cp's OEIS Frontend

The number of powers of 7 that divide n. - Amiram Eldar, Mar 29 2025

Row sums = A024786 starting with A024786(2): (1, 1, 3, 4, 8, 11, 19, 26, ...) = number of 2's in all partitions of n.

A173238 as an infinite lower triangular matrix * [1, 2, 3, ...] = A103650.

Let the triangle = M. Then Lim_{n->inf} M^n = (1, 1, 3, 4, 10, 13, 26, ...), the left-shifted vector considered as a sequence = A092119, the Euler transform of the ruler sequence, A001511.

Given P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), then P(x) = A(x)/A(x^2), where A(x) = polcoeff A092119: (1 + x + 3x^2 + 4x^3 + ...).

Conjectures: Given the infinite set of triangles with A000041 in every column shifted down 0, 1, 2, ... n times, row sums of n-th triangle (where A173238 = 2nd in the set) = the numbers of n's in all partitions of n. E.g., row sums = of A173238 = A024786, the numbers of 2's in all partitions of n.

Similarly, row sums of triangle A173239 with columns >0 shifted down thrice = numbers of 3's in all partitions of n, and so on. Refer to comments in A000041 regarding the numbers of 1's in partitions of n.

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Second conjecture: Given the infinite set of analogous triangles with columns shifted down 2, 3, 4, ..., k times, we let such triangles = T(k) and perform lim_{n->inf} T^n(k), obtaining the left-shifted vectors considered as sequences. The conjecture states that the infinite set of such left-shifted vectors = the Euler transform of the infinite set of Ruler functions starting with the ruler function for k=2 = A001511: (1, 2, 1, 3, 1, 2, 1, ...)

To obtain the k-th ruler functions, begin with the natural numbers, 1,...2,...3,...4,...5,...6,...7,...8,...9,...; then for k = 2 we get A001511: 1,...2,...1,...3,...1,...2,...1,...4,...1,...; by finding the highest exponent of k dividing n, then adding 1. Similarly, for k = 2, we obtain A051064: 1,...1,...2,...1,...1,...2,...1,...1,...3,...

Next, we obtain the Euler transforms of the ruler functions (e.g., Euler transform of A001511 = A092119: (1, 1, 3, 4, 10, 13, 26, ...), noting that A092119 is lim_{n->inf} A173238^n, the left-shifted vector.

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Third conjecture: Let P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + ...), and A(k)(x) = the Euler transform of k-th ruler sequence, (k=2,3,...). Then P(x) = A(k)(x) / A(k)(x^k).

Examples: for k=2, A(x) = A092119: (1, 1, 3, 4, 10, 13, ...), then P(x) = (1 + x + 2x^2 + 3x^3 + ...) = (1 + x + 3x^2 + 4x^3 + ...) / (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...). For k=3 relating to triangle A173238, the left-shifted vector = the Euler transform of A051064 = A(x) for k=3, then P(x) = A(x) / A(x^3).

The conjecture extends the analogous conclusions to all k.

From Gary W. Adamson, Feb 25 2010: (Start)

Proof of second conjecture received from Helmut Prodinger 02/28/10 with a summary by R. J. Mathar:

Consider Product_{n>=0} z^(t^n)/(1-z^(t^n)) = Sum_{k>=1} (1+v_t(k))z^k where v_t(n) is the number of trailing zeros in the t-ary expansion of n, and its Euler transform A(z) = product_{k >= 1} 1/(1-z^k)^{1+v_t(k)}, then A(z)/A(z^t) = product_{k >= 1} 1/(1-z^k) is the partition generating function.

Here is the proof: A(z)/A(z^t) = Product_{k>=1} (1-z^(tk))^(1+v_t(k))/(1-z^k)^(1+v_t(k))

= Product_{k>=1} (1-z^(tk))^(v_t(tk))/(1-z^k)^(1+v_t(k))

= Product_{k>=1} (1-z^k)^(v_t(k))/(1-z^k)^(1+v_t(k)) (*)

= Product_{k>=1} 1/(1-z^k)

as desired. Notice that for (*), that v_t(n)=0 if n is not divisible by t. [Helmut Prodinger, hproding(AT)sun.ac.za, Feb 28 2010] (End)

The sequence extends to negative n by defining a(n) = a(-n).

For k >= 1 numbers 1..k occur with the same periodic and mirror symmetries as in ruler function A051064, in which k occurs 3 times more frequently than k+1. Here k occurs 3/2 times more frequently than k+1, precisely 2^(k-1) times in every 3^k terms. 0 has asymptotic density 0. Taking a trisection shows some scale symmetry, again comparable to ruler functions, as illustrated in the example section.

The links include a pin plot of a(0..162) aligned above an inverted plot of A051064 (the emphatic marking of 0's is significant). Between each n_k where A051064(n_k) = k >= 2 and the nearest n_k' where A051064(n_k') > k (or n_k' = 0 if nearer), there are 2^(k-2) indices where k occurs in this sequence, forming a 2^(k-2)-tuple. The 2^(k-2)-tuples have identical patterns and each has symmetry about an n_(k-1) where A051064(n_(k-1)) = k-1.

For a given k, the tuples described above are periodic with two per fundamental period, and the closest pairs of these tuples jointly form the pattern of one of the equivalent tuples for k+1. These patterns relate to the nonperiodic pattern for 0's and to the Cantor set as follows.

Let S_k be the sequence of positive indices at which k occurs, with 3^(k-2) subtracted when k >= 2. Given its ruler-type symmetries, S_k k >= 2 is determined by its first 2^(k-2) terms, which are the same as the first 2^(k-2) terms of S_i for i > k. The limiting sequence as k goes to infinity is S_0, which is A191108. {A191108(i)/(2*3^k) | 1 <= i <= 2^k} is the set of center points of the intervals deleted at step k+1 when generating the Cantor ternary set. This leads to the following scaling property.

Define c: Z -> P(R) so that c(n) is the scaled and translated Cantor ternary set spanning [n-1, n+1], and let C_k be the union of c(n) for all integer n with a(n) = k. Clearly C_1 consists of a scaled Cantor set repeated with period 3. (The set's two nonempty thirds occur at alternating intervals of 4/3 and 5/3.) For k >= 1, C_k is C_1 scaled by 3^(k-1), consisting therefore of a scaled Cantor set repeated with period 3^k. C_0 is special: C_0 = (C_0)*3 = (C_0)/3 = -C_0. Specifically, (C_0)/2 is the closure of the Cantor ternary set under multiplication by 3 and by -1.

Take a Sierpinski arrowhead curve formed of unit edges numbered consecutively from 0 at its axis of symmetry and aligned with an infinite Sierpinski gasket so that each edge is contained in the boundary of either the plane sector occupied by the gasket or a triangular region of the gasket's complement. If a(n) = 0, the n-th edge is contained in the sector boundary, otherwise the relevant triangular region seems to have side 2^(a(n)-1). See A307672 for a fuller description. The conjectured formulas below (that use A094373) derive from summing areas of regions within the gasket. - Corrected by Peter Munn, Aug 09 2019

From Charlie Neder, Jul 05 2019: (Start)

For each n, define the "2-balanced ternary expansion" E(n) as follows:

- E(n) begins with 0 or 1, according to the parity of n.

- The following digits are +, 0, or - as in standard balanced ternary, except + and - correspond to +2 and -2, respectively.

For example, we have E(4) = 0+-, E(7) = 10-, and E(13) = 1+-.

Then a(n) is the distance from the end of the rightmost 0, counting the last digit as 1, or 0 if 0 never appears. (End)

Let A = set of numbers of form 6n + 1, B = numbers of form 6n - 1. Eliminating numbers of form 25 + 30s from A and those of form 35 + 30s from B we obtain sets A* and B*. Removing all terms of the sequence from the union of A* and B*, only prime numbers remain. - Hisanobu Shinya (ilikemathematics(AT)hotmail.com), Jul 14 2002

Divide n by a*b*c where a = 2^(A001511(n)-1), b = 3^(A051064(n)-1) and c = 5^(A055457(n) -1). Then the resulting sequence includes only primes and a(n). - Alford Arnold, Sep 08 2003

Composite numbers not divisible by 2, 3 or 5. - Lekraj Beedassy, Jun 30 2004

Composite numbers k such that k^4 mod 30 = 1. - Gary Detlefs, Dec 09 2012

Composite numbers congruent to 1, 7, 11, 13, -13, -11, -7, -1 (mod 30). Since asymptotically, 100% of integers are composite, we have a(n)/n ~ 30/phi(30) = 30/8 = 3.75. - Daniel Forgues, Mar 16 2013

"John [Conway] recommends the more refined partition [of the positive numbers]: 1, prime, trivially composite, or nontrivially composite. Here, a composite integer is trivially composite if it is divisible by 2, 3, or 5." See link to (van der Poorten, Thomsen, and Wiebe; 2006) pp. 73-74. - Daniel Forgues, Jan 30 2015, Feb 04 2015

For the eight congruences coprime to 30, we can use one byte to encode the "primality/non-primality (unit or composite)" for each [30*n, 30*(n+1)[, n >= 0, closed-open interval, either as little endian binary sequence {01111111, 11111011, 11110111, 01111110, ...}, or as big endian binary sequence {11111110, 11011111, 11101111, 01111110, ...}, which we may then express in base 10. - Daniel Forgues, Feb 05 2015

Row sums = A024787, the numbers of 3's in all partitions of n, where A024787 starts with offset 1: (0, 0, 1, 1, 2, 4, 6, 9, 15,...). Triangle A173239 row sums start with the first "1" of A024787.

Let the triangle = M as an infinite lower triangular matrix. Then Lim_{n->inf} = A173241, the Euler transform of A051064 (the ruler function for 3).

Let P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), then P(x) = A(x) / A(x^3), where A(x) = polcoeff A173241: (1 + x + 2x^2 + 4x^3 + 6x^4 + ...)

Refer to A173238 comments for three conjectures relating A000041 to the infinite set of generalized ruler function sequences.

The sequence extends to negative n by defining a(n) = a(-n).

Consider a Sierpinski arrowhead curve formed of unit edges numbered consecutively from 0 at its axis of symmetry and aligned with an infinite Sierpinski gasket so that each edge is contained in the boundary of either the plane sector occupied by the gasket or a triangular region of the gasket's complement. If a(n) = 0, the n-th edge is contained in the sector boundary, otherwise the relevant triangular region has side a(n). Every length 3 segment of these boundaries contains exactly one edge of the arrowhead curve. A191108 lists positive n such that edge n is contained in the plane sector boundary. See A307672 for further properties.

See the graphic (in the links) of the arrowhead curve aligned with the gasket. Note the even-indexed edges (colored red) are the edges contained in a triangular region boundary on the left side of the vector. - Peter Munn, Jul 29 2019

a(n) = 0 if A307744(n) = 0, otherwise a(n) = 2^(A307744(n) - 1). Thus, this sequence has the symmetries of A307744, which are similar to ruler functions (especially A051064) and described further in A307744.

When listening to this, set pitch modulus to 35 or 36.

Weigh transform of A051064.