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.

A028875 a(n) = n^2 - 5.

Original entry on oeis.org

-5, -4, -1, 4, 11, 20, 31, 44, 59, 76, 95, 116, 139, 164, 191, 220, 251, 284, 319, 356, 395, 436, 479, 524, 571, 620, 671, 724, 779, 836, 895, 956, 1019, 1084, 1151, 1220, 1291, 1364, 1439, 1516, 1595, 1676, 1759, 1844, 1931, 2020, 2111, 2204, 2299, 2396
Offset: 0

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Author

Patrick De Geest, Dec 11 1999

Keywords

Comments

a(n) gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 20 for b = 2*n. In general D = b^2 - 4*a*c > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
For n>2, a(n) represents the area of the triangle created by the three points defined with coordinates: (n-3,n-2), ((n-1)*n/2,n*(n+1)/2), and ((n+1)^2, (n+2)^2). - J. M. Bergot, May 22 2014

Crossrefs

Cf. A028877 (subset of primes).

Programs

Formula

From R. J. Mathar, Apr 28 2008: (Start)
G.f.: x^3*(4 - x - x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
a(n) = 2*n + a(n-1) + 5, with n>0, a(0)=4. - Vincenzo Librandi, Aug 05 2010
a(-n) = a(n). - Michael Somos, May 26 2014
E.g.f.: (x^2 + x - 5)*exp(x). - G. C. Greubel, Aug 19 2017
From Amiram Eldar, Nov 04 2020: (Start)
Sum_{n>=0} 1/a(n) = -(1 + sqrt(5)*Pi*cot(sqrt(5)*Pi))/10.
Sum_{n>=0} (-1)^n/a(n) = -(1 + sqrt(5)*Pi*cosec(sqrt(5)*Pi))/10. (End)
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=0} (1 - 1/a(n)) = sqrt(6/5)*sin(sqrt(6)*Pi)/sin(sqrt(5)*Pi).
Product_{n>=3} (1 + 1/a(n)) = sqrt(5)*Pi/(6*sin(sqrt(5)*Pi)). (End)

A028876 Numbers k such that k^2 - 5 is prime.

Original entry on oeis.org

4, 6, 8, 12, 14, 16, 22, 24, 32, 34, 36, 38, 42, 44, 46, 52, 58, 64, 72, 74, 78, 82, 94, 102, 112, 116, 122, 132, 144, 152, 164, 166, 168, 174, 176, 182, 184, 186, 188, 198, 204, 212, 222, 226, 232, 234, 236, 252, 262, 264, 278, 284, 288, 292, 298, 302, 318, 324
Offset: 1

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Author

Keywords

Crossrefs

Programs

A352804 a(n) = A028876(n)/2; numbers k such that 4*k^2 - 5 is prime.

Original entry on oeis.org

2, 3, 4, 6, 7, 8, 11, 12, 16, 17, 18, 19, 21, 22, 23, 26, 29, 32, 36, 37, 39, 41, 47, 51, 56, 58, 61, 66, 72, 76, 82, 83, 84, 87, 88, 91, 92, 93, 94, 99, 102, 106, 111, 113, 116, 117, 118, 126, 131, 132, 139, 142, 144, 146, 149, 151, 159, 162, 171, 177, 179
Offset: 1

Views

Author

Jianing Song, Apr 04 2022

Keywords

Comments

A028876 with common factor 2 removed.

Crossrefs

Cf. A028876, A028877 (the resulting primes), A352805.

Programs

  • Mathematica
    Select[Range[200], PrimeQ[4*#^2 - 5] &]
  • PARI
    isA352804(n)=isprime(4*n^2-5)

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.
		

Crossrefs

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.