A002314 -id:A002314 - cp's OEIS frontend

A155107 Numbers that are 23 or 30 (mod 53).

23, 30, 76, 83, 129, 136, 182, 189, 235, 242, 288, 295, 341, 348, 394, 401, 447, 454, 500, 507, 553, 560, 606, 613, 659, 666, 712, 719, 765, 772, 818, 825, 871, 878, 924, 931, 977, 984, 1030, 1037, 1083, 1090, 1136, 1143, 1189, 1196, 1242, 1249, 1295

Offset: 1

Vincenzo Librandi, Jan 20 2009

Also, numbers k such that k^2 == -1 (mod 53).

The first pair (a,b) is such that a+b=p=53, a*b=p*h+1, with h<=(p-1)/4; subsequent pairs are given as (a+kp, b+kp), k=1,2,3...

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

Cf. numbers n such that n^2 == -1 (mod p), where p is a prime of the form 4k+1: A047221 (p=5), A155086 (p=13), A155095 (p=17), A155096 (p=29), A155097 (p=37), A155098 (p=41), this sequence (p=53), A241406 (p=61), A241407 (p=73), A241520 (p=89), A241521 (p=97).

I:=[23,30,76]; [n le 3 select I[n] else Self(n-1)+Self(n-2)-Self(n-3): n in [1..50]]; // Vincenzo Librandi, Apr 24 2014

[-23*(-1)^n+53*Floor(n/2): n in [1..50]]; // Vincenzo Librandi, Apr 24 2014

Select[Range[1300], PowerMod[#, 2, 53] == 52 &] (* or *) LinearRecurrence[ {1, 1, -1}, {23, 30, 76}, 50] (* Harvey P. Dale, Nov 30 2011 *)
CoefficientList[Series[(23 + 7 x + 23 x^2)/((1 + x) (1 - x)^2), {x, 0, 100}], x] (* Vincenzo Librandi, Apr 24 2014 *)

A155107(n)=n\2*53-23*(-1)^n /* M. F. Hasler, Jun 16 2010 */

a(n) = 23*(-1)^(n+1) + 53*floor(n/2). - M. F. Hasler, Jun 16 2010
a(2k+1) = 53 k + a(1), a(2k) = 53 k - a(1), with a(1) = 23 = A002314(7) since 53 = A002144(7). - M. F. Hasler, Jun 16 2010
a(n) = a(n-2) + 53 for all n > 2. - M. F. Hasler, Jun 16 2010
From R. J. Mathar, Feb 19 2009: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) = 53*n/2 - 53/4 - 39*(-1)^n/4.
G.f.: x*(23 + 7*x + 23*x^2)/((1+x)*(1-x)^2). (End)

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

A177979 Smallest number k such that A002313(n) divides k^2+1.

Original entry on oeis.org

1, 2, 5, 4, 12, 6, 9, 23, 11, 27, 34, 22, 10, 33, 15, 37, 44, 28, 80, 19, 81, 14, 107, 89, 64, 16, 82, 60, 53, 138, 25, 114, 148, 136, 42, 104, 115, 63, 20, 143, 29, 179, 67, 109, 48, 208, 235, 52, 118, 86, 24, 77, 125, 35, 194, 154, 149, 106, 58, 26, 135, 96, 353, 87, 39

Offset: 1

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), May 16 2010

1 followed by A002314. [From R. J. Mathar, May 29 2010]

			1^2+1 = 2 which is divided by A002313(1) which adds 1 to the sequence.
2^2+1 = 5 is divided by A002313(2) which adds 2 to the sequence.
27^2+1 = 730 is divided by A002313(10) which adds 27 to the sequence.

Friedhelm Padberg: Elementare Zahlentheorie. Spektrum Akademischer Verlag, Berlin Heidelberg, 1996

Disentangled variables in the definition - R. J. Mathar, Jun 07 2010

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

A047818 a(n) is the least number m such that A002313(n)*m - 1 is a square.

Original entry on oeis.org

1, 1, 2, 1, 5, 1, 2, 10, 2, 10, 13, 5, 1, 10, 2, 10, 13, 5, 37, 2, 34, 1, 50, 34, 17, 1, 25, 13, 10, 65, 2, 41, 65, 53, 5, 29, 34, 10, 1, 50, 2, 74, 10, 26, 5, 85, 106, 5, 25, 13, 1, 10, 26, 2, 61, 37, 34, 17, 5, 1, 26, 13, 170, 10, 2, 5, 130, 58, 125, 106, 73, 130, 50, 26, 170

Offset: 1

N. J. A. Sloane

nonn

A002313 has the 4k+1 and 4k+2 primes.

Related to Stormer numbers.

			a(3) = 2 because A002313(3)=13 and 13*2-1 = 5^2.

J. Todd, A problem on arc tangent relations, Amer. Math. Monthly, 56 (1949), 517-528.

Cf. A002313, A002314, A005528.

a(n) = ((A002314(n-1))^2 + 1) / A002313(n).

Edited by Don Reble, Apr 13 2006

A180342 a(n) = the smallest number k such that the smallest prime factor of k^2 + 1 equals A002144(n).

Original entry on oeis.org

2, 34, 4, 46, 6, 50, 76, 194, 100, 144, 366, 10, 730, 324, 374, 254, 286, 266, 886, 274, 14, 794, 610, 546, 16, 456, 494, 334, 724, 964, 520, 526, 834, 664, 1596, 504, 3510, 20, 2720, 1234, 1120, 516, 566, 874, 810, 756, 1134, 2110, 1224, 24, 670, 726

Offset: 1

Michel Lagneau, Jan 18 2011

nonn

The sequence giving the smallest number k such that the greatest prime factor of k^2 + 1 equals A002144(n) is A002314.

			a(1) = 2 because 2^2 + 1 = 5 = A002144(1) ;
a(2) = 34 because 34^2 + 1= 13*89 = A002144(2) * 89 ;
a(3) = 4 because 4^2 + 1 = 17 = A002144(3) ;
a(4) = 46 because 46^2 + 1 = 29*73 = A002144(4) * 73.

Cf. A002522, A002144, and A002314.

with(numtheory):T:=array(1..200):k:=1:for p from 1 to 1000 do: if type(p,prime)=true
  and irem(p,4)=1 then T[k]:=p:k:=k+1:else fi:od:for q from 1 to k do:z:=T[q]:ind:=0:for n from 1 to 10000 while(ind=0) do: x:=n^2+1:y:=factorset(x):if z=y[1] then ind:=1:printf(`%d, `,n):else fi:od: od:

A197062 Successive records in A152676.

Original entry on oeis.org

3, 8, 13, 17, 31, 32, 50, 55, 75, 91, 98, 100, 105, 129, 162, 183, 241, 288, 311, 334, 381, 392, 413, 489, 553, 578, 615, 651, 670, 722, 726, 741, 844, 968, 1013, 1152, 1164, 1261, 1264, 1461, 1561, 1601, 1682, 1800, 1809, 1905, 1979, 2048, 2225, 2312, 2450

Offset: 1

Artur Jasinski, Oct 09 2011

On plot picture of A152676 is easy to see that all points occurred between some upper and lower limit curve. Numbers in this sequence are the nearest upper limit curve.

Cf. A002144, A002314, A152676, A152680.

aa = {}; max = 0; Do[If[Mod[Prime[n], 4] == 1, k = 1; While[! Mod[k^2 + 1, Prime[n]] == 0, k++]; If[Prime[n] - k > max, max = Prime[n] - k; AppendTo[aa, Prime[n] - k]]], {n, 1, 1000}]; aa

cp's OEIS Frontend

A155107 Numbers that are 23 or 30 (mod 53).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Magma

Mathematica

PARI

Formula

Extensions

A177979 Smallest number k such that A002313(n) divides k^2+1.

Views

Author

Keywords

Comments

Examples

References

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

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

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

A047818 a(n) is the least number m such that A002313(n)*m - 1 is a square.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

Extensions

A180342 a(n) = the smallest number k such that the smallest prime factor of k^2 + 1 equals A002144(n).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

A197062 Successive records in A152676.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica