A002314 -id:A002314 - cp's OEIS frontend

A152676 a(n) = A002144(n) - A002314(n).

3, 8, 13, 17, 31, 32, 30, 50, 46, 55, 75, 91, 76, 98, 100, 105, 129, 93, 162, 112, 183, 122, 144, 177, 241, 187, 217, 228, 155, 288, 203, 189, 213, 311, 269, 274, 334, 381, 266, 392, 254, 382, 348, 413, 301, 286, 489, 439, 483, 553, 516, 476, 578, 423, 487, 504

Offset: 1

Artur Jasinski, Dec 10 2008

nonn

For the four numbers {1, A002314(n), A152676(n), A152680(n)}, the multiplication table modulo A002144(n) is isomorphic with the Latin square
  1 2 3 4
  2 4 1 3
  3 1 4 2
  4 3 2 1
and is isomorphic with the multiplication table for {1,i,-i,-1} where i = sqrt(-1), A152680(n) is isomorphic with -1, A002314(n) with i or -i and A152676(n) vice versa -i or i.
1, A002314(n), A152676(n), A152680(n) are a subfield of the Galois Field [A002144(n)].
Let p = A002144(n), the n-th prime of the form 4k+1. Then a(n) and A002314(n) are the two square roots of -1 (mod p). Note that a(n) is also the multiplicative inverse of A002314(n) (mod p). - T. D. Noe, Feb 18 2010

T. D. Noe, Table of n, a(n) for n = 1..1000
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, pp. 1-21.

Cf. A002144, A002314, A152680.

aa = {}; Do[If[Mod[Prime[n], 4] == 1, k = 1; While[ ! Mod[k^2 + 1, Prime[n]] == 0, k++ ]; AppendTo[aa, Prime[n] - k]], {n, 1, 200}]; aa

A186814 a(n) = smallest number m such that A002144(n) divides gcd(A002314(n)^2+1,(A002314(n)+m)^2+1).

Original entry on oeis.org

1, 3, 9, 5, 25, 23, 7, 39, 19, 21, 53, 81, 43, 83, 63, 61, 101, 13, 143, 31, 169, 15, 55, 113, 225, 105, 157, 175, 17, 263, 89, 41, 77, 269, 165, 159, 271, 361, 123, 363, 75, 315, 239, 365, 93, 51, 437, 321, 397, 529, 439, 351, 543, 229, 333, 355, 449, 557, 625, 431, 517, 27, 583

Offset: 1

Michel Lagneau, Feb 27 2011

nonn

Sequence A002314 gives the minimal integer square root of -1 modulo p(n),where p(n) = n-th prime of form 4k+1.

			for n=1, k = A002314(1) = 2 => a(1) = 1, because 2^2+1 = 5 and (2+1)^2+1 = 2*5 ;
for n=2, k = A002314(2) = 5 => a(2) = 3, because 5^2+1 = 2*13 and (5+3)^2+1 = 5*13 ;
for n=3, k = A002314(3) = 4 => a(3) = 9, because 4^2+1 = 17 and (4+9)^2+1 = 2*5*17;
for n=4, k = A002314(4)= 12 => a(4)= 5, because 12^2+1
= 5*29 and (12+5)^2+1 = 2*5*29, and 29 divides
GCD(5*29, 2*5*29)=145.

Cf. A002314, A002144, A186710, A186713.

with(numtheory):T:=array(1..90):j:=1:for i from 1 to 250 do:x:=4*i+1:if type(x,prime)=true
  then T[j]:=x:j:=j+1:else fi:od:for p from 1 to j do:u:=T[p]:id:=0: for m from
  1 to 1000 while(id=0) do: z:=m^2+1:for d from 1 to u while(id=0) do: z1:=(m+d)^2+1:zz:=
  gcd(z,z1):if irem(zz,u)=0 then id:=1:printf(`%d, `,d):else fi:od:od:od:

A002144 Pythagorean primes: primes of the form 4*k + 1.

Original entry on oeis.org

5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101, 109, 113, 137, 149, 157, 173, 181, 193, 197, 229, 233, 241, 257, 269, 277, 281, 293, 313, 317, 337, 349, 353, 373, 389, 397, 401, 409, 421, 433, 449, 457, 461, 509, 521, 541, 557, 569, 577, 593, 601, 613, 617

Offset: 1

N. J. A. Sloane

Rational primes that decompose in the field Q(sqrt(-1)). - N. J. A. Sloane, Dec 25 2017
These are the prime terms of A009003.
-1 is a quadratic residue mod a prime p if and only if p is in this sequence.
Sin(a(n)*Pi/2) = 1 with Pi = 3.1415..., see A070750. - Reinhard Zumkeller, May 04 2002
If at least one of the odd primes p, q belongs to the sequence, then either both or neither of the congruences x^2 = p (mod q), x^2 = q (mod p) are solvable, according to Gauss reciprocity law. - Lekraj Beedassy, Jul 17 2003
Odd primes such that binomial(p-1, (p-1)/2) == 1 (mod p). - Benoit Cloitre, Feb 07 2004
Primes that are the hypotenuse of a right triangle with integer sides. The Pythagorean triple is {A002365(n), A002366(n), a(n)}.
Also, primes of the form a^k + b^k, k > 1. - Amarnath Murthy, Nov 17 2003
The square of a(n) is the average of two other squares. This fact gives rise to a class of monic polynomials x^2 + bx + c with b = a(n) that will factor over the integers regardless of the sign of c. See A114200. - Owen Mertens (owenmertens(AT)missouristate.edu), Nov 16 2005
Also such primes p that the last digit is always 1 for the Nexus numbers of form n^p - (n-1)^p. - Alexander Adamchuk, Aug 10 2006
The set of Pythagorean primes is a proper subset of the set of positive fundamental discriminants (A003658). - Paul Muljadi, Mar 28 2008
A079260(a(n)) = 1; complement of A137409. - Reinhard Zumkeller, Oct 11 2008
From Artur Jasinski, Dec 10 2008: (Start)
If we take 4 numbers: 1, A002314(n), A152676(n), A152680(n) then multiplication table modulo a(n) is isomorphic to the Latin square:
   1 2 3 4
   2 4 1 3
   3 1 4 2
   4 3 2 1
and isomorphic to the multiplication table of {1, i, -i, -1} where i is sqrt(-1), A152680(n) is isomorphic to -1, A002314(n) with i or -i and A152676(n) vice versa -i or i. 1, A002314(n), A152676(n), A152680(n) are subfield of Galois field [a(n)]. (End)
Primes p such that the arithmetic mean of divisors of p^3 is an integer. There are 2 sequences of such primes: this one and A002145. - Ctibor O. Zizka, Oct 20 2009
Equivalently, the primes p for which the smallest extension of F_p containing the square roots of unity (necessarily F_p) contains the 4th roots of unity. In this respect, the n = 2 case of a family of sequences: see n=3 (A129805) and n=5 (A172469). - Katherine E. Stange, Feb 03 2010
Subsequence of A007969. - Reinhard Zumkeller, Jun 18 2011
A151763(a(n)) = 1.
k^k - 1 is divisible by 4*k + 1 if 4*k + 1 is a prime (see Dickson reference). - Gary Detlefs, May 22 2013
Not only are the squares of these primes the sum of two nonzero squares, but the primes themselves are also. 2 is the only prime equal to the sum of two nonzero squares and whose square is not. 2 is therefore not a Pythagorean prime. - Jean-Christophe Hervé, Nov 10 2013
The statement that these primes are the sum of two nonzero squares follows from Fermat's theorem on the sum of two squares. - Jerzy R Borysowicz, Jan 02 2019
The decompositions of the prime and its square into two nonzero squares are unique. - Jean-Christophe Hervé, Nov 11 2013. See the Dickson reference, Vol. II, (B) on p. 227. - Wolfdieter Lang, Jan 13 2015
p^e for p prime of the form 4*k+1 and e >= 1 is the sum of 2 nonzero squares. - Jon Perry, Nov 23 2014
Primes p such that the area of the isosceles triangle of sides (p, p, q) for some integer q is an integer. - Michel Lagneau, Dec 31 2014
This is the set of all primes that are the average of two squares. - Richard R. Forberg, Mar 01 2015
Numbers k such that ((k-3)!!)^2 == -1 (mod k). - Thomas Ordowski, Jul 28 2016
This is a subsequence of primes of A004431 and also of A016813. - Bernard Schott, Apr 30 2022
In addition to the comment from Jean-Christophe Hervé, Nov 10 2013: All powers as well as the products of any of these primes are the sum of two nonzero squares. They are terms of A001481, which is closed under multiplication. - Klaus Purath, Nov 19 2023

			The following table shows the relationship between several closely related sequences:
Here p = A002144 = primes == 1 (mod 4), p = a^2+b^2 with a < b;
a = A002331, b = A002330, t_1 = ab/2 = A070151;
p^2 = c^2 + d^2 with c < d; c = A002366, d = A002365,
t_2 = 2ab = A145046, t_3 = b^2 - a^2 = A070079,
with {c,d} = {t_2, t_3}, t_4 = cd/2 = ab(b^2-a^2).
  ---------------------------------
   p  a  b  t_1  c   d t_2 t_3  t_4
  ---------------------------------
   5  1  2   1   3   4   4   3    6
  13  2  3   3   5  12  12   5   30
  17  1  4   2   8  15   8  15   60
  29  2  5   5  20  21  20  21  210
  37  1  6   3  12  35  12  35  210
  41  4  5  10   9  40  40   9  180
  53  2  7   7  28  45  28  45  630
  ...
a(7) = 53 = A002972(7)^2 + (2*A002973(7))^2 = 7^2 + (2*1)^2 = 49 + 4, and this is the only way. - _Wolfdieter Lang_, Jan 13 2015

David A. Cox, "Primes of the Form x^2 + n y^2", Wiley, 1989.
L. E. Dickson, "History of the Theory of Numbers", Chelsea Publishing Company, 1919, Vol I, page 386
L. E. Dickson, History of the Theory of Numbers, Carnegie Institution, Publ. No. 256, Vol. II, Washington D.C., 1920, p. 227.
G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, Cambridge, University Press, 1940, p. 132.
M. du Sautoy, The Music of the Primes, Fourth Estate / HarperCollins, 2003; see p. 76.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 241, 243.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 90.

Zak Seidov, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math.Series 55, Tenth Printing, 1972.
C. Banderier, Calcul de (-1/p).
J. Butcher, Mathematical Miniature 8: The Quadratic Residue Theorem, NZMS Newsletter, No. 75, April 1999.
Hing Lun Chan, Windmills of the minds: an algorithm for Fermat's Two Squares Theorem, arXiv:2112.02556 [cs.LO], 2021.
R. Chapman, Quadratic reciprocity.
A. David Christopher, A partition-theoretic proof of Fermat's Two Squares Theorem, Discrete Mathematics, Volume 339, Issue 4, 6 April 2016, Pages 1410-1411.
J. E. Ewell, A Simple Proof of Fermat's Two-Square Theorem, The American Mathematical Monthly, Vol. 90, No. 9 (Nov., 1983), pp. 635-637.
Bernard Frénicle de Bessy, Traité des triangles rectangles en nombres : dans lequel plusieurs belles propriétés de ces triangles sont démontrées par de nouveaux principes, Michalet, Paris (1676) pp. 0-116; see p. 44, Consequence II.
Bernard Frénicle de Bessy, Méthode pour trouver la solution des problèmes par les exclusions. Abrégé des combinaisons. Des Quarrez magiques, in "Divers ouvrages de mathématiques et de physique, par MM. de l'Académie royale des sciences", (1693) "Troisième exemple", pp. 17-26, see in particular p. 25.
A. Granville and G. Martin, Prime number races, arXiv:math/0408319 [math.NT], 2004.
D. & C. Hazzlewood, Quadratic Reciprocity.
Ernest G. Hibbs, Component Interactions of the Prime Numbers, Ph. D. Thesis, Capitol Technology Univ. (2022), see p. 33.
Lucas Lacasa, Bartolome Luque, Ignacio Gómez, and Octavio Miramontes, On a Dynamical Approach to Some Prime Number Sequences, Entropy 20.2 (2018): 131, also arXiv:1802.08349 [math.NT], 2018.
R. C. Laubenbacher and D. J. Pengelley, Eisenstein's Misunderstood Geometric Proof of the Quadratic Reciprocity Theorem, In: Anderson M, Katz V, Wilson R, eds. Who Gave You the Epsilon?: And Other Tales of Mathematical History. Spectrum. Mathematical Association of America; 2009:309-312.
R. C. Laubenbacher and D. J. Pengelley, Gauss, Eisenstein and the 'third' proof of the Quadratic Reciprocity Theorem, The Mathematical Intelligencer 16, 67-72 (1994).
K. Matthews, Serret's algorithm based Server.
Jorma K. Merikoski, Pentti Haukkanen, and Timo Tossavainen, The congruence x^n = -a^n (mod m): Solvability and related OEIS sequences, Notes. Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 516-529. See p. 521.
Carlos Rivera, Puzzle 968. Another property of primes 4m+1, The Prime Puzzles & Problems Connection.
D. Shanks, Review of "K. E. Kloss et al., Class number of primes of the form 4n+1", Math. Comp., 23 (1969), 213-214. [Annotated scanned preprint of review]
S. A. Shirali, A family portrait of primes-a case study in discrimination, Math. Mag. Vol. 70, No. 4 (Oct., 1997), pp. 263-272.
Rosemary Sullivan and Neil Watling, Independent divisibility pairs on the set of integers from 1 to n, INTEGERS 13 (2013) #A65.
Eric Weisstein's World of Mathematics, Wilson's Theorem.
Eric Weisstein's World of Mathematics, Pythagorean Triples.
Wikipedia, Quadratic reciprocity
Wolfram Research, The Gauss Reciprocity Law.
G. Xiao, Two squares.
D. Zagier, A One-Sentence Proof That Every Prime p == 1 (mod 4) Is a Sum of Two Squares, Am. Math. Monthly, Vol. 97, No. 2 (Feb 1990), p. 144. [From _Wolfdieter Lang_, Jan 17 2015 (thanks to Charles Nash)]
Index to sequences related to decomposition of primes in quadratic fields.

Cf. A002145, A002314, A002476, A002972, A002973, A003658, A004431, A007519, A010051, A016813, A076339, A094407.
Cf. A114200, A133870, A142925, A152676, A152680, A173330, A173331, A208177, A208178, A334912.
Cf. A004613 (multiplicative closure).
Apart from initial term, same as A002313.
For values of n see A005098.
Primes in A020668.

a002144 n = a002144_list !! (n-1)
a002144_list = filter ((== 1) . a010051) [1,5..]
-- Reinhard Zumkeller, Mar 06 2012, Feb 22 2011

[a: n in [0..200] | IsPrime(a) where a is 4*n + 1 ]; // Vincenzo Librandi, Nov 23 2014

a := []; for n from 1 to 500 do if isprime(4*n+1) then a := [op(a),4*n+1]; fi; od: A002144 := n->a[n];
# alternative
A002144 := proc(n)
    option remember ;
    local a;
    if n = 1 then
        5;
    else
        for a from procname(n-1)+4 by 4 do
            if isprime(a) then
                return a;
            end if;
        end do:
    end if;
end proc:
seq(A002144(n),n=1..100) ; # R. J. Mathar, Jan 31 2024

Select[4*Range[140] + 1, PrimeQ[ # ] &] (* Stefan Steinerberger, Apr 16 2006 *)
Select[Prime[Range[150]],Mod[#,4]==1&] (* Harvey P. Dale, Jan 28 2021 *)

PARI
```
select(p->p%4==1,primes(1000))
```

A002144_next(p=A2144[#A2144])={until(isprime(p+=4),);p} /* NB: p must be of the form 4k+1. Beyond primelimit, this is *much* faster than forprime(p=...,, p%4==1 && return(p)). */
A2144=List(5); A002144(n)={while(#A2144A002144_next())); A2144[n]}
\\ M. F. Hasler, Jul 06 2024

from sympy import prime
A002144 = [n for n in (prime(x) for x in range(1,10**3)) if not (n-1) % 4]
# Chai Wah Wu, Sep 01 2014

from sympy import isprime
print(list(filter(isprime, range(1, 618, 4)))) # Michael S. Branicky, May 13 2021

def A002144_list(n): # returns all Pythagorean primes <= n
    return [x for x in prime_range(5,n+1) if x % 4 == 1]
A002144_list(617) # Peter Luschny, Sep 12 2012

Odd primes of form x^2 + y^2, (x=A002331, y=A002330, with x < y) or of form u^2 + 4*v^2, (u = A002972, v = A002973, with u odd). - Lekraj Beedassy, Jul 16 2004
p^2 - 1 = 12*Sum_{i = 0..floor(p/4)} floor(sqrt(i*p)) where p = a(n) = 4*n + 1. [Shirali]
a(n) = A000290(A002972(n)) + A000290(2*A002973(n)) = A000290(A002331(n+1)) + A000290(A002330(n+1)). - Reinhard Zumkeller, Feb 16 2010
a(n) = A002972(n)^2 + (2*A002973(n))^2, n >= 1. See the Jean-Christophe Hervé Nov 11 2013 comment. - Wolfdieter Lang, Jan 13 2015
a(n) = 4*A005098(n) + 1. - Zak Seidov, Sep 16 2018
From Vaclav Kotesovec, Apr 30 2020: (Start)
Product_{k>=1} (1 - 1/a(k)^2) = A088539.
Product_{k>=1} (1 + 1/a(k)^2) = A243380.
Product_{k>=1} (1 - 1/a(k)^3) = A334425.
Product_{k>=1} (1 + 1/a(k)^3) = A334424.
Product_{k>=1} (1 - 1/a(k)^4) = A334446.
Product_{k>=1} (1 + 1/a(k)^4) = A334445.
Product_{k>=1} (1 - 1/a(k)^5) = A334450.
Product_{k>=1} (1 + 1/a(k)^5) = A334449. (End)
From Vaclav Kotesovec, May 05 2020: (Start)
Product_{k>=1} (1 + 1/A002145(k)) / (1 + 1/a(k)) = Pi/(4*A064533^2) = 1.3447728438248695625516649942427635670667319092323632111110962...
Product_{k>=1} (1 - 1/A002145(k)) / (1 - 1/a(k)) = Pi/(8*A064533^2) = 0.6723864219124347812758324971213817835333659546161816055555481... (End)
Sum_{k >= 1} 1/a(k)^s = (1/2) * Sum_{n >= 1 odd numbers} moebius(n) * log((2*n*s)! * zeta(n*s) * abs(EulerE(n*s - 1)) / (Pi^(n*s) * 2^(2*n*s) * BernoulliB(2*n*s) * (2^(n*s) + 1) * (n*s - 1)!))/n, s >= 3 odd number. - Dimitris Valianatos, May 21 2020
Legendre symbol (-1, a(n)) = +1, for n >= 1. - Wolfdieter Lang, Mar 03 2021

A209874 Least m > 0 such that the prime p=A002313(n+1) divides m^2+1.

Original entry on oeis.org

1, 2, 8, 4, 12, 6, 32, 30, 50, 46, 34, 22, 10, 76, 98, 100, 44, 28, 80, 162, 112, 14, 122, 144, 64, 16, 82, 60, 228, 138, 288, 114, 148, 136, 42, 104, 274, 334, 20, 266, 392, 254, 382, 348, 48, 208, 286, 52, 118, 86, 24, 516, 476, 578, 194, 154, 504, 106, 58, 26, 566, 96, 380, 670, 722, 62, 456, 582, 318, 526, 246, 520, 650, 726, 494, 324

Offset: 0

M. F. Hasler, Mar 11 2012

nonn

This yields the prime factors of numbers of the form N^2+1, cf. formula in A089120: For n=0,1,2,... check whether N = +/- a(n) [mod 2*A002313(n+1)], if so, then A002313(n+1) is a prime factor of N^2+1.
Obviously, p then divides (2kp +/- a(n))^2+1 for all k >=0 ; in particular it will be the least prime factor of such numbers if there is no earlier match.
Alternatively one could deal separately with the case of odd N, for which p=2 divides N^2+1, and even N, for which only Pythagorean primes A002144(n)=A002313(n+1) can be prime factors of N^2+1.

Cf. A002496, A014442, A085722, A144255, A089120, A193432.

Cf. A002314, A177979.

A209874(n)=if( n, 2*lift(sqrt(Mod(-1, A002144[n])/4)), 1)

/* for illustrative purpose: a(n) is the smaller of the 2 possible remainders mod 2*p of numbers N such that N^2+1 has p as smallest prime factor */ forprime( p=1,199, p>2 & p%4 != 1 & next; my(c=[]); for(i=1,9e9, factor(i^2+1)[1,1]==p |next; c=vecsort(concat(c,i%(2*p)),,8); #c==1 || print1(","c[1]) || break))

For n>0, A209874(n) = 2*sqrt(-1/4 mod A002144(n)), where sqrt(a mod p) stands for the positive x < p/2 such that x^2=a in Z/pZ.

A209874(n) = A209877(n)*2 for n>0.

A256011 Integers n with the property that the largest prime factor of n^2 + 1 is less than n.

Original entry on oeis.org

7, 18, 21, 38, 41, 43, 47, 57, 68, 70, 72, 73, 83, 99, 111, 117, 119, 123, 128, 132, 133, 142, 157, 172, 173, 174, 182, 185, 191, 192, 193, 200, 211, 212, 216, 233, 237, 239, 242, 251, 253, 255, 265, 268, 273, 278, 293, 294, 302, 305, 307, 313, 319, 322, 327

Offset: 1

Michael Kaltman, May 31 2015

nonn

Every Pythagorean prime, p, can be written as the sum of two positive integers, a and b, such that ab is congruent to 1 (mod p). Further: no number is the addend of two different primes, and the numbers that are NEVER addends are precisely the numbers in this list.
In particular: 5 = 2+3 and 2*3 = 6 == 1 (mod 5), 13 = 5+8 and 5*8 = 40 == 1 (mod 13), 17 = 4+13 and 4*13 = 52 == 1 (mod 17), 29 = 12+17 and 12*17 = 204 == 1 (mod 29), and so on.
Every integer greater than 1 is in exactly one of A002314, A152676, and the present sequence. - Michael Kaltman, May 11 2019

			7^2 + 1 = 50 = 2 * 5^2;
18^2 + 1 = 325 = 5^2 * 13;
21^2 + 1 = 442 = 2 * 13 * 17.

Giovanni Resta, Table of n, a(n) for n = 1..10000

Cf. A002144 (Pythagorean primes), A014442, A002314, A152676.

[k:k in [1..330]| Max(PrimeDivisors(k^2+1)) lt k]; // Marius A. Burtea, Jul 27 2019

select(n -> max(numtheory:-factorset(n^2+1))Robert Israel, Jun 09 2015

Select[Range[10^4], FactorInteger [#^2 + 1][[-1, 1]] < # &] (* Giovanni Resta, Jun 09 2015 *)

for(n=1,10^3,N=n^2+1;if(factor(N)[,1][omega(N)] < n,print1(n,", "))) \\ Derek Orr, Jun 08 2015

is(n)=my(f=factor(n^2+1)[,1]); f[#f]Charles R Greathouse IV, Jun 09 2015

A152680 a(n) = 4*A005098(n) = A002144(n) - 1.

Original entry on oeis.org

4, 12, 16, 28, 36, 40, 52, 60, 72, 88, 96, 100, 108, 112, 136, 148, 156, 172, 180, 192, 196, 228, 232, 240, 256, 268, 276, 280, 292, 312, 316, 336, 348, 352, 372, 388, 396, 400, 408, 420, 432, 448, 456, 460, 508, 520, 540, 556, 568, 576, 592, 600, 612, 616

Offset: 1

Artur Jasinski, Dec 10 2008

nonn

If we take the 4 numbers 1, A002314(n), A152676(n), A152680(n) then the multiplication table modulo A002144(n) is isomorphic with the Latin square
1 2 3 4
2 4 1 3
3 1 4 2
4 3 2 1
and isomorphic with the multiplication table of {1,I,-I,-1} where I is sqrt(-1), A152680(n) is isomorphic with -1, A002314(n) with I or -I and A152676(n) vice versa -I or I.
1, A002314(n), A152676(n), A152680(n) are subfields of the Galois Field [A002144(n)].
Numbers n such that A172019(n) + 1 = primes - 1. - Giovanni Teofilatto, Feb 02 2010

Cf. A002314, A152676, A152680.

aa = {}; Do[If[Mod[Prime[n], 4] == 1, AppendTo[aa, Prime[n] - 1]], {n, 1, 200}]; aa

A155095 Numbers k such that k^2 == -1 (mod 17).

Original entry on oeis.org

4, 13, 21, 30, 38, 47, 55, 64, 72, 81, 89, 98, 106, 115, 123, 132, 140, 149, 157, 166, 174, 183, 191, 200, 208, 217, 225, 234, 242, 251, 259, 268, 276, 285, 293, 302, 310, 319, 327, 336, 344, 353, 361, 370, 378, 387, 395, 404, 412, 421, 429, 438, 446, 455

Offset: 1

Vincenzo Librandi, Jan 20 2009

The first pair (a,b) is such that a+b=p, a*b=p*h+1, with h<=(p-1)/4; other pairs are given by(a+kp, b+kp), k=1,2,3...

Numbers congruent to {4, 13} mod 17. - Amiram Eldar, Feb 27 2023

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A002144, A155086, A155096, A155097, A155098.

Select[Range[500],PowerMod[#,2,17]==16&] (* or *) LinearRecurrence[ {1,1,-1},{4,13,21},60] (* Harvey P. Dale, Jun 25 2011 *)

A155095(n)=n\2*17-4*(-1)^n /* M. F. Hasler, Jun 16 2010 */

From M. F. Hasler, Jun 16 2010: (Start)
a(n) = 4*(-1)^(n+1) + 17*floor(n/2).
a(2k+1) = 17 k + a(1), a(2k) = 17 k - a(1), with a(1) = A002314(3) since 17 = A002144(3).
a(n) = a(n-2) + 17 for all n > 2. (End)
From Bruno Berselli, Sep 26 2010: (Start)
G.f.: x*(4+9*x+4*x^2)/((1+x)*(1-x)^2).
a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.
a(n) = (34*n + (-1)^n - 17)/4. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(9*Pi/34)*Pi/17. - Amiram Eldar, Feb 27 2023

Terms checked & minor edits by M. F. Hasler, Jun 16 2010

A155096 Numbers k such that k^2 == -1 (mod 29).

Original entry on oeis.org

12, 17, 41, 46, 70, 75, 99, 104, 128, 133, 157, 162, 186, 191, 215, 220, 244, 249, 273, 278, 302, 307, 331, 336, 360, 365, 389, 394, 418, 423, 447, 452, 476, 481, 505, 510, 534, 539, 563, 568, 592, 597, 621, 626, 650, 655, 679, 684, 708, 713, 737, 742, 766

Offset: 1

Vincenzo Librandi, Jan 20 2009

Numbers k such that k == 12 or 17 (mod 29). - Charles R Greathouse IV, Dec 27 2011

The first pair (a,b) is such that a+b=p, a*b=p*h+1, with h<=(p-1)/4; subsequent pairs are given as (a+kp, b+kp), k=1,2,3,...

			Let p = 29, a+b=29, a*b=29h+1, h<=7; for h=7, a+b=29, a*b=204, a=12, b=17; other pairs (12+29, 17+29) and so on.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A002144, A155086, A155095, A155097, A155098.

LinearRecurrence[{1,1,-1},{12,17,41},100] (* Vincenzo Librandi, Feb 29 2012 *)
Select[Range[800], PowerMod[#, 2, 29] == 28 &] (* Vincenzo Librandi, Apr 24 2014 *)
CoefficientList[Series[(12 + 5 x + 12 x^2)/((1 + x) (1 - x)^2), {x, 0, 30}], x] (* Vincenzo Librandi, May 03 2014 *)

A155096(n)=n\2*29-12*(-1)^n /* M. F. Hasler, Jun 16 2010 */

From M. F. Hasler, Jun 16 2010: (Start)
a(n) = 12*(-1)^(n+1) + 29 [n/2].
a(2k+1) = 29 k + a(1), a(2k) = 29 k - a(1), with a(1) = A002314(4) since 29 = A002144(4).
a(n) = a(n-2) + 29 for all n > 2. (End)
G.f.: x*(12 + 5*x + 12*x^2)/((1 + x)*(1 - x)^2). - Vincenzo Librandi, May 03 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(5*Pi/58)*Pi/29. - Amiram Eldar, Feb 27 2023

Terms checked & minor edits by M. F. Hasler, Jun 16 2010

A155097 Numbers k such that k^2 == -1 (mod 37).

Original entry on oeis.org

6, 31, 43, 68, 80, 105, 117, 142, 154, 179, 191, 216, 228, 253, 265, 290, 302, 327, 339, 364, 376, 401, 413, 438, 450, 475, 487, 512, 524, 549, 561, 586, 598, 623, 635, 660, 672, 697, 709, 734, 746, 771, 783, 808, 820, 845, 857, 882, 894, 919, 931, 956, 968

Offset: 1

Vincenzo Librandi, Jan 20 2009

Numbers k such that k == 6 or 31 (mod 37). - Charles R Greathouse IV, Dec 27 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A002144, A155086, A155095, A155096, A155098.

LinearRecurrence[{1,1,-1},{6,31,43},100] (* Vincenzo Librandi, Feb 29 2012 *)
Select[Range[1000],PowerMod[#,2,37]==36&] (* Harvey P. Dale, May 06 2012 *)
CoefficientList[Series[(6 + 25 x + 6 x^2)/((1 + x) (1 - x)^2), {x, 0, 30}], x] (* Vincenzo Librandi, May 03 2014 *)

A155097(n)=n\2*37-6*(-1)^n /* M. F. Hasler, Jun 16 2010 */

From M. F. Hasler, Jun 16 2010: (Start)
a(n) = 6*(-1)^(n+1) + 37*floor(n/2).
a(2k+1) = 37*k + a(1), a(2k) = 37*k - a(1), with a(1) = A002314(5) since 37 = A002144(5).
a(n) = a(n-2) + 37 for all n > 2. (End)
G.f.: x*(6 + 25*x + 6*x^2)/((1 + x)*(1 - x)^2). - Vincenzo Librandi, May 03 2014
Sum_{n>=1} (-1)^(n+1)/a(n) = cot(6*Pi/37)*Pi/37. - Amiram Eldar, Feb 26 2023

Terms checked, a(28) corrected, and minor edits by M. F. Hasler, Jun 16 2010

A155098 Numbers k such that k^2 == -1 (mod 41).

Original entry on oeis.org

9, 32, 50, 73, 91, 114, 132, 155, 173, 196, 214, 237, 255, 278, 296, 319, 337, 360, 378, 401, 419, 442, 460, 483, 501, 524, 542, 565, 583, 606, 624, 647, 665, 688, 706, 729, 747, 770, 788, 811, 829, 852, 870, 893, 911, 934, 952, 975, 993, 1016, 1034, 1057

Offset: 1

Vincenzo Librandi, Jan 20 2009

Numbers k such that k == 9 or 32 (mod 41). - Charles R Greathouse IV, Dec 27 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A002144, A155086, A155095, A155096, A155097.

LinearRecurrence[{1,1,-1},{9,32,50},100] (* Vincenzo Librandi, Feb 29 2012 *)
Select[Range[1100], PowerMod[#, 2, 41] == 40 &] (* Vincenzo Librandi, Apr 24 2014 *)

A155098(n)=n\2*41-9*(-1)^n /* M. F. Hasler, Jun 16 2010 */

From M. F. Hasler, Jun 16 2010: (Start)
a(n) = 9*(-1)^(n+1) + 41*floor(n/2).
a(2k+1) = 41*k + a(1), a(2k) = 41*k - a(1), with a(1) = A002314(6) since 41 = A002144(6).
a(n) = a(n-2) + 41 for all n > 2. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = cot(9*Pi/41)*Pi/41. - Amiram Eldar, Feb 26 2023

Terms checked & minor edits by M. F. Hasler, Jun 16 2010

cp's OEIS Frontend

A152676 a(n) = A002144(n) - A002314(n).

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A186814 a(n) = smallest number m such that A002144(n) divides gcd(A002314(n)^2+1,(A002314(n)+m)^2+1).

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A002144 Pythagorean primes: primes of the form 4*k + 1.

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A209874 Least m > 0 such that the prime p=A002313(n+1) divides m^2+1.

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A256011 Integers n with the property that the largest prime factor of n^2 + 1 is less than n.

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A152680 a(n) = 4*A005098(n) = A002144(n) - 1.

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A155095 Numbers k such that k^2 == -1 (mod 17).

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