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 1-6 of 6 results.

A355631 List of numbers k such that A355627(k) > 0.

Original entry on oeis.org

10, 11, 28, 29, 31, 32, 34, 35, 37, 82, 83, 85, 86, 88, 89, 91, 92, 94, 95, 97, 98, 100, 101, 103, 104, 106, 110, 112, 113, 115, 116, 118, 119, 121, 244, 245, 247, 248, 250, 251, 253, 254, 256, 257, 259, 260, 262, 263, 265, 266, 268, 269, 271, 272, 274, 275, 277, 278, 280
Offset: 1

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Author

Markus Sigg, Jul 15 2022

Keywords

Comments

A positive integer p is in this list iff it can be written as p = Product_{i = 1..k} (3 + 1/t_i) with a positive integer k and integers t_i >= 2.
The sequence does not contain multiples of 3, see comments in A355627.

Examples

			10 = (3 + 1/4) * (3 + 1/13). 100 = (3 + 1/2) * (3 + 1/7) * (3 + 1/34) * (3 + 1/1133). 280 = (3 + 1/4) * (3 + 1/7) * (3 + 1/22) * (3 + 1/2614) * (3 + 1/6831253).
		

Crossrefs

Programs

A355626 a(n) is the number of tuples (t_1, ..., t_n) with integers 2 <= t_1 <= ... <= t_n such that Product_{i = 1..n} (3 + 1/t_i) is an integer.

Original entry on oeis.org

0, 3, 80, 15222
Offset: 1

Views

Author

Markus Sigg, Jul 15 2022

Keywords

Comments

Bounds on the values t_i can be derived as shown in "Bounds on t_i when the product of factors (3 + 1/t_i) is given" in the Links section. These bounds allow to calculate a(3) = 80 and a(4) = 15222 with the PARI program in the link. It seems that determining a(5) would need stronger methods. The identity (3 + 1/82) * (3 + 1/6670) * (3 + 1/44484454) * (3 + 1/1978866618021814) * (3 + 1/3915913091921090597566167836053) = 244 shows that the values t_n can get quite large. Integer products of more factors (3 + 1/t_i) can have even much larger t_n.

Examples

			a(1) = 0: As 1/t_1 is not an integer for t_1 >= 2, there is no t_1 >= 2 with integer 3 + 1/t_1.
a(2) = 3: With p := (3 + 1/t_1) * (3 + 1/t_2) we have p > 9, so for integer p it is p >= 10. With p <= (3 + 1/t_1)^2 we get t_1 <= 6. Solving p = 10, p = 11, p = 12 with 2 <= t_1 <= 6 for t_2 shows that the only integer solutions are (t_1,t_2) = (4,13) and (t_1,t_2) = (5,8) for p = 10, and (t_1,t_2) = (2,7) for p = 11.
		

Crossrefs

Programs

  • PARI
    \\ See link. The program has code for sequences A355626, ..., A355631 and can handle the more general case of factors (s + 1/t_i) with s >= 2.

A355628 a(n) is the number of positive integers p that can be written as p = Product_{i = 1..n} (3 + 1/t_i) with integers t_i >= 2.

Original entry on oeis.org

0, 2, 7, 25, 96, 364, 1344, 4921
Offset: 1

Views

Author

Markus Sigg, Jul 15 2022

Keywords

Examples

			a(1) = 0 because (3 + 1/t_1) is not integer for t_1 >= 2.
a(2) = 2 because 10 = (3 + 1/4) * (3 + 1/13) and 11 = (3 + 1/2) * (3 + 1/7) and no other positive integer p can be written as p = (3 + 1/t_1) * (3 + 1/t_2) with integers t_1, t_2 >= 2, cf. A355626.
		

Crossrefs

Programs

A355629 a(n) is the number of tuples (t_1, ..., t_n) with integers 2 <= t_1 <= ... <= t_n such that 3^n + 1 = Product_{i = 1..n} (3 + 1/t_i).

Original entry on oeis.org

0, 2, 50, 9291
Offset: 1

Views

Author

Markus Sigg, Jul 15 2022

Keywords

Examples

			a(2) = 2: 3^2 + 1 = 10 can be expressed as (3 + 1/4) * (3 + 1/13) and as (3 + 1/5) * (3 + 1/8);
a(3) = 50:  There are 50 representations of 3^3 + 1 = 28 with 10 <= min(t_i) <= 23 and 38 <= max(t_i) <= 8773. A product with minimal t_1 and maximal t_3 is 28 = (3 + 1/10) * (3 + 1/94) * (3 + 1/8773), maximal t_1 and minimal t_3 occur in 28 = (3 + 1/23) * (3 + 1/25) * (3 + 1/38).
		

Crossrefs

A356210 is the same problem with target 2^n + 1 and factors (2 + 1/t_k).

Programs

A355630 a(n) is the largest integer that can be written as Product_{i = 1..n} (3 + 1/t_i) with integers t_i >= 2.

Original entry on oeis.org

11, 37, 121, 413, 1442, 5047, 16807, 58457, 204085, 709667, 2483663, 8068753, 30415033
Offset: 2

Views

Author

Markus Sigg, Jul 15 2022

Keywords

Comments

Obviously 3^n < a(n) < 3.5^n.

Examples

			11 = (3 + 1/2) * (3 + 1/7) is the largest integer p that can be written as p = (3 + 1/t_1) * (3 + 1/t_2) with integers t_1,t_2 >= 2 because any such integer p is smaller than 3.5^2 = 12.25 and there is no such representation for p = 12. Hence, a(2) = 11.
		

Crossrefs

Programs

A334127 Number of nonempty sets {p_1, p_2, ..., p_k} such that Product_{i=1..k} p_i divides Product_{i=1..k} (n + p_i), where the p_i are distinct primes.

Original entry on oeis.org

1, 3, 4, 7, 6, 19, 8, 17, 8, 25, 12, 105, 8, 35, 22, 24, 16, 59, 28, 77, 30, 26, 22, 159, 8, 117, 23, 161, 26, 787, 32, 69, 46, 57, 30, 534, 32, 69, 90, 137, 22, 707, 20, 266, 54, 73, 50, 423, 38, 626, 62, 229, 52, 1324, 220, 489, 130, 285, 58, 2943, 24, 119, 274, 171, 202, 12089, 46, 181, 158, 201, 66, 1999, 58, 391, 234, 917, 126, 451, 42, 1767, 73, 1034, 86, 34691, 81, 150, 142, 233, 94, 18319, 226, 477, 70, 477, 114, 4419, 54, 157, 234, 2252
Offset: 1

Views

Author

Jinyuan Wang, May 10 2020

Keywords

Comments

a(n) is always finite. Proof: let p_1 < p_2 < ... < p_k, we can prove p_k <= 2*n^2 + n. If p_k > 2*n^2 + n, then 2*p_k > p_k + n, thus p_k - n is in the set. If p_k - m*n is in the set and m < n, then 2*(p_k - m*n) > p_k + n, thus p_k - (m+1)*n is in the set. Therefore, p_k - m*n are in the set for 0 <= m <= n. Since p_k - n*n > n + 1, p_k - m*n can be divisible by n + 1 for some m <= n, which is a contradiction to the p_i being primes.

Examples

			For n = 3, {3}, {2, 3}, {2, 5} and {2, 3, 5} are such sets, thus a(3) = 4.
		

Crossrefs

Programs

  • PARI
    a(n, k=primepi(2*n^2+n)) = {my(c=-1, p=primes(k)); forsubset(k, v, if(vecprod(vector(#v, i, p[v[i]]+n))%vecprod(vector(#v, i, p[v[i]])) == 0, c++)); c; }

Extensions

Terms a(13) onward from Max Alekseyev, Apr 08 2025
Showing 1-6 of 6 results.