cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A352957 Triangle read by rows: Row n is the lexicographically earliest strictly monotonic completely additive sequence of length n.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 2, 3, 4, 0, 2, 3, 4, 5, 0, 3, 5, 6, 7, 8, 0, 3, 5, 6, 7, 8, 9, 0, 4, 6, 8, 9, 10, 11, 12, 0, 5, 8, 10, 11, 13, 14, 15, 16, 0, 5, 8, 10, 12, 13, 14, 15, 16, 17, 0, 5, 8, 10, 12, 13, 14, 15, 16, 17, 18, 0, 7, 11, 14, 16, 18, 19, 21, 22, 23, 24, 25
Offset: 1

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Author

Peter Munn, Apr 11 2022

Keywords

Comments

Each sequence consists of nonnegative integers indexed from 1.
Note in particular in the formula section, the lower bound, floor(n/k), for first differences between terms in a row. This follows (using the additive property) from the strict monotonicity of floor(n/k)+1 consecutive terms near the end of the row.
For any k, with increasing length n >= k, the first k terms of the sequences approach similarity with a real-valued logarithmic function defined on the integers. For example, the asymptote of T(n,3)/T(n,2) is log(3)/log(2), A020857.

Examples

			(For row 4.) A completely additive sequence requires T(4,1) = 0. Strict monotonicity requires T(4,4) > T(4,3) > T(4,2). So T(4,4) >= T(4,2) + 2. Using the additivity this becomes T(4,2) + T(4,2) >= T(4,2) + T(4,1) + 2. Subtracting T(4,2) and substituting 0 for T(4,1) we get T(4,2) >= 2. So from T(4,4) > T(4,3) > T(4,2), we see T(4,3) >= 3, T(4,4) >= 4. So row 4 = (0, 2, 3, 4) as it is strictly monotonic and completely additive and from the preceding arguments is seen to be the lexicographically earliest such.
Triangle starts:
0;
0, 1;
0, 1,  2;
0, 2,  3,  4;
0, 2,  3,  4,  5;
0, 3,  5,  6,  7,  8;
0, 3,  5,  6,  7,  8,  9;
0, 4,  6,  8,  9, 10, 11, 12;
0, 5,  8, 10, 11, 13, 14, 15, 16;
0, 5,  8, 10, 12, 13, 14, 15, 16, 17;
0, 5,  8, 10, 12, 13, 14, 15, 16, 17, 18;
0, 7, 11, 14, 16, 18, 19, 21, 22, 23, 24, 25;
0, 7, 11, 14, 16, 18, 19, 21, 22, 23, 24, 25, 26;
0, 7, 11, 14, 16, 18, 20, 21, 22, 23, 24, 25, 26, 27;
0, 8, 13, 16, 19, 21, 23, 24, 26, 27, 28, 29, 30, 31, 32;
0, 9, 14, 18, 21, 23, 25, 27, 28, 30, 31, 32, 33, 34, 35, 36;
		

Crossrefs

Cf. A020857.
Completely additive sequences, s, with primes p mapped to a function of s(p-1) and maybe s(p+1): A064097, A344443, A344444; and for functions of earlier terms, see A334200.
For completely additive sequences with primes p mapped to a function of p, see A001414.
For completely additive sequences with prime(k) mapped to a function of k, see A104244.
For completely additive sequences where some primes are mapped to 1, the rest to 0 (notably, some ruler functions) see the cross-references in A249344.

Formula

The definition specifies: T(n,j*k) = T(n,j) + T(n,k); for k > 1, T(n,k) > T(n,k-1).
T(n,1) = 0, otherwise T(n,k) >= T(n,k-1) + floor(n/k).
For prime p, T(p,p) = T(p-1,p-1) + 1, otherwise T(p,k) = T(p-1,k).
T(n,2) >= 2*floor(n/4) + floor(n/9).
T(n,3) >= ceiling( (3*T(n,2) + floor(n/9)) / 2).
T(11,k) = A344443(k).
For k <> 13, T(23,k) = A344444(k).

A382061 Numbers whose number of divisors is divisible by their number of unitary divisors.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 24, 26, 27, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 46, 47, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 72, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 88, 89, 91, 93, 94, 95, 96, 97
Offset: 1

Views

Author

Amiram Eldar, Mar 14 2025

Keywords

Comments

Numbers k such that A034444(k) | A000005(k).
The criterion according to which a number belongs to this sequence depends only on the prime signature of this number: if {e_1, e_2, ... } are the exponents in the prime factorization of k then k is a term if and only if A000005(k)/A344444(k) = Product_{i} (e_i + 1)/2 is an integer.
The exponentially odd numbers (A268335) are all terms, since their prime factorization has only odd exponents e_i, so (e_i + 1)/2 is an integer. This sequence first differs from A268335 at n = 53: a(53) = 72 = 2^3 * 3^2 is not a term of A268335. The next terms that are not in A268335 are 108, 200, 360, 392, 432, 500, ... .
All the squarefree numbers (A005117, which is a subsequence of A268335) are terms. These are the numbers k such that A034444(k) = A000005(k).
A number k is a term if and only if the powerful part of k, A057521(k), is a term. Therefore, the primitive terms of this sequence are the powerful terms, A382062.
The asymptotic density of this sequence is Sum_{n>=1} f(A382062(n)) = 0.72201619..., where f(n) = (n/zeta(2)) * Product_{prime p|n} (p/(p+1)).
The asymptotic density of a few subsequences can be evaluated more easily. For example:
1) Powerful numbers that are exponentially odd (A335988): When summing only over these numbers, the formula for the asymptotic density gives the density of the exponentially odd numbers: Product_{p prime} (1 - 1/(p*(p+1))) = 0.704442... (A065463).
2) Numbers of the form p^(2*k) * q^(2*m+1), where k and m >= 1, and p != q are primes: When summing only over these numbers, the density of the numbers whose powerful part is of this form is ((Sum_{p prime} p/((p^2-1)*(p+1))) * (Sum_{p prime} p^2/((p^4-1)*(p+1))) - Sum_{p prime} p^3/((p^2-1)^2*(p^2+1)*(p+1)^2)) / zeta(2) = 0.017174455422470834821... .

Crossrefs

Programs

  • Mathematica
    q[k_] := Divisible[DivisorSigma[0, k], 2^PrimeNu[k]]; Select[Range[100], q]
  • PARI
    isok(k) = {my(f = factor(k)); !(numdiv(f) % (1<
    				

Formula

2 is a term since A000005(2) = A034444(2) = 2, so 2 | 2.
24 is a term since A000005(24) = 8, A034444(24) = 4, and 4 | 8.

A344443 Completely additive with a(2)=5; for odd prime p, a(p) = ceiling((a(p-1) + a(p+1))/2).

Original entry on oeis.org

0, 5, 8, 10, 12, 13, 14, 15, 16, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24, 24, 25, 25, 25, 25, 26, 26, 26, 26, 27, 27, 27, 27, 27, 27, 28, 28, 28, 28, 28, 28, 28, 29, 29, 29, 29, 29, 30, 29, 30, 30, 30, 30, 30, 30, 30, 30, 31, 31, 31, 31, 31, 31
Offset: 1

Views

Author

Peter Munn, May 19 2021

Keywords

Comments

Monotonic until a(55) = 30 > 29 = a(56).
The length of a monotonic completely additive sequence using positive integers is limited by the size of the initial terms, and this constraint is related to the conflicts inherent in specifying musical scales. In either case, the route to a good compromise solution that avoids complexity can be seen as a search for rational approximations to ratios between logarithms of the first few prime numbers.
Here a(i)/a(j) is the intended approximation to log(i)/log(j), so by starting with a(2) = 5 we open the door to using 8/5 as an approximation to log(3)/log(2) = 1.58496... . This approximation underlies simple musical scales that divide an octave into 5, where a note of twice the frequency is 5 tones higher and a note of about 3 times the frequency is 8 tones higher. The most commonly used more complex musical scales divide an octave into 12, equivalent to starting a sequence like this with a(2) = 12 (see A344444).

Examples

			a(4) = a(2*2) = a(2) + a(2) from the definition of completely additive. So a(4) = 5 + 5 = 10.
3 is an odd prime number, so a(3) = ceiling((a(3-1) + a(3+1))/2). Using the values a(2) = 5 and a(4) = 10 that we already know, we get a(3) = ceiling((5 + 10)/2) = ceiling(7.5) = 8.
The sequence is defined as completely additive, so a(1) = 0, the identity element for addition. (To see this, note that "completely additive" implies a(2) = a(2*1) = a(2)+a(1), and solve the equation for a(1).)
		

Crossrefs

Equivalent sequence with a(2)=12, a(3)=19: A344444.
The first 11 terms match row 11 of A352957.
For other completely additive sequences see the references in A104244.

Programs

  • Mathematica
    a[1] = 0; a[n_] := a[n] = Plus @@ ((Last[#] * If[First[#] == 2, 5, Ceiling[(a[First[#] - 1] + a[First[#] + 1])/2]]) & /@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Jun 27 2021 *)
  • PARI
    A344443(n) = if(1==n,0, my(f=factor(n)); sum(k=1,#f~,f[k,2]*if(2==f[k,1],5,ceil((1/2)*(A344443(f[k,1]-1)+A344443(f[k,1]+1)))))); \\ Antti Karttunen, May 19 2021

Formula

a(n*k) = a(n) + a(k).
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