cp's OEIS Frontend

The sorted prime signature of n is row n of A124010.

Let gpf(m) = A006530(m) and let phi(m) = A000010(m) for m in A005117.

Row n contains m in A005117 such that A000720(A006530(m)) = n, sorted such that phi(m)/m increases as k increases.

Let m be the squarefree kernel A007947(m') of m'. We only consider squarefree m since phi(m)/m = phi(m')/m'. Let prime p | n and prime q be a nondivisor of n.

Since m is squarefree, we might encode the multiplicities of its prime divisors in a positional notation M that is finite at n significant digits. For example, m = 42 can be encoded reverse(A067255(42)) = 1,0,1,1 = 7^1 * 5^0 * 3^1 * 2^1. It is necessary to reverse row m of A067255 (hereinafter simply A067255(m)) so as to preserve zeros in M = A067255(m) pertaining to small nondivisor primes q < p. The code M is a series of 0's and 1's since m is squarefree. Then it is clear that row n contains all m such that A067255(m) has n terms, and there are 2^(n - 1) possible terms for n >= 1.

We may use an approach that generates the binary expansion of the range 2^(n - 1) < M < 2^n - 1, or we may append 1 to the reversed (n - 1)-tuples of {1, 0} to achieve codes M -> m for each row n, which is tantamount to ordering according to A059894.

Originally it was thought that the codes M were in order of the latter algorithm, and we could avoid sorting. Observation shows that the m still require sorting by the function phi(m)/m indeed to be in increasing order in row n. Still, the latter approach is slightly more efficient than the former in generating the sequence.

Every positive integer has a unique q-factorization (encoded by A324924) into factors q(i) = prime(i)/i, i > 0. For example:

11 = q(1) q(2) q(3) q(5)

50 = q(1)^3 q(2)^2 q(3)^2

360 = q(1)^6 q(2)^3 q(3)

Also the number of terminal subtrees with Matula-Goebel number k of the rooted tree with Matula-Goebel number n.

Sum of odd exponents in prime factorization of n minus the sum of even exponents in prime factorization of n.

An XOR-triangle is an inverted 0-1 triangle formed by choosing a top row and having each entry in subsequent rows be the XOR of the two values above it, i.e., A038554(n) applied recursively until we reach a single bit.

Let b(n) = n written in binary and let L(n) = 1 + floor(log_2(n)) = A070939(n). Let => be a single iteration of XOR across pairs of bits in b(n). Let t(n) be the XOR triangle initiated by b(n). Thus we may refer to any bit in t(n) by the address S(i,j) with 1 <= i <= L(n) and 1 <= j <= L(n) - j + 1.

We detect triangles of zeros, which are "voids" amid surrounding 1's or undefined "space" in the t(n) via run lengths of -1 in S(i,j) - S(i-1,j) for i > 1, and for i = 1, run lengths of zeros.

A334591(n) = length of row n.

From Michael De Vlieger, May 27 2020: (Start)

We can compactify row n by taking the product of prime(k)^T(n,k) for 1 <= k <= A334591(n), decoding the compactified row using A067255. This way, we can compactify the populations of zero-triangles for large n. Example: for n = 151, t(151) has 3 singleton zeros and 4 zero-triangles of side length k = 2. Thus row 151 has {3, 4}. 2^3 * 3^4 = 8 * 81 = 648. A067255(648) = {3, 4}.

A333625(m) = Product(prime(k)^T(m,k)) for m in A334556 (rotationally symmetrical XOR-triangles).

A334896(m) = Product(prime(k)^T(m,k)) for m in A334769 (rotationally symmetrical XOR-triangles with central zero-triangles).

(End)

a(n) = d_infinite(n, n-1) as defined in Kolossváry & Kolossváry link.

Subset of A055932.

For n > 1, subset of A360768, which is in turn a subset of A126706.

Conjecture: for n > 2, subset of A364702. - Michael De Vlieger, Oct 04 2024

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

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

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

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

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

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

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

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

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

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

Previous name: Numbers k such that there exists a solution to (p_m ^ a_m)*(p_m-1 ^ a_m-1)*...*(3^a_1)*(2^a_0) = (B^m)*a_m + (B^m-1)*a_m-1 + ... + (B^1)*a_1 + (B^0)*a_0 where k = (p_m ^ a_m)*(p_m-1 ^ a_m-1)*...*(3^a_1)*(2^a_0); a_m >= 1; a_(i= 0; p_0, p_1, ..., p_m are prime numbers; a_0, a_1, ..., a_m, B are integers.

B is the base in which we can express k as Sum_{i=0..m} B^i * a_i. B may also be seen as the variable in a polynomial, and k is then also an encoding of the polynomial (defined by the product of primes formula).

For k = (2^r)*3 we have B = (2^r)*3 - r.

A167221(n) is the smallest positive integer that yields a solution for k = a(n).

Negative B's can be obtained when the polynomial is an even function. This happens for instance when for k = 10, 100, 3125, ... - Michel Marcus, Aug 10 2022

From Peter Munn, Aug 13 2022: (Start)

Positive integers k such that k is a fixed point of a completely additive function f_B:N+ -> Z, B > 0, where f_B(prime(i+1)) = B^i for all i >= 0. Equivalently, since row B of A104244 is f_B, {a(n)} lists the columns of A104244 that contain their own column number.

If we require B to be negative instead, the sequence appears to start 10, 100, 3125, 1799875, 65610000, ... . Of these, 1799875 = 5^3 * 7 * 11^2 * 17 is the only k with only negative solutions (B = -11); the solutions for 65610000 are {4049, -4051}.

(End)

If p is the (k+1)-th prime and p is congruent to 1 modulo k, then p^p is a term with p^((p-1)/k) a solution for B. The list of such primes starts 3, 5, 7, 31, 97, 101, 331, ... . I suspect this list is infinite, meaning the greatest prime factor of the terms would be unbounded. - Peter Munn, Aug 15 2022

This sequence is a permutation of the natural numbers.

The powers of 2 are the fixed points of this sequence.

a(prime(i)) = A002110(i) for any i > 0.

a(i^2) = a(i)^2 for any i > 0.

a(A019565(n)) = A019565(A006068(n)) for any n >= 0.