A177979 -id:A177979 - cp's OEIS frontend

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

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.

A223702 Irregular triangle of numbers k such that A002313(n), the n-th prime not congruent to 3 mod 4 is the largest prime factor of k^2 + 1.

Original entry on oeis.org

1, 2, 3, 7, 5, 8, 18, 57, 239, 4, 13, 21, 38, 47, 268, 12, 17, 41, 70, 99, 157, 307, 6, 31, 43, 68, 117, 191, 302, 327, 882, 18543, 9, 32, 73, 132, 278, 378, 829, 993, 2943, 23, 30, 83, 182, 242, 401, 447, 606, 931, 1143, 1772, 6118, 34208, 44179, 85353, 485298

Offset: 1

T. D. Noe, Apr 03 2013

Note that primes of the form 4x+3 are not divisors.

			Irregular triangle:
   p | {k}
-----+---------------------------------
   2 | {1},
   5 | {2, 3, 7},
  13 | {5, 8, 18, 57, 239},
  17 | {4, 13, 21, 38, 47, 268},
  29 | {12, 17, 41, 70, 99, 157, 307},
  37 | {6, 31, 43, 68, 117, 191, 302, 327, 882, 18543},
  41 | {9, 32, 73, 132, 278, 378, 829, 993, 2943}
  ...

Andrew Howroyd, Table of n, a(n) for n = 1..811 (first 22 rows for primes up to 197)
Florian Luca, Primitive divisors of Lucas sequences and prime factors of x^2 + 1 and x^4 + 1, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae 31 (2004), pp. 19-24.
Filip Najman, Smooth values of some quadratic polynomials, Glasnik Matematicki Series III 45 (2010), pp. 347-355
Filip Najman, List of Publications Page (Adjacent to entry number 7 are links with a data file for the first 22 rows (=811 terms) of this sequence). [As of Dec 2024, the link has the incorrect URL. Should be https://web.math.pmf.unizg.hr/~fnajman/rezplus1.html]

Cf. A002313, A014442, A177979 (first terms), A185389 (last terms), A223705, A285283, A379346 (row lengths), A379347 (row sums).
Cf. A285282, A285523.
Cf. A223701, A223703, A223704 (related tables).

t = Table[FactorInteger[n^2 + 1][[-1,1]], {n, 10^5}]; Table[Flatten[Position[t, Prime[n]]], {n, 13}]

Definition amended by Andrew Howroyd, Dec 22 2024

A379346 Number of integers of the form k^2 + 1 whose greatest prime factor is A002313(n), the n-th prime not congruent to 3 mod 4.

Original entry on oeis.org

1, 3, 5, 6, 7, 10, 9, 16, 17, 20, 26, 36, 36, 40, 46, 47, 56, 67, 73, 86, 97, 107

Offset: 1

Andrew Howroyd, Dec 22 2024

See A223702 for additional information.

Florian Luca, Primitive divisors of Lucas sequences and prime factors of x^2 + 1 and x^4 + 1, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae 31 (2004), pp. 19-24.
Filip Najman, Smooth values of some quadratic polynomials, Glasnik Matematicki Series III 45 (2010), pp. 347-355.

Row lengths of A223702.
First differences of A285283.
Cf. A002313, A177979, A185389.

A209877 a(n) = A209874(n)/2: Least m > 0 such that 4*m^2 = -1 modulo the Pythagorean prime A002144(n).

Original entry on oeis.org

1, 4, 2, 6, 3, 16, 15, 25, 23, 17, 11, 5, 38, 49, 50, 22, 14, 40, 81, 56, 7, 61, 72, 32, 8, 41, 30, 114, 69, 144, 57, 74, 68, 21, 52, 137, 167, 10, 133, 196, 127, 191, 174, 24, 104, 143, 26, 59, 43, 12, 258, 238, 289, 97, 77, 252, 53, 29, 13, 283, 48, 190, 335, 361, 31, 228, 291, 159, 263, 123, 260, 325, 363, 247, 162

Offset: 1

M. F. Hasler, Mar 14 2012

nonn

Also: Square root of -1/4 in Z/pZ, for Pythagorean primes p=A002144(n).
Also: Least m>0 such that the Pythagorean prime p=A002144(n) divides 4(kp +/- m)^2+1 for all k>=0.
In practice these can also be determined by searching the least N^2+1 whose least prime factor is p=A002144(n): For given p, all of these N will have a(n) or p-a(n) as remainder mod 2p.

			a(1)=1 since A002144(1)=5 and 4*1^2+1 is divisible by 5; as a consequence 4*(5k+/-1)^2+1 = 100k^2 +/- 40k + 5 is divisible by 5 for all k.
a(2)=4 since A002144(2)=13 and 4*4^2+1 = 65 is divisible by 13, while 4*1^1+1=5, 4*2^2+1=17 and 4*3^2+1=37 are not. As a consequence, 4*(13k+/-4)^2+1 = 13(...)+4*4^1+1 is divisible by 13 for all k.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A209874.
Cf. A002496, A014442, A085722, A144255, A089120, A193432.
Cf. A002314, A177979.

f:= proc(p) local m;
   if not isprime(p) then return NULL fi;
   m:= numtheory:-msqrt(-1/4, p);
   min(m,p-m);
end proc:
map(f, [seq(i,i=5..1000,4)]); # Robert Israel, Mar 13 2018

f[p_] := Module[{r}, r /. Solve[4 r^2 == -1, r, Modulus -> p] // Min];
f /@ Select[4 Range[300] + 1, PrimeQ] (* Jean-François Alcover, Jul 27 2020 *)

apply(p->lift(sqrt(Mod(-1,p)/4)), A002144)

A223705 Least number k such that prime(n) is the largest divisor of k^2 + 1, or 0 if there is no such k.

Original entry on oeis.org

1, 0, 2, 0, 0, 5, 4, 0, 0, 12, 0, 6, 9, 0, 0, 23, 0, 11, 0, 0, 27, 0, 0, 34, 22, 10, 0, 0, 33, 15, 0, 0, 37, 0, 44, 0, 28, 0, 0, 80, 0, 19, 0, 81, 14, 0, 0, 0, 0, 107, 89, 0, 64, 0, 16, 0, 82, 0, 60, 53, 0, 138, 0, 0, 25, 114, 0, 148, 0, 136, 42, 0, 0, 104, 0, 0

Offset: 1

T. D. Noe, Apr 03 2013

nonn

Note that a(n) = 0 for prime(n) = 3 (mod 4). If the zeros are removed, A002314 (with 1 prepended) and A177979 are produced.

Cf. A223701-A223707 (related sequences).

nn = 100; t = Table[0, {nn}]; Do[If[Mod[Prime[n], 4] == 3, t[[n]] = -1], {n, nn}]; n = 0; While[Times @@ t == 0, n++; s = FactorInteger[n^2 + 1][[-1, 1]]; p = PrimePi[s]; If[p <= nn && t[[p]] == 0, t[[p]] = n]]; Do[If[Mod[Prime[n], 4] == 3, t[[n]] = 0], {n, nn}]; t

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

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A223702 Irregular triangle of numbers k such that A002313(n), the n-th prime not congruent to 3 mod 4 is the largest prime factor of k^2 + 1.

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A379346 Number of integers of the form k^2 + 1 whose greatest prime factor is A002313(n), the n-th prime not congruent to 3 mod 4.

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A209877 a(n) = A209874(n)/2: Least m > 0 such that 4*m^2 = -1 modulo the Pythagorean prime A002144(n).

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A223705 Least number k such that prime(n) is the largest divisor of k^2 + 1, or 0 if there is no such k.

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