login
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A212354 a(n) is the second smallest positive incongruent solutions of the congruence x^2 + (x+1)^2 == 0 (mod prime), where prime = A002144(n) (Pythagorean primes). 1
3, 10, 10, 20, 21, 36, 41, 55, 59, 61, 59, 55, 92, 105, 118, 96, 92, 126, 171, 152, 105, 175, 188, 152, 136, 175, 168, 254, 215, 300, 215, 242, 242, 197, 238, 331, 365, 210, 337, 406, 343, 415, 402, 254, 358, 403, 296, 337, 327, 300, 554, 538, 595, 405 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

The companion sequence is A212353.

See the comments on A212353 for the proof of two incongruent solutions of this congruence for each prime A002144(n). One takes the smallest positive representatives in each case as A212353(n) and a(n), with A212353(n) < a(n).

All positive solutions of this congruence are provided by the two sequences with entries u(n,k) = A212353(n) + k*A002144(n) and v(n,k) = a(n) + k*A002144(n), n >= 1, k >= 0. For the cases p = 5, 13 and 17 see A047219, A212160 and A212161, respectively, where the even-indexed numbers are the u(n,k) and the odd-indexed ones the v(n,k) (bisection).

r2(n) := 2*a(n) + 1 >= A002144(n) iff r(n) := 2*A212353(n) + 1 <= A002144(n)- 2. r2(n)^2 == +1 (Modd A002144(n)) but only r(n) belongs to the relevant restricted residue class. See A206549. Note that floor(r2(n)^2/A002144(n)) is odd. The same holds for r2 replaced by r.

LINKS

Table of n, a(n) for n=1..54.

FORMULA

a(n) is the second smallest positive incongruent solution of the congruence x^2 + (x+1)^2 = 2*x^2 + 2*x + 1  == 0 (mod A002144(n)), where A002144 lists the primes 1 modulo 4.

a(n) = A002144(n) - 1 - A212353(n), n >= 1.

EXAMPLE

n=1: a(1)=3 because 3^2 + 4^2 = 25 == 0 (mod 5). The other solution is (5-1) - 3 = 1 = A212353(1).

n=3: a(3)=10 because 10^2 + 11^2 = 221 = 13*17 == 0 (mod 17). (17-1) - 10 = 6 = A212353(3).

n=14: a(14)=105 because p=A002144(14) = 113 = A027862(5), and 105^2 + 106^2 = 197*113 == 0 (mod 113). (113-1) - 105 = 7 = A212353(14).

The first pair of solutions [u(n)=A212353(n), v(n)=a(n)], n >= 1, are [1, 3], [2, 10], [6, 10], [8, 20], [15, 21], [4, 36], [11, 41], [5, 55], [13, 59], [27, 61], ...

CROSSREFS

Cf. A212353, A206549.

Sequence in context: A213214 A009030 A168331 * A129489 A104702 A106596

Adjacent sequences:  A212351 A212352 A212353 * A212355 A212356 A212357

KEYWORD

nonn

AUTHOR

Wolfdieter Lang, May 10 2012

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified June 21 08:08 EDT 2021. Contains 345358 sequences. (Running on oeis4.)