A323359 -id:A323359 - cp's OEIS frontend

A323386 a(n) = b(n+1)/b(n) - 1 where b(1)=2 and b(k) = b(k-1) + lcm(floor(sqrt(2)*k),b(k-1)).

1, 1, 5, 7, 1, 3, 11, 1, 7, 5, 1, 1, 19, 7, 11, 1, 5, 13, 1, 29, 31, 1, 11, 1, 1, 19, 13, 41, 1, 43, 1, 23, 1, 1, 1, 13, 53, 1, 1, 19, 59, 1, 31, 1, 13, 1, 67, 23, 1, 1, 73, 1, 19, 1, 79, 1, 41, 83, 1, 43, 29, 89, 1, 13, 31, 47, 1, 97, 1, 1, 101, 103

Offset: 1

Pedja Terzic, Jan 13 2019

nonn

Conjectures:
This sequence consists only of 1's and primes.
Every odd prime of the form floor(sqrt(2)*m) is a term of this sequence.
At the first appearance of each prime of the form floor(sqrt(2)*m), it is the next prime after the largest prime that has already appeared.

Cf. A135506, A008578, A323359, A323388.

Generator(n)={b1=2; list=[]; for(k=2, n, b2=b1+lcm(floor(sqrt(2)*k), b1); a=b2/b1-1; list=concat(list,a); b1=b2); print(list)}

A323388 a(n) = b(n+1)/b(n) - 1 where b(1)=3 and b(k) = b(k-1) + lcm(floor(sqrt(3)*k), b(k-1)).

Original entry on oeis.org

1, 5, 1, 1, 5, 1, 13, 5, 17, 19, 1, 11, 1, 5, 1, 29, 31, 1, 17, 1, 19, 13, 41, 43, 1, 23, 1, 1, 17, 53, 1, 19, 29, 1, 31, 1, 13, 67, 23, 71, 1, 37, 1, 1, 79, 1, 83, 1, 43, 1, 1, 13, 31, 1, 1, 1, 1, 1, 103, 1, 107, 109, 1, 1, 1, 29, 1, 1, 1, 61, 1, 1

Offset: 1

Pedja Terzic, Jan 13 2019

nonn

Conjectures:
1. This sequence consists only of 1's and primes.
2. Every odd prime of the form floor(sqrt(3)*m) greater than 3 is a term of this sequence.
3. At the first appearance of each prime of the form floor(sqrt(3)*m), it is larger than any prime that has already appeared.
The 2nd and 3rd conjectures are proved at the Mathematics Stack Exchange link. - Sungjin Kim, Jul 17 2019

Michel Marcus, Table of n, a(n) for n = 1..10000
Mathematics Stack Exchange, Generating primes of the form floor(sqrt(3)*n)

Cf. A184796 (primes of the form floor(sqrt(3)*m)).

Cf. A135506, A323359, A323386.

Generator(n)={b1=3; list=[]; for(k=2, n, b2=b1+lcm(floor(sqrt(3)*k), b1); a=b2/b1-1; list=concat(list,a); b1=b2); print(list)}

lista(nn)={my(b1=3, b2, va=vector(nn)); for(k=2, nn+1, b2=b1+lcm(sqrtint(3*k^2), b1); va[k-1]=b2/b1-1; b1=b2); va}; \\ Michel Marcus, Aug 20 2022

cp's OEIS Frontend

A323386 a(n) = b(n+1)/b(n) - 1 where b(1)=2 and b(k) = b(k-1) + lcm(floor(sqrt(2)*k),b(k-1)).

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A323388 a(n) = b(n+1)/b(n) - 1 where b(1)=3 and b(k) = b(k-1) + lcm(floor(sqrt(3)*k), b(k-1)).

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