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.

User: Mohammed Bouras

Mohammed Bouras's wiki page.

Mohammed Bouras has authored 12 sequences. Here are the ten most recent ones:

A372763 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(n+5))))).

Original entry on oeis.org

13, 19, 5, 31, 37, 43, 7, 11, 61, 67, 73, 79, 17, 1, 97, 103, 109, 23, 11, 127, 1, 139, 29, 151, 157, 163, 1, 1, 181, 1, 193, 199, 41, 211, 1, 223, 229, 47, 241, 1, 1, 1, 53, 271, 277, 283, 1, 59, 1, 307, 313, 1, 1, 331, 337, 1, 349, 71, 1, 367, 373, 379, 1, 1, 397, 1, 409, 83, 421
Offset: 3

Views

Author

Mohammed Bouras, May 12 2024

Keywords

Comments

Conjecture 1: The sequence contains only 1's and the primes.
Conjecture 2: Except for 2 and 3, all primes appear in the sequence once.
Conjecture: Record values correspond to A045375(m), m > 2. - Bill McEachen, Aug 03 2024

Examples

			For n=3, 1/(2 - 3/(3 + 5)) = 8/13, so a(3)=13.
For n=4, 1/(2 - 3/(3 - 4/(4 + 5))) = 23/19, so a(4)=19.
For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(5 + 5)))) = 13/5, so a(5)=5.
For n=6, 1/(2 - 3/(3 - 4/(4 - 5/(5 - 6/(6 + 5))))) = 227/31, so a(6)=31.

Links

Crossrefs

Cf. A051403, A356360, A369797. A370726.

Formula

a(n) = (6n - 5)/gcd(6n - 5, A051403(n-2) + 5*A051403(n-3)).

A372761 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(n+4))))).

Original entry on oeis.org

11, 4, 7, 13, 31, 1, 41, 23, 17, 1, 61, 1, 71, 19, 1, 43, 1, 1, 101, 53, 37, 29, 1, 1, 131, 1, 47, 73, 151, 1, 1, 83, 1, 1, 181, 1, 191, 1, 67, 103, 211, 1, 1, 113, 1, 59, 241, 1, 251, 1, 1, 1, 271, 1, 281, 1, 97, 1, 1, 1, 311, 79, 107, 163, 331, 1, 1, 173, 1
Offset: 3

Views

Author

Mohammed Bouras, May 12 2024

Keywords

Comments

Conjecture 1: Except for 4, the sequence contains only 1's and the primes.
Conjecture 2: Except for 3 and 5, all odd primes appear in the sequence once.
Conjecture: Record values correspond to A030430 (except a(6) = 13). - Bill McEachen, Aug 03 2024

Examples

			For n=3, 1/(2 - 3/(3 + 4)) = 7/11, so a(3)=11.
For n=4, 1/(2 - 3/(3 - 4/(4 + 4))) = 5/4, so a(4)=4.
For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(5 + 4)))) = 19/7, so a(5)=7.
For n=6, 1/(2 - 3/(3 - 4/(4 - 5/(5 - 6/(6 + 4))))) = 101/13, so a(6)=13.

Links

Crossrefs

Cf. A051403, A356360, A369797. A370726.

Formula

a(n) = (5n - 4)/gcd(5n - 4, A051403(n-2) + 4*A051403(n-3)).

A370726 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(n+3))))).

Original entry on oeis.org

3, 13, 17, 7, 5, 29, 11, 37, 41, 1, 7, 53, 19, 61, 1, 23, 73, 1, 1, 1, 89, 31, 97, 101, 1, 109, 113, 1, 1, 1, 43, 1, 137, 47, 1, 149, 1, 157, 1, 1, 1, 173, 59, 181, 1, 1, 193, 197, 67, 1, 1, 71, 1, 1, 1, 229, 233, 79, 241, 1, 83, 1, 257, 1, 1, 269, 1, 277
Offset: 3

Views

Author

Mohammed Bouras, Feb 28 2024

Keywords

Comments

Conjecture: The sequence contains only 1's and the primes.
Conjecture: Record values correspond to A002144 (n>3). - Bill McEachen, May 21 2024

Examples

			For n=3, 1/(2 - 3/(3 + 3)) = 2/3, so a(3)=3.
For n=4, 1/(2 - 3/(3 - 4/(4 + 3))) = 17/13, so a(4)=13.
For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(5 + 3)))) = 49/17, so a(5)=17.

Links

Crossrefs

Cf. A051403, A356360, A369797.

Formula

a(n) = (4n - 3)/gcd(4n - 3, A051403(n-2) + 3*A051403(n-3)).

A369797 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(n+2))))).

Original entry on oeis.org

7, 5, 13, 2, 19, 11, 5, 1, 31, 17, 37, 1, 43, 23, 1, 1, 1, 29, 61, 1, 67, 1, 73, 1, 79, 41, 1, 1, 1, 47, 97, 1, 103, 53, 109, 1, 1, 59, 1, 1, 127, 1, 1, 1, 139, 71, 1, 1, 151, 1, 157, 1, 163, 83, 1, 1, 1, 89, 181, 1, 1, 1, 193, 1, 199, 101, 1, 1, 211
Offset: 3

Views

Author

Mohammed Bouras, Feb 25 2024

Keywords

Comments

Conjecture: The sequence contains only 1's and the primes.
Conjecture: The sequence of record values is A002476. - Bill McEachen, Mar 24 2024
a(n) = 1 positions appear to correspond to A334919(m) - 1, m > 2. - Bill McEachen, Aug 05 2024

Examples

			For n=3, 1/(2 - 3/(3 + 2)) = 5/7, so a(3)=7.
For n=4, 1/(2 - 3/(3 - 4/(4 + 2))) = 7/5, so a(4)=5.
For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(5 + 2)))) = 41/13, so a(5)=13.

Links

Crossrefs

Cf. A051403, A356360.

Programs

  • Python
    from math import gcd, factorial
def A369797(n): return (a:=3*n-2)//gcd(a,a*sum(factorial(k) for k in range(n-2))+n*factorial(n-2)>>1) # Chai Wah Wu, Feb 26 2024

Formula

a(n) = (3n - 2)/gcd(3n - 2, A051403(n-2) + 2*A051403(n-3)).

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).

A363482 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-5))))).

Original entry on oeis.org

13, 23, 7, 49, 13, 83, 103, 5, 149, 1, 29, 233, 53, 23, 67, 373, 59, 1, 499, 109, 593, 643, 139, 107, 1, 863, 71, 197, 1049, 223, 1, 179, 53, 1399, 59, 1553, 71, 1, 257, 1, 1973, 2063, 431, 173, 67, 349, 2543, 1, 2749, 571, 2963, 439, 1, 3299, 683, 3533, 281, 151, 557, 1, 4153
Offset: 3

Views

Author

Mohammed Bouras, Jun 04 2023

Keywords

Comments

Conjecture: Except for 49, every term of this sequence is either a prime or 1.
The conjecture holds through n=10000. The set of ordered primes appear to match A038901. - Bill McEachen, Jul 08 2025

Examples

			For n=3, 1/(2 - 3/(-5)) = 5/13, so a(3) = 13.
For n=4, 1/(2 - 3/(3 - 4/(-5))) = 19/23, so a(4) = 23.
For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(-5)))) = 11/7, so a(5) = 7.

Links

Crossrefs

Cf. A006530, A038901, A051403.
Cf. A362086, A363102.

Programs

  • PARI
    lf(n) = sum(k=0, n-1, k!); \\ A003422
f(n) = (n+2)*lf(n+1)/2; \\ A051403
a(n) = (n^2 + 3*n - 5)/gcd(n^2 + 3*n - 5, 5*f(n-3) + n*f(n-4)); \\ Michel Marcus, Jun 06 2023

Formula

a(n) = (n^2 + 3*n - 5)/gcd(n^2 + 3*n - 5, 5*A051403(n-3) + n*A051403(n-4)).
Except for n=6, if gpf(n^2 + 3*n - 5) > n, then we have:
a(n) = gpf(n^2 + 3*n - 5), where gpf = "greatest prime factor".
If a(n) = a(m) and n < m < a(n), then we have:
a(n) = n + m + 3.
a(n) divides gcd(n^2 + 3*n - 5, m^2 + 3*m - 5).

A362086 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-3))))).

Original entry on oeis.org

3, 17, 9, 13, 53, 23, 29, 107, 43, 17, 179, 23, 79, 269, 101, 113, 29, 139, 1, 503, 61, 199, 647, 233, 251, 809, 17, 103, 43, 1, 373, 1187, 419, 443, 61, 1, 173, 1637, 191, 601, 1889, 659, 53, 127, 751, 1, 2447, 283, 883, 2753, 953, 1, 181, 1063, 367, 263, 131
Offset: 3

Views

Author

Mohammed Bouras, May 28 2023

Keywords

Comments

Conjecture: Except for 9, every term of this sequence is either a prime or 1.
Conjecture: Record values correspond to A248697 (n>3). - Bill McEachen, Mar 06 2024

Examples

			For n=3, 1/(2 - 3/(-3)) = 1/3, so a(3) = 3.
For n=4, 1/(2 - 3/(3 - 4/(-3))) = 13/17, so a(4) = 17.
For n=5, 1/(2 - 3/(3 - 4/(4 - 5/(-3)))) = 13/9, so a(5) = 9.
a(4) = a(12) = 4 + 12 + 1 = 17.
a(7) = a(45) = 7 + 45 + 1 = 53.

Links

Crossrefs

Cf. A006530, A014209, A051403, A248697.

Formula

a(n) = (n^2 + n - 3)/gcd(n^2 + n - 3, 3*A051403(n-3) + n*A051403(n-4)).
If gpf(n^2 + n - 3) > n, then we have:
a(n) = gpf(n^2 + n - 3), where gpf = "greatest prime factor".
If a(n) = a(m) and n < m < a(n), then we have:
a(n) = n + m + 1.
a(n) divides gcd(n^2 + n - 3, m^2 + m - 3).

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.

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.

A357127 a(n) = A081257(n) if A081257(n) > n, otherwise a(n) = 1.

Original entry on oeis.org

7, 13, 7, 31, 43, 19, 73, 13, 37, 19, 157, 61, 211, 241, 1, 307, 1, 127, 421, 463, 1, 79, 601, 31, 37, 757, 271, 67, 1, 331, 151, 1123, 397, 97, 43, 67, 1483, 223, 547, 1723, 139, 631, 283, 109, 103, 61, 181, 1, 2551, 379, 919, 409, 2971, 79, 103, 3307, 163, 3541, 523, 97, 3907, 109, 73, 613
Offset: 2

Views

Author

Mohammed Bouras, Sep 13 2022

Keywords

Comments

All the primes in this sequence appear exactly twice.
The new primes encountered seem to match the terms of A256148 for n>1. Bill McEachen, Oct 13 2022

Examples

			a(2) = a(a(2) - 2 - 1) = a(7 - 2 - 1) = a(4).
a(3) = a(9) = 3 + 9 + 1 = 13.
a(5) = a(25) = gcd(5^2 + 5 + 1, 25^2 + 25 + 1) = 31.

Crossrefs

Cf. A081257, A081256, A108768, A256148.

Programs

  • Python
    from sympy import primefactors
def A357127(n): return m if (m:=max(primefactors(n*(n+1)+1))) > n else 1 # Chai Wah Wu, Oct 15 2022

Formula

Conjecture 1: If a(n) != 1, then a(n) = a(a(n) - n - 1).
Conjecture 2: If n != m and a(n) = a(m), then
a(n) = gcd(n^2 + n + 1, m^2 + m + 1) = n + m + 1.

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