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 1-10 of 14 results. Next

A369054 Number of representations of n as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1
Offset: 0

Views

Author

Antti Karttunen, Jan 20 2024

Keywords

Comments

Number of solutions to n = x', where x' is the arithmetic derivative of x (A003415), and x is a product of three odd primes (not all necessarily distinct, A046316).
See the conjecture in A369055.

Examples

			a(27) = 1 as 27 can be expressed in exactly one way in the form (p*q + p*r + q*r), with p, q, r all being 3 in this case, as 27 = (3*3 + 3*3 + 3*3).
a(311) = 5 as 311 = (3*5 + 3*37 + 5*37) = (3*7 + 3*29 + 7*29) = (3*13 + 3*17 + 13*17) = (5*7 + 5*23 + 7*23) = (7*11 + 7*13 + 11*13). Expressed in the terms of arithmetic derivatives, of the A099302(311) = 8 antiderivatives of 311 [366, 430, 494, 555, 609, 663, 805, 1001], only the last five are products of three odd primes: 555 = 3*5*37, 609 = 3*7*29, 663 = 3*13*17, 805 = 5*7*23, 1001 = 7 * 11 * 13.
		

Crossrefs

Cf. A369055 [quadrisection, a(4n-1)], and its trisections A369460 [= a((12*n)-9)], A369461 [= a((12*n)-5)], A369462 [= a((12*n)-1)].
Cf. A369251 (positions of terms > 0), A369464 (positions of 0's).
Cf. A369063 (positions of records), A369064 (values of records).
Cf. A369241 [= a(2^n - 1)], A369242 [= a(n!-1)], A369245 [= a(A006862(n))], A369247 [= a(3*A057588(n))].

Programs

  • PARI
    \\ Use this for building up a list up to a certain n. We iterate over weakly increasing triplets of odd primes:
    A369054list(up_to) = { my(v = [3,3,3], ip = #v, d, u = vector(up_to)); while(1, d = ((v[1]*v[2]) + (v[1]*v[3]) + (v[2]*v[3])); if(d > up_to, ip--, ip = #v; u[d]++); if(!ip, return(u)); v[ip] = nextprime(1+v[ip]); for(i=1+ip,#v,v[i]=v[i-1])); };
    v369054 = A369054list(100001);
    A369054(n) = if(!n,n,v369054[n]);
    
  • PARI
    \\ Use this for computing the value of arbitrary n. We iterate over weakly increasing pairs of odd primes:
    A369054(n) = if(3!=(n%4),0, my(v = [3,3], ip = #v, r, c=0); while(1, r = (n-(v[1]*v[2])) / (v[1]+v[2]); if(r < v[2], ip--, ip = #v; if(1==denominator(r) && isprime(r),c++)); if(!ip, return(c)); v[ip] = nextprime(1+v[ip]); for(i=1+ip,#v,v[i]=v[i-1])));

Formula

a(n) = Sum_{i=1..A002620(n)} A369058(i)*[A003415(i)==n], where [ ] is the Iverson bracket.
For n >= 2, a(n) <= A099302(n).

A369252 Arithmetic derivative applied to the numbers of the form p*q*r where p,q,r are (not necessarily distinct) odd primes.

Original entry on oeis.org

27, 39, 51, 55, 75, 71, 87, 75, 91, 111, 103, 123, 95, 119, 147, 131, 119, 151, 183, 151, 135, 195, 167, 155, 231, 147, 199, 191, 187, 255, 167, 267, 211, 291, 195, 215, 247, 191, 263, 215, 327, 251, 247, 363, 203, 375, 311, 271, 255, 239, 411, 231, 311, 343, 299, 231, 435, 359, 331, 447, 311, 263, 391, 483, 263
Offset: 1

Views

Author

Antti Karttunen, Jan 22 2024

Keywords

Comments

The table showing the possible modulo 3 combinations for p, q, r and the sum ((p*q) + (p*r) + (q*r)):
| p | q | r | sum ((p*q) + (p*r) + (q*r)) (mod 3)
--+------+------+------+----------------------------------------
| 0 | 0 | 0 | 0, p=q=r=3, sum is 27.
--+------+------+------+----------------------------------------
| 0 | 0 | +/-1 | 0, p=q=3, r > 3.
--+------+------+------+----------------------------------------
| 0 | +1 | +1 | +1
--+------+------+------+----------------------------------------
| 0 | -1 | -1 | +1
--+------+------+------+----------------------------------------
| 0 | -1 | +1 | -1
--+------+------+------+----------------------------------------
| 0 | +1 | -1 | -1
--+------+------+------+----------------------------------------
| +1 | +1 | +1 | 0
--+------+------+------+----------------------------------------
| -1 | -1 | -1 | 0
--+------+------+------+----------------------------------------
| -1 | +1 | +1 | -1, regardless of the order, thus x3.
--+------+------+------+----------------------------------------
| +1 | -1 | -1 | -1, regardless of the order, thus x3.
--+------+------+------+----------------------------------------
Notably a(n) is a multiple of 3 only when A046316(n) is either a multiple of 9, or all primes p, q and r are either == +1 (mod 3) or all are == -1 (mod 3), and the case a(n) == +1 (mod 3) is only possible when A046316(n) is a multiple of 3, but not of 9, and furthermore, it is required that r == q (mod 3). See how these combinations affects sequences like A369241, A369245, A369450, A369451, A369452.
For n=1..9 the number of terms of the form 3k, 3k+1 and 3k+2 in range [1..10^n-1] are:
6, 2, 1,
39, 22, 38,
291, 209, 499,
2527, 1884, 5588,
23527, 17020, 59452,
227297, 156240, 616462,
2232681, 1453030, 6314288,
22119496, 13661893, 64218610,
220098425, 129624002, 650277572.
It seems that 3k+2 terms are slowly gaining at the expense of 3k+1 terms when n grows, while the density of the multiples of 3 might converge towards a limit.

Crossrefs

Cf. A369251 (same sequence sorted into ascending order, with duplicates removed).
Cf. A369464 (numbers that do not occur in this sequence).
Cf. also the trisections of A369055: A369460, A369461, A369462 and their partial sums A369450, A369451, A369452, also A369241, A369245.
Only terms of A004767 occur here.

Formula

a(n) = A003415(A046316(n)).

A369239 Number of integers whose arithmetic derivative is larger than 1 and equal to the n-th partial sum of primorial numbers.

Original entry on oeis.org

0, 1, 2, 1, 2, 1, 2, 1, 27, 0, 319, 1
Offset: 1

Views

Author

Antti Karttunen, Jan 18 2024

Keywords

Comments

Note how there are generally less solutions for even n than for odd n. This is explained by the fact that A143293(2n) == 1 (mod 4) and A143293(2n+1) == 3 (mod 4) and the arithmetic derivative A003415 of a product of any three odd primes (A046316) is always of the form 4k+3, therefore the solution set counted by a(2n) does not have any solutions from A046316 that contribute the majority of the solutions counted by a(2n+1). See also A369055.
a(13) >= 1 as there are solutions like 5744093403180469, 12538540924097819, etc., probably thousands or even more in total.
a(14) >= 1 [see examples].

Examples

			a(12) = 1 as there is a unique solution k such that k' = A143293(12) = 7628001653829, that k being 318745032938881 = 71*173*307*1259*67139. It's also the first solution with more than four prime factors.
a(14) >= 1, because as A143293(14)-2 = 13394639596851069-2 = 13394639596851067 is a prime, we have at least one solution, with A003415(2*13394639596851067) = A003415(26789279193702134) = 2+13394639596851067 = A143293(14).
For more examples, see A369240.
		

Crossrefs

Programs

  • PARI
    \\ See the attached program.

A369056 Numbers k of the form 4m+3 for which there is no representation as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r.

Original entry on oeis.org

3, 7, 11, 15, 19, 23, 31, 35, 43, 47, 59, 63, 67, 79, 83, 99, 107, 115, 127, 139, 143, 159, 163, 171, 175, 179, 207, 219, 223, 227, 235, 243, 259, 279, 283, 295, 303, 307, 319, 323, 339, 347, 367, 379, 387, 399, 403, 415, 427, 443, 463, 499, 515, 523, 531, 547, 559, 571, 579, 595, 603, 619, 639, 643, 655, 659, 675
Offset: 1

Views

Author

Antti Karttunen, Jan 20 2024

Keywords

Comments

Numbers k in A004767 for which A369054(k) = 0.
Numbers k of the form 4m-1 such that they are not arithmetic derivative (A003415) of any term of A046316.
Question: Is it possible that this sequence might be finite (although very long)? See comments in A369055.

Crossrefs

Setwise difference A004767 \ A369251.
Subsequences: A369248 (terms that are multiples of 3), A369249 (primes in this sequence).
Cf. also A369250 (4m+3 primes missing from this sequence).

Programs

  • Maple
    N:= 1000: # for terms <= N
    S:= {seq(i,i=3..N,4)}:
    P:= select(isprime, [seq(i,i=3..N/3,2)]):
    for i from 1 to nops(P) do
      p:= P[i];
      for j from i to nops(P) do
        q:= P[j];
        if 2*p*q + q^2 > N then break fi;
        for k from j to nops(P) do
          r:= P[k];
          v:= p*q + p*r + q*r;
          if v > N then break fi;
          S:= S minus {v};
    od od od:
    sort(convert(S,list)); # Robert Israel, Apr 17 2024
  • PARI
    isA369056(n) = ((3==(n%4)) && !A369054(n)); \\ Needs also program from A369054.

A369248 Numbers of the form 12*m+3 for which there is no representation as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r.

Original entry on oeis.org

3, 15, 63, 99, 159, 171, 207, 219, 243, 279, 303, 339, 387, 399, 531, 579, 603, 639, 675, 699, 747, 783, 819, 879, 891, 963, 1059, 1107, 1143, 1179, 1215, 1227, 1299, 1323, 1359, 1467, 1527, 1563, 1611, 1659, 1731, 1779, 1791, 1803, 1899, 1923, 1971, 1983, 2007, 2019, 2115, 2235, 2319, 2403, 2427, 2487, 2499, 2547
Offset: 1

Views

Author

Antti Karttunen, Jan 22 2024

Keywords

Crossrefs

Intersection of A017557 and A369056, multiples of 3 in the latter.

Programs

A369251 Numbers that have at least one representation as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r.

Original entry on oeis.org

27, 39, 51, 55, 71, 75, 87, 91, 95, 103, 111, 119, 123, 131, 135, 147, 151, 155, 167, 183, 187, 191, 195, 199, 203, 211, 215, 231, 239, 247, 251, 255, 263, 267, 271, 275, 287, 291, 299, 311, 315, 327, 331, 335, 343, 351, 355, 359, 363, 371, 375, 383, 391, 395, 407, 411, 419, 423, 431, 435, 439, 447, 451, 455, 459
Offset: 1

Views

Author

Antti Karttunen, Jan 22 2024

Keywords

Comments

By necessity all terms are of the form 4m+3 (in A004767).

Crossrefs

Complement of A369464.
Sequence A369252 sorted into ascending order, with duplicates removed.
Setwise difference A004767 \ A369056.
Subsequence of A239433.
Cf. A369250 (primes in this sequence).

Programs

  • PARI
    isA369251(n) = if(3!=(n%4),0, my(v = [3,3], ip = #v, r); while(1, r = (n-(v[1]*v[2])) / (v[1]+v[2]); if(r < v[2], ip--, ip = #v; if(1==denominator(r) && isprime(r), return(1))); if(!ip, return(0)); v[ip] = nextprime(1+v[ip]); for(i=1+ip,#v,v[i]=v[i-1])));

Formula

{k | A369054(k) > 0}.

A369461 Number of representations of 12n-5 as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 2, 0, 0, 1, 1, 1, 0, 0, 2, 0, 1, 0, 0, 0, 0, 1, 2, 1, 0, 0, 3, 0, 0, 0, 2, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 3, 0, 0, 1, 1, 2, 0, 0, 2, 1, 1, 0, 1, 0, 0, 1, 2, 0, 0, 1, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 1, 2, 1, 0, 1, 1, 0, 0, 0, 4, 0, 1, 1, 0, 0, 0, 1, 2, 0, 1, 0, 0
Offset: 1

Views

Author

Antti Karttunen, Jan 23 2024

Keywords

Comments

The sequence seems to contain an infinite number of zeros. See A369451 for the cumulative sum, and comments there.
Question: Are there any sections of this sequence, with parameters k >= 2, 0 <= i < k, for which a((k*n)-i) = 0 for all n >= 1? - Antti Karttunen, Nov 20 2024

Crossrefs

Trisection of A369055.
Cf. A017605, A369054, A369451 (partial sums), A369460, A369462.

Programs

  • PARI
    A369054(n) = if(3!=(n%4),0, my(v = [3,3], ip = #v, r, c=0); while(1, r = (n-(v[1]*v[2])) / (v[1]+v[2]); if(r < v[2], ip--, ip = #v; if(1==denominator(r) && isprime(r),c++)); if(!ip, return(c)); v[ip] = nextprime(1+v[ip]); for(i=1+ip,#v,v[i]=v[i-1])));
    A369461(n) = A369054((12*n)-5);

Formula

a(n) = A369054(A017605(n-1)) = A369054((12*n)-5).
a(n) = A369055((3*n)-1).

A369462 Number of representations of 12n-1 as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 1, 0, 2, 1, 0, 1, 2, 0, 2, 1, 2, 0, 1, 1, 3, 1, 1, 2, 5, 0, 1, 0, 2, 2, 2, 1, 4, 1, 3, 0, 3, 1, 2, 2, 3, 0, 2, 1, 8, 1, 1, 1, 4, 2, 2, 3, 3, 0, 4, 0, 4, 1, 1, 4, 3, 1, 3, 1, 6, 2, 3, 0, 5, 3, 1, 2, 6, 2, 6, 2, 2, 0, 1, 1, 5, 1, 2, 1, 10, 1, 3, 1, 3, 4, 2, 1, 6, 3, 6, 1, 4, 1, 3, 1, 5, 2, 3, 0
Offset: 1

Views

Author

Antti Karttunen, Jan 23 2024

Keywords

Comments

See A369452 for the cumulative sum, and comments there.
Question: Is there only a finite number of 0's in this sequence? See discussion at A369055 and see A369463 for empirical data.

Crossrefs

Trisection of A369055.
Cf. A017653, A369054, A369252, A369452 (partial sums), A369460, A369461, A369463 (= (12*i)-1, where i are the indices of zeros in this sequence).

Programs

  • PARI
    A369054(n) = if(3!=(n%4),0, my(v = [3,3], ip = #v, r, c=0); while(1, r = (n-(v[1]*v[2])) / (v[1]+v[2]); if(r < v[2], ip--, ip = #v; if(1==denominator(r) && isprime(r),c++)); if(!ip, return(c)); v[ip] = nextprime(1+v[ip]); for(i=1+ip,#v,v[i]=v[i-1])));
    A369462(n) = A369054((12*n)-1);

Formula

a(n) = A369054(A017653(n-1)) = A369054(12*n - 1).
a(n) = A369055(3*n).

A369460 Number of representations of 12n-9 as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r.

Original entry on oeis.org

0, 0, 1, 1, 1, 0, 2, 1, 0, 1, 1, 1, 2, 0, 0, 1, 2, 0, 0, 3, 0, 2, 1, 0, 1, 0, 2, 2, 0, 1, 2, 1, 0, 0, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 0, 1, 1, 1, 0, 2, 0, 2, 3, 0, 2, 3, 0, 1, 0, 2, 1, 1, 0, 2, 1, 0, 1, 1, 0, 3, 1, 2, 1, 0, 0, 3, 2, 1, 1, 2, 0, 1, 3, 2, 1, 1, 2, 1, 0, 2, 2, 3, 0, 1, 2, 0, 4, 1, 0, 2, 1, 0, 0, 2, 2
Offset: 1

Views

Author

Antti Karttunen, Jan 23 2024

Keywords

Comments

See A369450 for the cumulative sum, and comments there.

Crossrefs

Trisection of A369055.
Cf. A369054, A369248 (gives the positions of 0's in this sequence when nine is added and divided by 12), A369450 (partial sums), A369461, A369462.

Programs

  • PARI
    A369054(n) = if(3!=(n%4),0, my(v = [3,3], ip = #v, r, c=0); while(1, r = (n-(v[1]*v[2])) / (v[1]+v[2]); if(r < v[2], ip--, ip = #v; if(1==denominator(r) && isprime(r),c++)); if(!ip, return(c)); v[ip] = nextprime(1+v[ip]); for(i=1+ip,#v,v[i]=v[i-1])));
    A369460(n) = A369054((12*n)-9);

Formula

a(n) = A369055((3*n)-2).

A369241 Number of representations of 2^n - 1 as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 1, 0, 2, 1, 3, 0, 2, 2, 6, 0, 13, 1, 13, 0, 15, 0, 24, 1, 49, 4, 47, 0, 156, 6, 129, 0, 441, 1, 616
Offset: 0

Views

Author

Antti Karttunen, Jan 21 2024

Keywords

Comments

Any solutions for odd cases must have p = 3, with q and r > 3, because A000225(2n-1) == 1 (mod 3), while on even n, 2^n - 1 is a multiple of 3. This explains why the odd bisection grows much more sluggishly than the even bisection.
Question 2: Is there an infinite number of 0's in this sequence? See also comments in A369055.

Crossrefs

Programs

  • PARI
    A369054(n) = if(3!=(n%4),0, my(v = [3,3], ip = #v, r, c=0); while(1, r = (n-(v[1]*v[2])) / (v[1]+v[2]); if(r < v[2], ip--, ip = #v; if(1==denominator(r) && isprime(r),c++)); if(!ip, return(c)); v[ip] = nextprime(1+v[ip]); for(i=1+ip,#v,v[i]=v[i-1])));
    search_for_3k1_cases(n) = if(3!=(n%4), 0, my(p = 5, q, c=0); while(1, q = (n-(3*p)) / (3+p); if(q < p, return(c), if(1==denominator(q) && isprime(q), c++; write("b369241_by_solutions_of_odd_bisection_to.txt", n, " ", 3*p*q))); p = nextprime(1+p)));
    A369241(n) = if(n%2, search_for_3k1_cases((2^n)-1), A369054((2^n)-1));

Formula

a(n) = A369054(A000225(n)).
For n >= 2, a(n) = A369055(2^(n-2)).
Showing 1-10 of 14 results. Next