A028871 -id:A028871 - cp's OEIS frontend

A302442 Number of primes of the form b^2-2 for b <= 10^n.

5, 26, 157, 1153, 8888, 72928, 615643, 5328644, 47034083, 420950239

Offset: 1

Seiichi Manyama, Apr 08 2018

From Jacques Tramu, Sep 13 2018: (Start)
Table C(i) = a(i)/pi(10^i) = a(i)/A000720(10^i)
a(1)  =         5         C(1)  = 1.25000000
a(2)  =        26         C(2)  = 1.04000000
a(3)  =       157         C(3)  = 0.93452381
a(4)  =      1153         C(4)  = 0.93816111
a(5)  =      8888         C(5)  = 0.92660550
a(6)  =     72928         C(6)  = 0.92904278
a(7)  =    615643         C(7)  = 0.92636541
a(8)  =   5328644         C(8)  = 0.92487818
a(9)  =  47034083         C(9)  = 0.92500224
a(10) = 420950239         C(10) = 0.92505860
(End)

			a(1) = 5 because there are 5 primes of the form b^2-2 for b <= 10 : 2, 7, 23, 47 and 79.

Number of primes of the form b^2+m for b <= 10^n: A302443 (m=-3), this sequence (m=-2), A206709 (m=1), A302434 (m=2), A302435 (m=3).

Cf. A028871.

{a(n) = sum(k=0, 10^n, isprime(k^2-2))}

from sympy import isprime
def aupton(terms):
  s, alst = 0, []
  for n in range(1, terms+1):
    s += sum(isprime(b**2-2) for b in range(10**(n-1), 10**n))
    alst.append(s)
  return alst
print(aupton(6)) # Michael S. Branicky, May 26 2021

a(10) from Jacques Tramu, Sep 14 2018

A309998 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 529)^2 = y^2.

Original entry on oeis.org

0, 276, 287, 740, 759, 1587, 3059, 3120, 5687, 5796, 10580, 19136, 19491, 34440, 35075, 62951, 112815, 114884, 202011, 205712, 368184, 658812, 670871, 1178684, 1200255, 2147211, 3841115, 3911400, 6871151, 6996876, 12516140, 22388936, 22798587, 40049280, 40782059, 72950687

Offset: 1

Mohamed Bouhamida, Aug 26 2019

For the generic case x^2 + (x + p^2)^2 = y^2 with p = m^2 - 2 a prime number in A028871, m>=5, (0; p^2) and (2*m^3 + 2*m^2 - 4*m - 4; m^4 + 2*m^3 - 4*m - 4) are solutions.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,6,-6,0,0,0,-1,1).

Cf. A207059.

Rest@ CoefficientList[Series[x^2*(276 + 11 x + 453 x^2 + 19 x^3 + 828 x^4 - 184 x^5 - 5 x^6 - 151 x^7 - 5 x^8 - 184 x^9)/((1 - x) (1 - 6 x^5 + x^10)), {x, 0, 36}], x] (* Michael De Vlieger, Sep 29 2019 *)

concat(0, Vec(x^2*(276 + 11*x + 453*x^2 + 19*x^3 + 828*x^4 - 184*x^5 - 5*x^6 - 151*x^7 - 5*x^8 - 184*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)) + O(x^40))) \\ Colin Barker, Aug 27 2019

a(n) = 6*a(n-5) - a(n-10) + 1058 with a(0) = 0, a(1) = 276, a(2) = 287, a(3) = 740, a(4) = 759, a(5) = 1587, a(6) = 3059, a(7) = 3120, a(8) = 5687, a(9) = 5796.
From Colin Barker, Aug 27 2019: (Start)
G.f.: x^2*(276 + 11*x + 453*x^2 + 19*x^3 + 828*x^4 - 184*x^5 - 5*x^6 - 151*x^7 - 5*x^8 - 184*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)).
a(n) = a(n-1) + 6*a(n-5) - 6*a(n-6) - a(n-10) + a(n-11) for n>11.
(End)

A063531 Numbers k such that sigma(k) + 1 is a square.

Original entry on oeis.org

2, 7, 8, 14, 15, 23, 32, 33, 35, 47, 54, 56, 57, 60, 72, 78, 79, 84, 87, 92, 95, 120, 123, 124, 128, 138, 143, 154, 165, 167, 174, 184, 190, 196, 213, 223, 235, 242, 252, 253, 258, 267, 295, 312, 315, 319, 323, 327, 348, 359, 375, 378, 380, 393, 412, 423, 439

Offset: 1

Labos Elemer, Aug 02 2001

nonn

Numbers k such that A000203(k) = -1 + m^2 for some m.

			If k = p(p+2) is a product of twin primes (from A037074), then sigma(k) + 1 = 1 + (p+1)(p+3) = (p+2)^2, square of the larger twin. Other solutions can be either special primes = m^2 - 2 or composites like 120: sigma(120) = 120 + 60 + ... + 1 = 360 = 19^2 - 1. Square number solution is, e.g., 196: sigma(196) = 399 = 20^2 - 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harry J. Smith)

Cf. A000203, A028871, A037074, A063530, A088580.

Select[Range[500],IntegerQ[Sqrt[DivisorSigma[1,#]+1]]&] (* Harvey P. Dale, Jul 02 2021 *)

{ n=0; for (a=1, 10^9, if (issquare(sigma(a) + 1), write("b063531.txt", n++, " ", a); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 25 2009

Minor edits from Franklin T. Adams-Watters, Aug 29 2009

A146329 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 4.

Original entry on oeis.org

6, 7, 8, 14, 20, 23, 24, 28, 32, 33, 34, 42, 47, 48, 52, 55, 60, 62, 69, 72, 75, 78, 79, 80, 95, 98, 110, 119, 120, 126, 133, 135, 136, 138, 140, 141, 142, 156, 167, 168, 174, 180, 189, 194, 205, 210, 213, 215, 219, 220, 222, 223, 224, 248, 252, 254, 272, 287, 288

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

For primes in this sequence see A028871 - {2}.

			a(2) = 7 because continued fraction of (1 + sqrt(7))/2 = 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, ... has period (1,1,4,1) length 4.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000290, A078370, A146326-A146345, A028871, A146348-A146360.

isA146329 := proc(n) RETURN(A146326(n) = 4) ; end:
for n from 2 to 400 do if isA146329(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

cf4Q[n_]:=Module[{s=(1+Sqrt[n])/2},If[IntegerQ[s],1,Length[ ContinuedFraction[ s][[2]]]]==4]; Select[Range[300],cf4Q] (* Harvey P. Dale, Dec 14 2017 *)

39, 68, 150, 155, etc. removed by R. J. Mathar, Sep 06 2009

A259764 Least prime p such that prime(p*n)-1 is a square, or 0 if no such p exists.

Original entry on oeis.org

3, 13, 41, 3, 11, 2, 241, 181, 5, 2927, 5, 523, 2, 4967, 3, 421, 33053, 8447, 17107, 20747, 1811, 5743, 20407, 99643, 165443, 769, 21269, 46099, 3121, 9883, 16301, 523, 10771, 41603, 17, 7, 48383, 455353, 711317, 1637, 3, 105397, 43, 12071, 186113, 56437, 303157, 211, 25951, 178817

Offset: 1

Zhi-Wei Sun, Jul 04 2015

nonn

Conjecture: a(n) > 0 for all n > 0.
This is stronger than the conjecture in A259731. It implies the well-known conjecture that there are infinitely many primes of the form x^2-1 with x an integer.
I also conjecture that for any positive integer n there exists a prime p such that prime(p*n)+2 is a square.

			a(1) = 3 since 3 is prime and prime(3*1)-1 = 2^2 is a square.
a(2) = 13 since 13 is prime and prime(13*2)-1 = 10^2 is a square.

Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Zhi-Wei Sun, Table of n, a(n) for n = 1..500
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.

Cf. A000040, A000290, A002496, A028871, A259731.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[k=0;Label[bb];k=k+1;If[SQ[Prime[Prime[k]*n]-1],Goto[aa],Goto[bb]];Label[aa];Print[n," ",Prime[k]];Continue,{n,1,50}]

A332000 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 47^2)^2 = y^2.

Original entry on oeis.org

0, 752, 1820, 2231, 3995, 6627, 10575, 16511, 18840, 28952, 44180, 67116, 101664, 115227, 174135, 262871, 396539, 597891, 676940, 1020276, 1537464, 2316536, 3490100, 3950831, 5951939, 8966331, 13507095, 20347127, 23032464, 34695776, 52264940, 78730452, 118597080, 134248371

Offset: 1

Mohamed Bouhamida, Feb 04 2020

For the generic case x^2 + (x + p^2)^2 = y^2 with p = m^2 - 2 a (prime) number in A028871, m>=7 (means p>=47), the first five consecutive solutions are: (0; p^2), (2*m^3+2*m^2-4*m-4; m^4+2*m^3-4*m-4), (4*m^3+8*m^2+8*m; m^4+4*m^3+12*m^2+8*m+4), (3*m^4-20*m^3+44*m^2-40*m+12; 5*m^4-28*m^3+60*m^2-56*m+20), (3*m^4-10*m^3+2*m^2+20*m-16; 5*m^4-14*m^3+28*m-20) and the other solutions are defined by: (X(n); Y(n))= (3*X(n-5)+2*Y(n-5)+p^2; 4*X(n-5)+3*Y(n-5)+2*p^2).

X(n) = 6*X(n-5) - X(n-10) + 2*p^2, and Y(n) = 6*Y(n-5) - Y(n-10) (can be easily proved using X(n) = 3*X(n-5) + 2*Y(n-5) + p^2, and Y(n) = 4*X(n-5) + 3*Y(n-5) + 2*p^2).

			For p=47 (m=7) the first five (5) consecutive solutions are (0, 2209), (752, 3055), (1820, 4421), (2231, 4969), (3995, 7379).

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,6,-6,0,0,0,-1,1).

Cf. A028871 (Primes of the form m^2 - 2).

Solutions x to x^2+(x+p^2)^2=y^2: A118554 (p=7), A207059 (p=17), A309998 (p=23), A331265 (p=31), this sequence (p=47).

I:=[0, 752, 1820, 2231, 3995, 6627, 10575, 16511, 18840, 28952]; [n le 10 select I[n] else 6*Self(n-5) - Self(n-10)+4418: n in [1..100]];

LinearRecurrence[{1, 0, 0, 0, 6, -6, 0, 0, 0, -1, 1}, {0, 752, 1820, 2231, 3995, 6627, 10575, 16511, 18840, 28952, 44180}, 40] (* Jean-François Alcover, Feb 08 2020 *)

concat(0, Vec(x^2*(752 + 1068*x + 411*x^2 + 1764*x^3 + 2632*x^4 - 564*x^5 - 472*x^6 - 137*x^7 - 472*x^8 - 564*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)) + O(x^40))) \\ Colin Barker, Feb 04 2020

a(n) = 6*a(n-5) - a(n-10) + 4418 for n >= 11; a(1)=0, a(2)=752, a(3)=1820, a(4)=2231, a(5)=3995, a(6)=6627, a(7)=10575, a(8)=16511, a(9)=18840, a(10)=28952.
From Colin Barker, Feb 04 2020: (Start)
G.f.: x^2*(752 + 1068*x + 411*x^2 + 1764*x^3 + 2632*x^4 - 564*x^5 - 472*x^6 - 137*x^7 - 472*x^8 - 564*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)).
a(n) = a(n-1) + 6*a(n-5) - 6*a(n-6) - a(n-10) + a(n-11) for n>11.
(End)

A363102 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-2))))).

Original entry on oeis.org

7, 7, 23, 17, 47, 31, 79, 7, 17, 71, 167, 97, 223, 127, 41, 23, 359, 199, 439, 241, 31, 41, 89, 337, 727, 1, 839, 449, 137, 73, 1087, 577, 1223, 647, 1367, 103, 1, 47, 73, 881, 1847, 967, 1, 151, 2207, 1151, 2399, 1249, 113, 193, 401, 1, 3023, 1567, 191, 41, 71, 257, 3719, 113, 3967, 89, 103, 311

Offset: 3

Mohammed Bouras, May 19 2023

nonn

Conjecture 1: The sequence contains only 1's and primes.
Conjecture 2: All prime numbers appear either twice (same as A356247 and A357127) or three times.
Similar terms of A164314.
Conjecture: Record values correspond to A028871(m), m > 1. - Bill McEachen, Mar 06 2024
a(n) = 1 positions appear to correspond to A060515(m), m > 2. - Bill McEachen, Aug 05 2024

			a(5) = (5^2 - 2)/gcd(5^2 - 2, 2*A051403(5-3) + 5*A051403(5-4))= 23.
a(6) = a(11) = 6 + 11 = 17.
a(7) = a(40) = 7 + 40 = 47.

Bill McEachen, Table of n, a(n) for n = 3..10002
Mohammed Bouras, The Distribution Of Prime Numbers And Continued Fractions, (ppt) (2022)

Cf. A008865, A051403, A059772, A164314, A356247, A357127.

a051403(n) = (n+2)*sum(k=0, n, k!)/2;
a(n) = (n^2 - 2)/gcd(n^2 - 2, 2*a051403(n-3) + n*a051403(n-4)); \\ Michel Marcus, May 24 2023

a(n) = (n^2 - 2)/gcd(n^2 - 2, 2*A051403(n-3) + n*A051403(n-4)).
a(n) = A164314(n) if A164314(n) > n.
If a(n) = a(m) and n < m < a(n), then a(n) = n + m.

A163158 Primes of the form f^2-2, where f is a Fibonacci number.

Original entry on oeis.org

2, 7, 23, 167, 439, 3023, 7919, 54287, 974167, 2550407, 32522920134767, 3372041405099481407, 9839618880490124200692486211717007, 724995932728680612729658820311719934835368079

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jul 21 2009

nonn

A basic heuristic suggests that this sequence is infinite with about k * log_phi(n) members below n.

Indices of associated Fibonacci numbers are 3, 4, 5, 7, 8, 10, 11, 13, 16, 17, 34, 46, 83, 109, 113, 158, 181, 203, 350, 490, 491, 565, 1024, 1114, 2800, 4222, 4847, 4961, 11507, 12554, ...

			2^2-2=2, 3^2-2=7, 5^2-2=23

Cf. A000045, A028871

f[n_]:=Fibonacci[n]^2-2; lst={};Do[If[PrimeQ[f[n]],AppendTo[lst,f[n]]],{n,6!}];lst

Comments from Charles R Greathouse IV, Nov 09 2009

A225078 Numbers n such that n^2+1 and (n+1)^2-2 are both prime.

Original entry on oeis.org

1, 2, 4, 6, 14, 20, 26, 36, 54, 74, 116, 120, 126, 130, 134, 160, 176, 204, 210, 230, 236, 256, 264, 284, 300, 314, 340, 386, 420, 440, 466, 490, 496, 544, 594, 636, 644, 714, 750, 760, 784, 816, 930, 950, 986, 1070, 1124, 1140, 1146, 1156, 1174, 1176, 1210

Offset: 1

César Aguilera, Apr 26 2013

nonn

Prime limits of the Legendré conjecture for a given n.

			n=2; n+1=3 ;n^2+1=5 and (n+1)^2-2=7.
n=490; n+1=491; n^2+1=240101 and (n+1)^2-2=241079.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Legendre's Conjecture
Wikipedia, Legendre's conjecture

Cf. A002522, A008865, A010051, A014085, A002496, A028871.

import Data.Function (on)
import Data.List (elemIndices)
a225078 n = a225078_list !! (n-1)
a225078_list = elemIndices 1 $
   zipWith ((*) `on` a010051') a002522_list a008865_list
-- Reinhard Zumkeller, May 06 2013

Select[Range[2000], PrimeQ[#^2 + 1] && PrimeQ[(# + 1)^2 - 2] &] (* T. D. Noe, May 06 2013 *)

A241528 Primes p such that p + 1234567890 is also prime where 1234567890 is the first pandigital number with digits in order.

Original entry on oeis.org

17, 23, 37, 59, 131, 139, 157, 199, 241, 311, 353, 359, 397, 433, 479, 547, 673, 691, 769, 877, 937, 947, 953, 1051, 1091, 1097, 1181, 1297, 1301, 1409, 1451, 1471, 1489, 1531, 1609, 1619, 1697, 1709, 1861, 1879, 1889, 1913, 1951, 2053, 2063, 2089, 2099, 2113

Offset: 1

K. D. Bajpai, Apr 25 2014

nonn

			17 is prime and appears in the sequence because 17 + 1234567890 = 1234567907, which is also prime.
23 is prime and appears in the sequence because 23 + 1234567890 = 1234567913, which is also prime.
19 is prime but not included in the sequence since 19 + 1234567890 = 1234567909 = (59107)*(20887), which is not prime.

K. D. Bajpai, Table of n, a(n) for n = 1..10000

Cf. A000040, A002496, A028871, A050289, A056899, A234812, A235640, A240587.

KD := proc() local a,k; k:=ithprime(n);a:=k+1234567890; if isprime(a) then RETURN (k); fi; end: seq(KD(), n=1..1000);

lst={}; Do[p=Prime[n]; If[PrimeQ[p+1234567890], AppendTo[lst,p]],{n,1,1000}]; lst
(* For the b-file *)  c=0; k=Prime[n]; a=k+1234567890; Do[If[PrimeQ[a], c++; Print[c," ",k]],{n,1,10^5}]
Select[Prime[Range[400]],PrimeQ[#+1234567890]&] (* Harvey P. Dale, Nov 18 2021 *)

s=[]; forprime(p=2, 3000, if(isprime(p+1234567890), s=concat(s, p))); s \\ Colin Barker, Apr 25 2014

cp's OEIS Frontend

A302442 Number of primes of the form b^2-2 for b <= 10^n.

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A309998 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 529)^2 = y^2.

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A063531 Numbers k such that sigma(k) + 1 is a square.

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A146329 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 4.

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Maple

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A259764 Least prime p such that prime(p*n)-1 is a square, or 0 if no such p exists.

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A332000 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 47^2)^2 = y^2.

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Magma

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A363102 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-2))))).

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