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.

Showing 1-4 of 4 results.

A307467 The number of points, corresponding to the first n primes, and placed on the unit circle according to an algorithm using the data from A077218 (in the spirit of Ulam's spiral, and described in the COMMENTS section below), which lie on the closed arc of the unit circle from 0 to 45 degrees.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 4, 4, 5, 5, 6, 7
Offset: 1

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Author

Bernard X. Russo, Apr 09 2019

Keywords

Comments

Algorithm:
1. Place a dot, representing p_1, on the unit circle at (1,0), or equivalently, at angle 0 degrees, and label this point as the complex number lambda_1 = 1 = e^(i*0).
2. Place a dot, representing p_2, on the unit circle at (-1,0), or equivalently, at angle 180 degrees, and label this point as the complex number lambda_2 = -1 = e^(i*Pi).
3. For n > 2, supposing that dots representing p_1, ..., p_n have already been placed on the unit circle, place a dot, representing p_{n+1}, on the unit circle as follows: Starting at the dot representing p_n, proceed counterclockwise around the circle until you reach a dot representing one of p_1, p_2, ..., p_{n-1}.
4. Starting at this latter dot, proceed counterclockwise until you reach another dot representing one of p_1, p_2, ..., p_n.
5. Continue this process until you have traversed a(n) (of A077218) successive arcs connecting dots.
6. Proceed in a counterclockwise direction one more time to the next dot and place the dot which will represent p_{n+1} at the midpoint of the arc just traversed. Label this point as the complex number lambda_{n+1} = e^(i*theta_{n+1}).
The sequence is of interest since the points seem to cluster in the mentioned arc as well as in the closed arc from 180 degrees to 225 degrees. The sequences have been constructed by James Farrington up to n=256, which suggests the clustering. The precise clustering properties are formulated as questions in a file which is referenced below under LINKS.

Examples

			Following this procedure, we would place a dot representing p_3 on the unit circle at (0,-1), or equivalently, at angle 270 degrees, and we would place a dot representing p_4 on the unit circle at the midpoint of the arc connecting the point (-1,0) to the point (0,-1), that is, at (-1/root2,-1/root2), or equivalently, at angle 225 degrees.
Thus lambda_3 = -i = e^(3*Pi/2) and lambda_4 = e^(5*Pi/4).
		

Crossrefs

Uses A077218 in algorithm.

A052297 Number of distinct prime factors of all composite numbers between n-th and (n+1)st primes.

Original entry on oeis.org

0, 1, 2, 3, 2, 4, 2, 5, 5, 3, 6, 5, 3, 5, 6, 7, 3, 7, 6, 2, 8, 4, 8, 9, 5, 3, 6, 2, 6, 14, 5, 8, 3, 11, 3, 9, 7, 6, 8, 8, 3, 13, 2, 6, 3, 14, 15, 5, 3, 7, 9, 3, 11, 8, 9, 9, 3, 9, 6, 3, 13, 16, 7, 3, 6, 16, 8, 13, 3, 6, 9, 10, 9, 9, 6, 8, 11, 6, 12, 14, 4, 14, 2, 10, 7, 8, 11, 6, 4, 6, 16, 10, 6, 13
Offset: 1

Views

Author

Labos Elemer, Feb 09 2000

Keywords

Comments

From Lei Zhou, Mar 18 2014: (Start)
This is also the number of primes such that the (n+1)-th prime (mod i-th prime) is smaller than the (n+1)-th prime (mod n-th prime) for 1 <= i < n.
Proof: We denote the n-th prime number as P_n. Suppose P_(n+1) mod P_i = k; we can write P_(n+1) = m*P_i + k. Setting l = P_(n+1) - P_n, the composite numbers between P_n and P_(n+1) will be consecutively m*P_i + C, where C = k-l+1, k-l+2, ..., k-1. If k < l, there must be a value at which C equals zero since k-1 > 0 and k-l+1 <= 0, so P_i is a factor of a composite number between P_n and P_(n+1). If k >= l, all C values are greater than zero, thus P_i cannot be a factor of a composite number between P_n and P_(n+1). (End)

Examples

			n=30, p(30)=113, the next prime is 127. Between them are 13 composites: {114, 115, ..., 126}. Factorizing all and collecting prime factors, the set {2,3,5,7,11,13,17,19,23,29,31,41,59,61} is obtained, consisting of 14 primes, so a(30)=14.
		

Crossrefs

Programs

  • Mathematica
    Length[Union[Flatten[Table[Transpose[FactorInteger[n]][[1]],{n, First[#]+ 1, Last[#]-1}]]]]&/@Partition[Prime[Range[100]],2,1] (* Harvey P. Dale, Jan 19 2012 *)

A308487 a(n) is the least prime p such that the total number of prime factors, with multiplicity, of the numbers between p and the next prime is n.

Original entry on oeis.org

3, 11, 59, 71, 239, 7, 13, 103, 97, 79, 127, 73, 23, 31, 61, 157, 373, 383, 251, 89, 359, 401, 683, 701, 139, 337, 283, 241, 211, 631, 1471, 199, 1399, 661, 113, 619, 1511, 509, 293, 953, 317, 773, 1583, 863, 2423, 1831, 2251, 1933, 1381, 4057, 2803, 523, 1069, 2861, 1259, 1759, 3803, 4159, 4703
Offset: 2

Views

Author

J. M. Bergot and Robert Israel, May 31 2019

Keywords

Comments

a(n) <= A164291(n).

Examples

			a(8) = 13 because between 13 and the next prime, 17, are 14 with 2 prime factors, 15 with 2, 16 with 4 (counted with multiplicity), for a total of 2+2+4=8, and this is the first prime for which the total of 8 occurs.
		

Crossrefs

Programs

  • Maple
    N:= 100: # to get a(2)..a(N)
    V:= Array(2..N): count:= 0:
    q:= 3:
    while count < N-1 do
      p:= q;
      q:= nextprime(q);
      v:= add(numtheory:-bigomega(t),t=p+1..q-1);
      if v > N or V[v] > 0 then next fi;
      V[v]:= p; count:= count+1;
    od:
    convert(V,list);
  • Mathematica
    Module[{nn=60,pfm},pfm=Table[{p,Total[PrimeOmega[Range[Prime[p]+1,Prime[ p+1]-1]]]},{p,2,1000}];Prime[#]&/@Table[SelectFirst[pfm,#[[2]]==n&],{n,2,nn}]][[All,1]] (* Harvey P. Dale, Aug 25 2022 *)
  • PARI
    count(start, end) = my(i=0); for(k=start+1, end-1, i+=bigomega(k)); i
    a(n) = forprime(p=1, , if(count(p, nextprime(p+1))==n, return(p))) \\ Felix Fröhlich, May 31 2019

Formula

A077218(A000720(a(n))) = n.

A361806 Sum of distinct prime factors of all composite numbers between n-th and (n+1)st primes.

Original entry on oeis.org

0, 2, 5, 10, 5, 17, 5, 28, 30, 10, 45, 42, 12, 44, 47, 76, 10, 72, 57, 5, 97, 51, 117, 150, 28, 22, 83, 5, 65, 321, 66, 131, 28, 298, 10, 108, 172, 145, 109, 205, 10, 276, 5, 127, 16, 441, 582, 130, 24, 80, 232, 10, 276, 195, 270, 256, 10, 218, 187, 52, 388, 701, 162
Offset: 1

Views

Author

Karl-Heinz Hofmann, Mar 26 2023

Keywords

Examples

			a(6): 6th prime = 13 and the (6+1)th prime = 17; the composites between are {14,15,16} and the distinct prime factors of this set are {2,7,3,5} (no duplicates allowed); so a(6) = 2 + 7 + 3 + 5 = 17.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Plus @@ Union@ (Join @@ (FactorInteger[#][[;; , 1]] & /@ Range[Prime[n] + 1, Prime[n + 1] - 1])); Array[a, 65] (* Amiram Eldar, Mar 27 2023 *)
  • PARI
    a(n) = my(list=List()); for(i=prime(n)+1, prime(n+1)-1, my(f=factor(i)[,1]); for (k=1, #f, listput(list, f[k]))); vecsum(Set(list)); \\ Michel Marcus, Mar 27 2023
  • Python
    from sympy import primefactors, sieve
    def A361806(n):
        primeset = []
        for composites in range (sieve[n]+1, sieve[n+1]):
            for p in primefactors(composites): primeset.append(p)
        return(sum(set(primeset)))
    

Formula

a(n) = A008472(A061214(n)).
Showing 1-4 of 4 results.