A002144 -id:A002144 - cp's OEIS frontend

A134441 Last two digits of primes of form 4n+1 (A002144), excluding 5. Leading 0's omitted.

1, 9, 13, 17, 21, 29, 33, 37, 41, 49, 53, 57, 61, 69, 73, 77, 81, 89, 93, 97

Offset: 1

Lekraj Beedassy, Jan 19 2008

Select[Union[Mod[#,100]&/@Select[4Range[200]+1,PrimeQ]], #!=5&]  (* Harvey P. Dale, Apr 02 2011 *)

A140385 Consecutive area triples (y^2-x^2,4xy,p=x^2+y^2) associated with the Pythagorean primes A002144.

Original entry on oeis.org

3, 8, 5, 5, 24, 13, 15, 16, 17, 21, 40, 29, 35, 24, 37, 9, 80, 41, 45, 56, 53, 11, 120, 61, 55, 96, 73, 39, 160, 89, 65, 144, 97, 99, 40, 101, 91, 120, 109, 15, 224, 113, 105, 176, 137, 51, 280, 149, 85, 264, 157, 165, 104, 173, 19, 360, 181, 95, 336, 193, 195, 56, 197, 221

Offset: 1

Juri-Stepan Gerasimov, Jun 13 2008

Triples A070079(i), 8*A070151(i), A002144(i), i=1,2,3... in compound order.

			(3,8,5) followed by (5,24,13) followed by (15,16,17) ...

Edited by R. J. Mathar, Jun 16 2008

A174492 a(n) = the smallest k such that k^2+1 = p*A002144(n)^2, p prime of A002144 .

Original entry on oeis.org

18, 70, 540, 800, 1486, 2984, 500, 6760, 776, 4060, 5604, 4030, 5744, 1710, 1744, 46146, 186174, 162886, 62064, 32150, 37416, 16610, 26884, 15006, 130026, 58724

Offset: 1

Michel Lagneau, Jan 25 2011

nonn

A002144 are the primes of the form 4q + 1.

			a(1) = 18 because 18^2 + 1 = 13*A002144(1) ^2 = 13*5^2 ;
a(2) = 70 because 70^2 + 1 = 29*A002144(2) ^2 = 29*13^2 ;
a(3) = 540 because 540^2 + 1 = 1009*A002144(3) ^2 = 1009*17^2 .

Cf. A002144 A059321.

with(numtheory):nn:=400:T:=array(1..nn):k:=1:for x from 1 to nn do: p:=4*x+1:if
  type(p,prime)=true then T[k]:=p:k:=k+1:else fi:od:for n from 1 to k do: ind:=0:for
  m from 1 to 500000 while(ind=0) do:y:=m^2+1:x:= factorset(y) : n1:=nops(x):n2
  :=bigomega(y):if n1=2 and n2 = 3 and x[1]=T[n] and ind=0 then ind:=1:printf(`%d,
  `,m):else fi:od:od:

A199692 Subsequence of Pythagorean primes (A002144): each square is used only once.

Original entry on oeis.org

5, 41, 73, 149, 157, 277, 313, 613, 617, 709, 929, 1117, 1201, 1741, 1753, 1873, 2381, 2897, 2969, 3469, 3613, 3637, 3697, 4513, 4597, 5101, 5953, 6173, 6857, 6869, 7577, 8089, 8581, 9661, 10369, 10513, 11717, 12097, 12653, 12973, 13001, 13649, 14461, 15313

Offset: 1

Zak Seidov, Nov 09 2011

nonn

			First Pythagorean prime A002144(1)=5=1+4, "uses" squares 1^2 and 2^2.
Next Pythagorean prime which does not "use" 1^2 and 2^2 is A002144(6)=41=4^2+5^2, hence a(2)=41.
Next Pythagorean prime which does not "use" 1^2, 2^2, 4^2 and 5^2 is A002144(9)=73=3^2+8^2, hence a(3)=73.
Next Pythagorean prime which does not "use" 1^2, 2^2, 3^2, 4^2, 5^2 and 8^2 is A002144(16)=149=7^2+10^2, hence a(4)=149.

Zak Seidov, Table of n, a(n) for n = 1..2000

Cf. A002144.

A212353 a(n) is the smallest positive solutions of the congruence x^2 + (x+1)^2 == 0 (mod prime), where prime = A002144(n) (Pythagorean primes).

Original entry on oeis.org

1, 2, 6, 8, 15, 4, 11, 5, 13, 27, 37, 45, 16, 7, 18, 52, 64, 46, 9, 40, 91, 53, 44, 88, 120, 93, 108, 26, 77, 12, 101, 94, 106, 155, 134, 57, 31, 190, 71, 14, 89, 33, 54, 206, 150, 117, 244, 219, 241, 276, 38, 62, 17, 211, 243, 74, 277, 307, 325, 67, 306, 176, 43

Offset: 1

Wolfdieter Lang, May 10 2012

nonn

The companion sequence is A212354.
There are at most two incongruent solutions of this congruence due to the degree. The fact that there are precisely two such solutions for each prime of the form 4*k+1 (see A002144) is due to the reduction of this problem to one of quadratic residues, namely to X^2 == -1 (mod 2p), with p a prime (see the Nagell reference, given in A210848, pp. 132-3, especially theorem 77), adapted to the quadratic form f(x) = 2*x^2 + 2*x + 1, with discriminant D=-4. This congruence with composite modulus has exactly two incongruent solutions because X^2 == -1 (mod 2) has only the solution +1 modulo 2 (odd numbers), and X^2 == -1 (mod p) has (at least one) solution if the Legendre symbol (-1/p) = +1 (i.e., if -1 is a quadratic residue modulo p). Now (-1/p) = (-1)^(p-1)/2 (see, e.g., the Niven-Zuckerman-Montgomery reference given in A001844, Theorem 3.2 (1), p. 132). Hence there is a solution modulo p iff p == 1 (mod 4). Call the smallest positive one X0, with 0 < X0 < p-1. Then one also has the incongruent solution X1 := p-X0. This implies that there are precisely two incongruent solution of the original congruence modulo 2*p for each 1 (mod 4) prime (see, e.g., Nagell's book, pp. 83-4, Theorem 46). If u is a solution for p = A002144(n) (the existence of u has just been proved) then also the companion v := p-1-u satisfies this congruence, and v is incongruent to u modulo p.
Note that x^2 + (x+1)^2 = 4*T(x) + 1, with the triangular numbers A000217.
The primes with x^2 +(x+1)^2 = prime (necessarily from A002144) are found under A027862. The corresponding x values are found under A027861. These x values explain the positions n' where a(n') is smaller than a(n'-1) (for n'>=6): determine k with x=A027861(k), and then n' from A027862(k) = A002144(n'). Note that a(n') = x for such values n'. E.g., n'=6 with a(6)=4:  x=4=A027861(3), p=41=A027862(3) = A002144(6). These values n' are n' = 1, 2, 6, 8, 14, 19, 30, ...
All positive solutions of this congruence are provided by the two sequences with entries u(n,k) = a(n) + k*A002144(n) and v(n,k) = A212354(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).
2*a(n) + 1 = A206549(n), the smallest positive nontrivial solution of X^2 == +1 (Modd A002144(n)). For the next larger solution 2*A212354(n) + 1 >= p, hence it does not belong to the restricted residue system Modd A002144(n).

			n=1: a(1)=1 because 1^2 + 2^2 = 5 == 0 (mod 5). The companion solution is (5-1) - 1 = 3 = A212354(1).
n=3: a(3)=6 because 6^2 + 7^2 = 85 = 5*17 == 0 (mod 17). The companion is (17-1) - 6 = 10 =  A212354(3).
n=14: a(14)=7 because p=A002144(14) = 113 = A027862(5), and 49^2 + 50^2 = 113. The companion is (113-1) - 7 = 105 = A212354(14).

Cf. A047219(1)=a(1), A212160(1)=a(2), A212161(1)=a(3), A212354 (companions), A206549.

a(n) is the smaller of the two smallest positive incongruent solutions 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 (primes of the form 4*k+1). For the proof of the existence of a(n) see a comment above. The next larger incongruent companion solution is A212354(n), n >= 1.

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).

Original entry on oeis.org

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

Offset: 1

Wolfdieter Lang, May 10 2012

nonn

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.

			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], ...

Cf. A212353, A206549.

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.

A216924 Consecutive Pythagorean primes p = A002144(r) and q = A002144(r+1) such that q - p > log(p)^2. The number a(n) is the n-th value of p.

Original entry on oeis.org

5, 17, 113, 197, 461, 881, 1493, 1801, 39581, 50593, 78989, 180797, 183089, 241601, 250501, 268297, 339841, 485209, 492421, 618637, 919421, 1264337, 1561829, 1637813, 1994101, 2116129, 2191633, 2243909, 2314373, 3254929, 3422917, 3440621, 4468889, 4855297, 4874717, 5059321, 5526613, 6118769, 7856441, 9199153

Offset: 1

Thomas Ordowski, Sep 20 2012

nonn

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A182315, A002144 (Pythagorean primes).

t = {}; p = 5; Do[While[q = p; While[p = NextPrime[p]; Mod[p, 4] == 3]; p - q < Log[q]^2]; AppendTo[t, q], {25}]; t (* T. D. Noe, Sep 21 2012 *)

r=1;v=List();p=5;forprime(q=11,1e7,if(q%4>1,next);if(q-p>r, r=log(p)^2\1; if(q-p>r,print1(p", ");listput(v,p)));p=q); Vec(v) \\ Charles R Greathouse IV, Sep 21 2012

a(22)-a(40) from Charles R Greathouse IV, Sep 21 2012

A305728 Numbers of the form 216*p^3, where p is a Pythagorean prime (A002144).

Original entry on oeis.org

27000, 474552, 1061208, 5268024, 10941048, 14886936, 32157432, 49027896, 84027672, 152273304, 197137368, 222545016, 279726264, 311665752, 555412248, 714516984, 835896888, 1118386872, 1280824056, 1552836312, 1651400568, 2593941624, 2732256792, 3023464536, 3666512088

Offset: 1

Bruno Berselli, Jun 22 2018

nonn

No term can be written as x^2 + y^2 + z^9.

Colin Barker, Table of n, a(n) for n = 1..1000
W. Jagy and I. Kaplansky, Sums of Squares, Cubes and Higher Powers, Experimental Mathematics, vol. 4 (1995), p. 171 (see Theorem under Higher powers).

Cf. A002144, A016911, A080169.

[216*p^3: p in PrimesUpTo(300) | IsOne(p mod 4)];

P := select(p -> isprime(p), [seq(n, n=5..1000, 4)]):
seq((6*p)^3, p in P); # Peter Luschny, Jun 22 2018

P = Select[Range[5, 300, 4], PrimeQ];
A305728 = (6P)^3 (* Jean-François Alcover, Jun 22 2018 *)

first(n) = my(res=List()); forprime(p=5, oo, if(p%4 == 1, listput(res,(6*p)^3); n--; if(n==0, return(res)))) \\ David A. Corneth, Jun 27 2018

A307152 a(n) = floor((A002144(n)+19)/24).

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7, 7, 8, 8, 8, 9, 10, 10, 10, 11, 12, 12, 12, 13, 13, 14, 14, 15, 15, 16, 17, 17, 17, 17, 18, 18, 19, 19, 20, 22, 22, 23, 24, 24, 24, 25, 25, 26, 26

Offset: 1

N. J. A. Sloane, Mar 31 2019

nonn

This sequence arises in several different contexts [Kramer].
The number of occurrences of k in the sequence is A296021(6*k) - A296021(6*k-6). - Robert Israel, Mar 31 2019
Original name was: "Floor( (q+19)/24 ) where q is a prime == 1 (mod 4)." - Robert Israel, Apr 07 2019

Robert Israel, Table of n, a(n) for n = 1..10000
Jürg Kramer, On the linear independence of certain theta-series, Mathematische Annalen 281.2 (1988): 219-228. See page 226.

Cf. A002144, A296021.

map(t -> floor((t+19)/24), select(isprime, [seq(i,i=1..1000,4)])); # Robert Israel, Mar 31 2019

Table[Floor[(q + 19)/24], {q, Select[Range[1, 650, 4], PrimeQ]}] (* Michael De Vlieger, Mar 31 2019 *)

A330406 a(n) is the smallest prime q such that q^((p-1)/2) == -1 (mod p), where p = A002144(n) is the n-th prime congruent to 1 mod 4.

Original entry on oeis.org

2, 2, 3, 2, 2, 3, 2, 2, 5, 3, 5, 2, 2, 3, 3, 2, 2, 2, 2, 5, 2, 2, 3, 7, 3, 2, 2, 3, 2, 5, 2, 5, 2, 3, 2, 2, 2, 3, 7, 2, 5, 3, 5, 2, 2, 3, 2, 2, 3, 5, 3, 7, 2, 3, 3, 2, 2, 5, 2, 2, 2, 2, 2, 3, 7, 2, 2, 3, 2, 2, 2, 3, 2, 3, 3, 5, 2, 3, 3, 2, 11, 2, 2, 5, 3, 2, 2, 2, 3, 2, 2, 11, 5, 2, 3, 11, 2, 3, 2, 2, 7, 2, 3, 5, 2, 7, 3, 2, 2

Offset: 1

Nicholas C. Singer, Dec 13 2019

nonn

Subset of A053760 corresponding to p == 1 (mod 4).

A002144(n) = p is a sum of two integer squares (Fermat): p = a^2 + b^2. To find a and b, calculate gcd(p, A330406(n)^((p-1)/4)+i) = a + bi in the Gaussian integers.

			Let p = A002144(30)= 313. Then (p-1)/2 = 156. Now 2^156 == 3^156 == 1 (mod p) but 5^156 == -1 (mod p).  Thus A330406(30)=5.

Cf. A002144, A053760.

Map[Block[{q = 2}, While[PowerMod[q, (# - 1)/2, #] != # - 1, q = NextPrime@ q]; q] &, Select[4 Range[350] + 1, PrimeQ]] (* Michael De Vlieger, Dec 29 2019 *)

A002144 = select(p->p%4==1, primes(2200));
A330406 = vector(1000); for(i=1, 1000, my(p=A002144[i]); forprime(j=1, 20, my(x=Mod(j, p)^((p-1)/2)); if(x+1, , A330406[i]=j; break)))
A330406

cp's OEIS Frontend

A134441 Last two digits of primes of form 4n+1 (A002144), excluding 5. Leading 0's omitted.

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A140385 Consecutive area triples (y^2-x^2,4xy,p=x^2+y^2) associated with the Pythagorean primes A002144.

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A174492 a(n) = the smallest k such that k^2+1 = p*A002144(n)^2, p prime of A002144 .

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A199692 Subsequence of Pythagorean primes (A002144): each square is used only once.

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A212353 a(n) is the smallest positive solutions of the congruence x^2 + (x+1)^2 == 0 (mod prime), where prime = A002144(n) (Pythagorean primes).

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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).

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A216924 Consecutive Pythagorean primes p = A002144(r) and q = A002144(r+1) such that q - p > log(p)^2. The number a(n) is the n-th value of p.

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A305728 Numbers of the form 216*p^3, where p is a Pythagorean prime (A002144).

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A307152 a(n) = floor((A002144(n)+19)/24).

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A330406 a(n) is the smallest prime q such that q^((p-1)/2) == -1 (mod p), where p = A002144(n) is the n-th prime congruent to 1 mod 4.