the function f(n) = noting the distinct prime divisors p of n by writing "1" in the (PrimePi(n) - 1)-th place, e.g, f(6) = f(12) = "11". This function is akin to A054841(n) except we don't note the multiplicity e of p in n, rather merely note "1" if e > 0.

This sequence decodes f(n) by reversing the digits.

If we decode f(n) without reversal, we have A007947(n), since f(n) sets any multiplicity e > 1 of prime divisor p of n to 1.

All terms except a(1) are of the form 2x with x odd. a(1) = 1, since f(1) = "0" and stands unaffected in reversal and decoding, and any zeros to the right of all 1's are lost in reversal. Thus f(15) = "110" reversed becomes "011" -> "11" decoded equals 2 * 3 = 6. Because we lose leading zeros, we always have 1 in position 1, which decoded is interpreted as the factor 2.

a(p) for p prime = 2, since primes are written via f(p) as 1 in the (PrimePi(p)-1)-th place. There is only one 1 in this number (similar to a perfect power of ten decimally) and when it is reversed, the number loses all leading zeros to become "1" -> 2. This also applies to prime powers p^e, since e is rendered as 1 by f(p^e), i.e., f(p^e) = f(p).

a(n) notes the distinct prime divisors p of n by writing "1" in the (pi(n)-1)-th place. Zeros hold the places of primes q less than the greatest prime divisor p that do not divide n. Thus a(n) consists of 1's and 0's like a binary number where each bit value, instead of representing 2^k, represents prime(k + 1).

a(n) = A054841(n) with all nonzero digits converted to 1's.

a(n) = a(A007947(n)), that is, a number n shares a value of a(n) with the largest squarefree divisor A007947(n). Thus a(18) = a(6) = 11.

a(p) = 1 in the leftmost place followed by (pi(p)-1) zeros.

This function is akin to A054841(n) except we don't note the multiplicity e of p in n, rather merely note "1" if e > 0.

Unlike A054841(1024) = 10, there are no overflows in a(n) into the next place that encodes prime(p+1) due to "carry". 1024 = 2^10, thus a(1024) = a(2^e) = 1, with e >= 1 = 1.

a(n) = terms t of row m of A288784 such that A002110(m) < t < 2*A002110(m).

The only odd term is 3; the only other term not ending in 10, 30, 70, or 90 in decimal is 42.

All terms t in row m have A001221(t) = m and at least one prime q coprime to t such that q < A006530(t).

Consider "tier" m and primorial p_m# = A002110(m), let "distension" i = pi(A006530(T(m, k))) - m and let "depth" j = m - pi(A053669(T(m, k))) + 1. Distension is the difference in the index of gpf(T(m, k)) and pi(m), while depth is the difference between the index of the least prime totative of T(m, k) and pi(m) + 1. We can calculate the maximum distension i given m and j via i_max = A020900(m - j + 1) - m - j + 1. This enables us to use permutations of 0 and 1 values in the notation A054841 and produce a(n) with some efficiency.

The most efficient method of generating a(n) is via f(x) = A287352(x), i.e., subtracting 1 from all values in row x of A287352. We use a pointer variable to direct increment on f(p_m#) = a constant array of m 1's, until we have exhausted producing terms p_m# < t < 2*p_m#. This enables the generation of T(m, k) for 1 <= m <= 100.