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.

A356360 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(n+1))))).

Original entry on oeis.org

5, 7, 3, 11, 13, 1, 17, 19, 1, 23, 1, 1, 29, 31, 1, 1, 37, 1, 41, 43, 1, 47, 1, 1, 53, 1, 1, 59, 61, 1, 1, 67, 1, 71, 73, 1, 1, 79, 1, 83, 1, 1, 89, 1, 1, 1, 97, 1, 101, 103, 1, 107, 109, 1, 113, 1, 1, 1, 1, 1, 1, 127, 1, 131, 1, 1, 137, 139, 1, 1, 1, 1, 149, 151, 1, 1, 157, 1, 1, 163, 1, 167
Offset: 3

Views

Author

Mohammed Bouras, Oct 15 2022

Keywords

Comments

Conjecture: The sequence contains only 1's and the primes.
Similar continued fraction to A356247.
Same as A128059(n), A145737(n-1) and A097302(n-2) for n > 5.
a(n) = 1 positions appear to correspond to A104275(m), m > 2. Conjecture: all odd primes are seen in order after 11. - Bill McEachen, Aug 05 2024

Links

Crossrefs

Cf. A051403, A051417, A097302, A128059, A145737, A356247.
Cf. A104275.

Programs

  • Python
    from math import gcd, factorial
def A356360(n): return (a:=(n<<1)-1)//gcd(a, a*sum(factorial(k) for k in range(n-2))+n*factorial(n-2)>>1) # Chai Wah Wu, Feb 26 2024

Formula

For n >= 3, the formula of the continued fraction is as follows:
(A051403(n-2) + A051403(n-3))/(2n - 1) = 1/(2-3/(3-4/(4-5/(...(n-1)-n/(n+1))))).
a(n) = (2n - 1)/gcd(2n - 1, A051403(n-2) + A051403(n-3)).
From the conjecture: Except for n = 5, a(n)= 2n - 1 if 2n-1 is prime, 1 otherwise.

A363102 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-2))))).

Original entry on oeis.org

7, 7, 23, 17, 47, 31, 79, 7, 17, 71, 167, 97, 223, 127, 41, 23, 359, 199, 439, 241, 31, 41, 89, 337, 727, 1, 839, 449, 137, 73, 1087, 577, 1223, 647, 1367, 103, 1, 47, 73, 881, 1847, 967, 1, 151, 2207, 1151, 2399, 1249, 113, 193, 401, 1, 3023, 1567, 191, 41, 71, 257, 3719, 113, 3967, 89, 103, 311
Offset: 3

Views

Author

Mohammed Bouras, May 19 2023

Keywords

Comments

Conjecture 1: The sequence contains only 1's and primes.
Conjecture 2: All prime numbers appear either twice (same as A356247 and A357127) or three times.
Similar terms of A164314.
Conjecture: Record values correspond to A028871(m), m > 1. - Bill McEachen, Mar 06 2024
a(n) = 1 positions appear to correspond to A060515(m), m > 2. - Bill McEachen, Aug 05 2024

Examples

			a(5) = (5^2 - 2)/gcd(5^2 - 2, 2*A051403(5-3) + 5*A051403(5-4))= 23.
a(6) = a(11) = 6 + 11 = 17.
a(7) = a(40) = 7 + 40 = 47.

Links

Crossrefs

Cf. A008865, A051403, A059772, A164314, A356247, A357127.

Programs

  • PARI
    a051403(n) = (n+2)*sum(k=0, n, k!)/2;
a(n) = (n^2 - 2)/gcd(n^2 - 2, 2*a051403(n-3) + n*a051403(n-4)); \\ Michel Marcus, May 24 2023

Formula

a(n) = (n^2 - 2)/gcd(n^2 - 2, 2*A051403(n-3) + n*A051403(n-4)).
a(n) = A164314(n) if A164314(n) > n.
If a(n) = a(m) and n < m < a(n), then a(n) = n + m.

A356684 a(n) = (n-1)*a(n-1) - n*a(n-2), with a(1) = a(2) = -1.

Original entry on oeis.org

-1, -1, 1, 7, 23, 73, 277, 1355, 8347, 61573, 523913, 5024167, 53479135, 624890417, 7946278813, 109195935523, 1612048228547, 25439293045885, 427278358483537, 7609502950269503, 143217213477235783, 2840152418116022377
Offset: 1

Views

Author

Mohammed Bouras, Aug 22 2022

Keywords

Links

Crossrefs

Cf. A051403, A356247.

Programs

  • Mathematica
    a[1]=a[2]=-1; a[n_]:=a[n]=(n-1)a[n-1]-n a[n-2]; Array[a,22] (* Stefano Spezia, Aug 23 2022 *)
  • PARI
    a(n) = if(n<=2, -1, 1/2*((n-1)*sum(k=0,n-3,k!) + n*(n-2)*sum(k=0,n-4,k!))) \\ Jianing Song, Oct 15 2022, following the formula above

Formula

a(n) = A051403(n-3) + n*A051403(n-4).
a(n)/(n^2 - n - 1) = 1/(2-3/(3-4/(4-5/(...(n-1)-n/(n-(n+1)))))), for n >= 2.

A363347 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-4))))).

Original entry on oeis.org

11, 5, 31, 11, 59, 19, 19, 29, 139, 41, 191, 1, 251, 71, 29, 89, 79, 109, 479, 131, 571, 31, 61, 181, 41, 1, 179, 239, 1019, 271, 1151, 61, 1291, 1, 1439, 379, 1, 419, 1759, 461, 1931, 101, 2111, 1, 1, 599, 499, 59, 2699, 701, 71, 151, 101, 811
Offset: 3

Views

Author

Mohammed Bouras, May 28 2023

Keywords

Comments

Conjecture 1: Every term of this sequence is either a prime or 1.
Conjecture 2: The sequence contains all prime numbers which end with a 1 or 9.
Conjecture 3: Except for 5, the primes all appear exactly twice.
Conjecture: The sequence of record values is A028877. - Bill McEachen, May 20 2024

Examples

			For n=3, 1/(2 - 3/(-4)) = 4/11, so a(3) = 11.
For n=4, 1/(2 - 3/(3 - 4/(-4))) = 4/5, so a(4) = 5.
For n=5, 1/(2 - 3/(3 - 4/(4 -5/(-4)))) = 47/31, so a(5) = 31.
a(3) = a(6) = 3 + 6 + 2 = 11.
a(5) = a(24) = 5 + 24 + 2 = 31.
a(7) = a(50) = 7 + 50 + 2 = 59.

Links

Crossrefs

Cf. A006530, A051403, A229525, A356247, A028877.

Formula

a(n) = (n^2 + 2*n - 4)/gcd(n^2 + 2*n - 4, 4*A051403(n-3) + n*A051403(n-4)).
a(n) = gpf(n^2 + 2*n - 4) if gpf(n^2 + 2*n - 4) > n, otherwise a(n) = 1 (where gpf(n) denotes the greatest prime factor of n).
If n != m and a(n) = a(m) != 1, then we have:
a(n) = n + m + 2.
a(n) = gcd(n^2 + 2*n - 4, m^2 + 2*m - 4).
Showing 1-4 of 4 results.

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