A056899 -id:A056899 - cp's OEIS frontend

A261655 Squares equal to the difference between two successive primes of the form k^2+2 in the order in which they appear in A056899.

1, 144, 1296, 3600, 176400, 156816, 2985984, 921600, 2702736, 11696400, 18974736, 21566736, 17740944, 5992704, 125888400, 7290000, 8363664, 12027024, 63680400, 210830400, 13838400, 72590400, 15116544, 15397776, 67568400, 128595600, 80784144, 93315600, 33039504

Offset: 1

Michel Lagneau, Aug 28 2015

nonn

Note that all terms after the first are divisible by 144: for n>1 the sequence b(n) = sqrt(a(n)/144) is 1, 3, 5, 35, 33, 144, 80, 137, 285, 363, 387, 351, 204, 935, 225, 241, 289, ..., see A261659.

The proof that all terms are == 0 (mod 144) is simple once you realize that the primes == 11 (mod 72), see comment in A056899. - Robert G. Wilson v, Sep 03 2015

			A056899(2)- A056899(1) = 3-2 = 1^2;
A056899(5)- A056899(4) = 227-83 = 144 = 12^2;
A056899(14)- A056899(13) = 12323-11027 = 1296 = 36^2.

Robert G. Wilson v, Table of n, a(n) for n = 1..1100

Cf. A056899, A216330, A261659.

q:=2:for n from 1 to 10^7 do:p:=n^2+2:if isprime(p) then x:=p-q:q:=p: z:=sqrt(x):if z=floor(z) then printf(`%d, `, x):else fi:fi:od:

Select[ Differences[ Select[ Range[0, 1000000], PrimeQ[#^2 + 2] &]^2], IntegerQ@ Sqrt@# &] (* or *)
k = 1; p = 3; lst = {1}; While[k < 10000001, q = (6k +3)^2 + 2; If[ PrimeQ@ q, If[ IntegerQ@ Sqrt[q - p], AppendTo[lst, q - p]]; p = q]; k++] (* Robert G. Wilson v, Sep 03 2015 *)

A089921 Duplicate of A056899.

Original entry on oeis.org

2, 3, 11, 83, 227, 443, 1091, 1523, 2027, 3251, 6563, 9803, 11027, 12323, 13691

Offset: 1

dead

A224519 For n >= 4, a(n) = (A056899(n) - A056899(n-1))/72, where A056899 lists the primes of the form k^2 + 2.

Original entry on oeis.org

1, 2, 3, 9, 6, 7, 17, 46, 45, 17, 18, 19, 20, 90, 106, 260, 37, 77, 40, 41, 42, 132, 190, 50, 51, 105, 222, 58, 119, 61, 62, 127, 335, 70, 71, 145, 74, 75, 310, 326, 169, 531, 92, 93, 189, 490, 101, 735, 442, 113, 345, 235, 854, 510, 660, 271, 414, 710, 438

Offset: 4

Michel Lagneau, Apr 09 2013

nonn

For n >= 3, all prime of the form n^2 + 2 (A056899) are equal to 11 modulo 72.
Observation: this sequence contains couples of consecutive numbers: (2,3), (6,7), (17, 18), (18, 19), (19, 20), (40, 41), (41, 42), ..., (1238, 1239), (1272, 1273), ...
Conjecture: the number of couples such that a(n+1) = a(n) + 1 is infinite.
Consequence: there exists an infinity of triples with 3 successive primes p1 < p2 < p3 of the form n^2 + 2 such that p2 = (p1 + p3)/2 - 36.
Proof: if the conjecture is true, a(n+1) - a(n) = 1 =>
(1)   p2 - p1 = 72a
(2)   p3 - p2 = 72(a+1)
and (2) - (1) => p2 = (p1 + p3)/2 - 36.
The triples of primes are (11, 83, 227), (83, 227, 443), (1091, 1523, 2027), (11027, 12323, 13691), ...

			a(5) = (A056899(5) - A056899(4))/72 = (227 - 83)/72 = 2.

Cf. A056899, A059100.

with(numtheory): T:=array(1..100):k:=0:for n from 3 to 2000 do: if type(n^2+2,prime)=true then k:=k+1:T[k]:=n^2+1:else fi:od:for i from 1 to k do: printf(`%d, `,(T[i+1]-T[i])/72):od:

A059100 a(n) = n^2 + 2.

Original entry on oeis.org

2, 3, 6, 11, 18, 27, 38, 51, 66, 83, 102, 123, 146, 171, 198, 227, 258, 291, 326, 363, 402, 443, 486, 531, 578, 627, 678, 731, 786, 843, 902, 963, 1026, 1091, 1158, 1227, 1298, 1371, 1446, 1523, 1602, 1683, 1766, 1851, 1938, 2027, 2118, 2211, 2306, 2403, 2502, 2603

Offset: 0

Henry Bottomley, Feb 13 2001

Let s(n) = Sum_{k>=1} 1/n^(2^k). Then I conjecture that the maximum element in the continued fraction for s(n) is n^2 + 2. - Benoit Cloitre, Aug 15 2002
Binomial transformation yields A081908, with A081908(0)=1 dropped. - R. J. Mathar, Oct 05 2008
1/a(n) = R(n)/r with R(n) the n-th radius of the Pappus chain of the symmetric arbelos with semicircle radii r, r1 = r/2 = r2. See the MathWorld link for Pappus chain (there are two of them, a left and a right one. In this case these two chains are congruent). - Wolfdieter Lang, Mar 01 2013
a(n) is the number of election results for an election with n+2 candidates, say C1, C2, ..., and C(n+2), and with only two voters (each casting a single vote) that have C1 and C2 receiving the same number of votes. See link below. - Dennis P. Walsh, May 08 2013
This sequence gives the set of values such that for sequences b(k+1) = a(n)*b(k) - b(k-1), with initial values b(0) = 2, b(1) = a(n), all such sequences are invariant under this transformation: b(k) = (b(j+k) + b(j-k))/b(j), except where b(j) = 0, for all integer values of j and k, including negative values. Examples are: at n=0, b(k) = 2 for all k; at n=1, b(k) = A005248; at n=2, b(k) = 2*A001541; at n=3, b(k)= A057076; at n=4, b(k) = 2*A023039. This b(k) family are also the transformation results for all related b'(k) (i.e., those with different initial values) including non-integer values. Further, these b(k) are also the bisections of the transformations of sequences of the form G(k+1) = n * G(k) + G(k-1), and those bisections are invariant for all initial values of g(0) and g(1), including non-integer values. For n = 1 this g(k) family includes Fibonacci and Lucas, where the invariant bisection is b(k) = A005248. The applicable bisection for this transformation of g(k) is for the odd values of k, and applies for all n. Also see A000032 for a related family of sequences. - Richard R. Forberg, Nov 22 2014
Also the number of maximum matchings in the n-gear graph. - Eric W. Weisstein, Dec 31 2017
Also the Wiener index of the n-dipyramidal graph. - Eric W. Weisstein, Jun 14 2018
Numbers of the form n^2+2 have no factors that are congruent to 7 (mod 8). - Gordon E. Michaels, Sep 12 2019
For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {n, 2n}]. - Magus K. Chu, Sep 10 2022

			For n = 2, a(2) = 6 since there are 6 election results in a 4-candidate, 2-voter election that have candidates c1 and c2 tied. Letting <i,j> denote voter 1 voting for candidate i and voter 2 voting for candidate j, the six election results are <1,2>, <2,1>, <3,3>, <3,4>, <4,3>, and <4,4>. - _Dennis P. Walsh_, May 08 2013

Harry J. Smith, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 52, 56.
Hesam Dashti, A New Upper Bound on the Length of Shortest Permutation Strings; An Algorithm for Generating Permutation Strings, arXiv:1009.5053 [math.CO], 2010. - _Jonathan Vos Post_, Sep 28 2010
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Alexander Soifer, Coffee Hour and the Conway-Soifer Cover-Up, In: How Does One Cut a Triangle? (2009), pp. 147-156. See also here
Dennis P. Walsh, Notes on a tied election.
Eric Weisstein's World of Mathematics, Dipyramidal Graph.
Eric Weisstein's World of Mathematics, Gear Graph.
Eric Weisstein's World of Mathematics, Matching.
Eric Weisstein's World of Mathematics, Maximum Independent Edge Set.
Eric Weisstein's World of Mathematics, Near-Square Prime.
Eric Weisstein's World of Mathematics, Pappus Chain.
Eric Weisstein's World of Mathematics, Wiener Index.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000032, A000196, A000290, A001541, A002522, A005248, A008865, A056105, A056109, A056899, A057076, A069987, A081908, A114964, A156798, A166464.
Apart from initial terms, same as A010000.
2nd row/column of A295707.

a059100 = (+ 2) . (^ 2)
a059100_list = scanl (+) (2) [1, 3 ..]
-- Reinhard Zumkeller, Feb 09 2015

with(combinat, fibonacci):seq(fibonacci(3, i)+1, i=0..49); # Zerinvary Lajos, Mar 20 2008

Table[n^2 + 2, {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, Dec 15 2008 *)
LinearRecurrence[{3, -3, 1}, {2, 3, 6}, 50] (* Vincenzo Librandi, Feb 15 2012 *)
Range[0, 20]^2 + 2 (* Eric W. Weisstein, Dec 31 2017 *)
CoefficientList[Series[(-2 + 3 x - 3 x^2)/(-1 + x)^3, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 31 2017 *)

a(n) = { n^2+2 } \\ Harry J. Smith, Jun 24 2009

[lucas_number1(3,n,-2) for n in range(0, 50)] # Zerinvary Lajos, May 16 2009

G.f.: (2 - 3*x + 3*x^2)/(1 - x)^3. - R. J. Mathar, Oct 05 2008
a(n) = ((n - 2)^2 + 2*(n + 1)^2)/3. - Reinhard Zumkeller, Feb 13 2009
a(n) = A000196(A156798(n) - A000290(n)). - Reinhard Zumkeller, Feb 16 2009
a(n) = 2*n + a(n-1) - 1 with a(0) = 2. - Vincenzo Librandi, Aug 07 2010
a(n+3) = (A166464(n+5) - A166464(n))/20. - Paul Curtz, Nov 07 2012
From Paul Curtz, Nov 07 2012: (Start)
a(3*n) mod 9 = 2.
a(3*n+1) = 3*A056109(n).
a(3*n+2) = 3*A056105(n+1). (End)
Sum_{n >= 1} 1/a(n) = Pi * coth(sqrt(2)*Pi) / 2^(3/2) - 1/4. - Vaclav Kotesovec, May 01 2018
From Amiram Eldar, Jan 29 2021: (Start)
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(2)*Pi*(csch(sqrt(2)*Pi)))/4.
Product_{n>=0} (1 + 1/a(n)) = sqrt(3/2)*csch(sqrt(2)*Pi)*sinh(sqrt(3)*Pi).
Product_{n>=0} (1 - 1/a(n)) = csch(sqrt(2)*Pi)*sinh(Pi)/sqrt(2). (End)
E.g.f.: exp(x)*(2 + x + x^2). - Stefano Spezia, Aug 07 2024

A146326 Length of the period of the continued fraction of (1+sqrt(n))/2.

Original entry on oeis.org

0, 2, 2, 0, 1, 4, 4, 4, 0, 2, 2, 2, 1, 4, 2, 0, 3, 6, 6, 4, 2, 6, 4, 4, 0, 2, 2, 4, 1, 2, 8, 4, 4, 4, 2, 0, 3, 6, 6, 8, 5, 4, 10, 6, 2, 8, 4, 4, 0, 2, 2, 4, 1, 6, 4, 2, 6, 6, 6, 4, 3, 4, 2, 0, 3, 6, 10, 6, 4, 6, 8, 4, 9, 6, 4, 8, 2, 4, 4, 4, 0, 2, 2, 2, 1, 6, 2, 8, 7, 2, 8, 8, 2, 12, 4, 8, 9, 4, 2, 0

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

First occurrence of n in this sequence see A146343.
Records see A146344.
Indices where records occurred see A146345.
a(n) =0 for n = k^2 (A000290).
a(n) =1 for n = 4 k^2 + 4 k + 5 (A078370). For primes see A005473.
a(n) =2 for n in A146327. For primes see A056899.
a(n) =3 for n in A146328. For primes see A146348.
a(n) =4 for n in A146329. For primes see A028871 - {2}.
a(n) =5 for n in A146330. For primes see A146350.
a(n) =6 for n in A146331. For primes see A146351.
a(n) =7 for n in A146332. For primes see A146352.
a(n) =8 for n in A146333. For primes see A146353.
a(n) =9 for n in A143577. For primes see A146354.
a(n)=10 for n in A146334. For primes see A146355.
a(n)=11 for n in A146335. For primes see A146356.
a(n)=12 for n in A146336. For primes see A146357.
a(n)=13 for n in A333640. For primes see A146358.
a(n)=14 for n in A146337. For primes see A146359.
a(n)=15 for n in A146338. For primes see A146360.
a(n)=16 for n in A146339. For primes see A146361.
a(n)=17 for n in A146340. For primes see A146362.

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

R. J. Mathar, Table of n, a(n) for n = 1..20000.

Cf. A003285, A146343, A146344, A146345.

Cf. also A000290, A078370, A146327-A146342; A005473, A056899, A146348-A146362.

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: seq(A146326(n),n=1..100) ; # R. J. Mathar, Sep 06 2009

Table[cf = ContinuedFraction[(1 + Sqrt[n])/2]; If[Head[cf[[-1]]] === List, Length[cf[[-1]]], 0], {n, 100}]
f[n_] := Length@ ContinuedFraction[(1 + Sqrt[n])/2][[-1]]; Array[f, 100] (* Robert G. Wilson v, Apr 11 2017 *)

a(n) = 0 iff n is a square (A000290). - Robert G. Wilson v, Apr 11 2017

a(39) and a(68) corrected by R. J. Mathar, Sep 06 2009

A054269 Length of period of continued fraction for sqrt(prime(n)).

Original entry on oeis.org

1, 2, 1, 4, 2, 5, 1, 6, 4, 5, 8, 1, 3, 10, 4, 5, 6, 11, 10, 8, 7, 4, 2, 5, 11, 1, 12, 6, 15, 9, 12, 6, 9, 18, 9, 20, 17, 18, 4, 5, 14, 21, 16, 13, 1, 20, 26, 4, 2, 5, 11, 12, 17, 14, 1, 12, 3, 24, 21, 13, 18, 5, 14, 16, 17, 11, 34, 19, 14, 7, 15, 4, 20, 5, 30, 8, 9, 21, 1, 21, 18, 37, 16

Offset: 1

N. J. A. Sloane, May 05 2000

The following sequences (allowing offset of first term) all appear to have the same parity: A034953, triangular numbers with prime indices; A054269, length of period of continued fraction for sqrt(p), p prime; A082749, difference between the sum of next prime(n) natural numbers and the sum of next n primes; A006254, numbers n such that 2n-1 is prime; A067076, 2n+3 is a prime. - Jeremy Gardiner, Sep 10 2004
Note that primes of the form n^2+1 (A002496) have a continued fraction whose period length is 1; odd primes of the form n^2+2 (A056899) have length 2; odd primes of the form n^2-2 (A028871) have length 4. - T. D. Noe, Nov 03 2006
For an odd prime p, the length of the period is odd if p=1 (mod 4) or even if p=3 (mod 4). - T. D. Noe, May 22 2007

T. D. Noe, Table of n, a(n) for n = 1..10000
A. I. Gliga, On continued fractions of the square root of prime numbers

Cf. A003285, A130272 (primes at which the period length sets a new record).

with(numtheory): for i from 1 to 150 do cfr := cfrac(ithprime(i)^(1/2), 'periodic','quotients'); printf(`%d,`, nops(cfr[2])) od:

Table[p=Prime[n]; Length[Last[ContinuedFraction[Sqrt[p]]]],{n,100}] (* T. D. Noe, May 22 2007 *)
Length[ContinuedFraction[Sqrt[#]][[2]]]&/@Prime[Range[100]] (* Harvey P. Dale, Sep 28 2024 *)

More terms from James Sellers, May 05 2000

A049423 Primes of the form k^2 + 3.

Original entry on oeis.org

3, 7, 19, 67, 103, 199, 487, 787, 1447, 2503, 2707, 3847, 4099, 4903, 5479, 5779, 8467, 8839, 11239, 12547, 14887, 16903, 17959, 19603, 21319, 23719, 24967, 25603, 29587, 31687, 47527, 52903, 58567, 59539, 61507, 65539, 75079, 81799, 88807

Offset: 1

Paul Jobling (paul.jobling(AT)whitecross.com)

Note that all terms after the first are congruent to 7 modulo 12.

			19 is prime and is equal to 4^2 + 3, so 19 is a term.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Near-Square Prime

Cf. A002496, A056899. Note that apart from first term, all of (a(n)-7)/12 have to be terms of A001082 for a(n) to be prime.

[n: n in PrimesUpTo(175000) | IsSquare(n-3)];  // Bruno Berselli, Apr 05 2011

[a: n in [0..300] | IsPrime(a) where a is n^2+3]; // Vincenzo Librandi, Dec 08 2011

Intersection[Table[n^2+3,{n,0,10^2}],Prime[Range[9*10^3]]] ...or... For[i=3,i<=3,a={};Do[If[PrimeQ[n^2+i],AppendTo[a,n^2+i]],{n,0,100}];Print["n^2+",i,",",a];i++ ] (* Vladimir Joseph Stephan Orlovsky, Apr 29 2008 *)
Select[Table[n^2+3,{n,0,198000}],PrimeQ] (* Vincenzo Librandi, Dec 08 2011 *)

list(lim)=my(v=List(),t); forstep(k=0,sqrtint(lim\1-3),2, if(isprime(t=k^2+3), listput(v,t))); Vec(v) \\ Charles R Greathouse IV, Nov 06 2024

Primes m such that m-3 is a square.
For n>0, a(n) = 36*A056902(n-1)^2 + 24*A056902(n-1) + 7. - Henry Bottomley, Jul 06 2000
a(n) = 3 + (2*A097697(n-1))^2. - R. J. Mathar, Aug 07 2008
a(n) >> n^2 log n. - Charles R Greathouse IV, Nov 06 2024

A056900 Numbers n where 36n^2+36n+11 is prime.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 9, 13, 16, 17, 18, 19, 20, 24, 28, 36, 37, 39, 40, 41, 42, 45, 49, 50, 51, 53, 57, 58, 60, 61, 62, 64, 69, 70, 71, 73, 74, 75, 79, 83, 85, 91, 92, 93, 95, 100, 101, 108, 112, 113, 116, 118, 125, 129, 134, 136, 139, 144

Offset: 0

Henry Bottomley, Jul 05 2000

36m^2+36m+11=(6m+3)^2+2, i.e. two more than the square of odd multiples of 3. 36m^2+36m+11=72*(m*(m+1)/2)+11, i.e. eleven more than seventy-two times triangular numbers.

			a(3)=3 because 36*3^2+36*3+11=443 which is prime

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

This sequence (with the formula above) generates all except the first two terms of the sequence of primes of the form k^2+2, A056899.

Cf. A091199.

[n: n in [0..200]| IsPrime(36*n^2+36*n+11)]; // Vincenzo Librandi, Jul 14 2012

lst={};Do[If[PrimeQ[36*n^2+36*n+11],AppendTo[lst,n]],{n,0,100}];lst (* Vincenzo Librandi, Jul 14 2012 *)
Select[Range[0,200],PrimeQ[36#^2+36#+11]&] (* Harvey P. Dale, Sep 19 2020 *)

a(n) =A002024((A056899(n+2)-11)/72)

a(n) = A091199(n+1) - 1. - Jeppe Stig Nielsen, May 14 2017

A056905 Primes of the form k^2 + 5.

Original entry on oeis.org

5, 41, 149, 1301, 2309, 5189, 6089, 9221, 13001, 15881, 26249, 28229, 39209, 41621, 60521, 66569, 86441, 112901, 116969, 138389, 171401, 186629, 207941, 213449, 242069, 254021, 266261, 285161, 304709, 331781, 345749, 352841, 389381, 443561

Offset: 1

Henry Bottomley, Jul 07 2000

Except for a(0), a(n) mod 180 = 41 or 149 since k must be a multiple of 6 without being a multiple of 30 for k^2+5 to be prime.

			a(2)=149 since 12^2 + 5 = 149, which is prime.

Vincenzo Librandi, Table of n, a(n) for n = 1..1800
Eric Weisstein's World of Mathematics, Near-Square Prime

Cf. A002496, A056899, A049423, A005473.

[a: n in [0..700] | IsPrime(a) where a is n^2+5]; // Vincenzo Librandi, Nov 30 2011

Select[Table[n^2+5,{n,0,7000}],PrimeQ] (* Vincenzo Librandi, Nov 30 2011 *)

is(n) = ispseudoprime(n) && issquare(n-5) \\ Felix Fröhlich, May 25 2018

a(n) = 36 * A056906(n) + 5.

A056909 Primes of the form k^2+6.

Original entry on oeis.org

7, 31, 127, 367, 631, 967, 1231, 3727, 4231, 6247, 7927, 8287, 11887, 17167, 21031, 22807, 30631, 34231, 39607, 48847, 72367, 108247, 109567, 126031, 160807, 185767, 198031, 231367, 235231, 261127, 265231, 279847, 290527, 323767, 354031, 366031, 373327, 421207

Offset: 1

Henry Bottomley, Jul 07 2000

a(n) mod 120 = 7 or 31 for all n.

			a(2)=127 since 11^2+6=127 which is prime.

Vincenzo Librandi, Table of n, a(n) for n = 1..2400

Cf. A002496, A056899, A049423, A005473, A056905.

[a: n in [0..700] | IsPrime(a) where a is n^2+6]; // Vincenzo Librandi, Nov 30 2011

Intersection[Table[n^2+6,{n,0,10^2}],Prime[Range[9*10^3]]] (* or *) For[i=6,i<=6,a={};Do[If[PrimeQ[n^2+i],AppendTo[a,n^2+i]],{n,0,100}];Print["n^2+",i,",",a];i++ ] (* Vladimir Joseph Stephan Orlovsky, Apr 29 2008 *)
Select[Table[n^2+6,{n,0,7000}],PrimeQ] (* Vincenzo Librandi, Nov 30 2011 *)

list(lim)=my(v=List(),t); forstep(k=1,sqrtint(lim\1-6),2, if(isprime(t=k^2+6), listput(v,t))); Vec(v) \\ Charles R Greathouse IV, Nov 06 2024

a(n) = 36*A056910(n)^2 + 12*A056910(n) + 7.

a(n) >> n^2 log n. - Charles R Greathouse IV, Nov 06 2024

cp's OEIS Frontend

A261655 Squares equal to the difference between two successive primes of the form k^2+2 in the order in which they appear in A056899.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

A089921 Duplicate of A056899.

Views

Author

Keywords

A224519 For n >= 4, a(n) = (A056899(n) - A056899(n-1))/72, where A056899 lists the primes of the form k^2 + 2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

A059100 a(n) = n^2 + 2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Sage

Formula

A146326 Length of the period of the continued fraction of (1+sqrt(n))/2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A054269 Length of period of continued fraction for sqrt(prime(n)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A049423 Primes of the form k^2 + 3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Magma

Mathematica

PARI

Formula

A056900 Numbers n where 36n^2+36n+11 is prime.

Views