A146326 -id:A146326 - cp's OEIS frontend

A146345 Indices in A146326 where records occur.

1, 2, 6, 18, 31, 43, 94, 106, 151, 211, 331, 394, 526, 694, 751, 886, 919, 1114, 1324, 1726, 1759, 1831, 2011, 2311, 2326, 2671, 3019, 3691, 3754, 3931, 4174, 4951, 4999, 5119, 6211, 6406, 7606, 8254, 8719, 8779, 9244, 9619, 9739, 10399, 10651, 12919, 13126

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

Robert G. Wilson v and Amiram Eldar, Table of n, a(n) for n = 1..416 (terms 1..218 from Robert G. Wilson v)

Cf. A000290, A078370, A146326-A146345, A146348-A146360, A146363.

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: read("transforms") ; a26 := [seq(A146326(n),n=1..1400)] ; RECORDS(a26)[2] ; # R. J. Mathar, Sep 06 2009

f[n_] := Length@ContinuedFraction[(1 + Sqrt[n])/2][[-1]]; mx = -1; k = 1; lst = {}; While[k < 14000, a = f@k; If[a > mx, mx = a; AppendTo[lst, k]]; k++]; lst (* Robert G. Wilson v, Apr 11 2017 *)

19 replaced by 18, 394 inserted, 4 more terms added by R. J. Mathar, Sep 06 2009

More terms from Robert G. Wilson v, Apr 11 2017

A146344 Records in A146326.

Original entry on oeis.org

0, 2, 4, 6, 8, 10, 12, 18, 20, 26, 34, 42, 48, 50, 52, 54, 60, 66, 72, 76, 80, 84, 94, 96, 102, 104, 114, 122, 126, 130, 140, 148, 152, 156, 158, 178, 190, 192, 196, 202, 204, 206, 210, 228, 234, 248, 258, 268, 276, 294, 322, 332, 348, 352, 374, 376, 380, 398

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

Amiram Eldar, Table of n, a(n) for n = 1..416

Cf. A000290, A078370, A146326-A146345, A146348-A146360, A146363.

146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: read("transforms") ; a26 := [seq(A146326(n),n=1..1000)] ; RECORDS(a26)[1] ; # R. J. Mathar, Sep 06 2009

f[n_] := Length @ ContinuedFraction[(1 + Sqrt[n])/2][[-1]]; fmax = -1; seq = {}; Do[f1 = f[n]; If[f1 > fmax, fmax = f1; AppendTo[seq, f1]], {n, 1, 10^4}]; seq (* Amiram Eldar, Apr 02 2020 *)

a(n) = A146326(A146345(n)). - Amiram Eldar, Apr 02 2020

42 inserted by R. J. Mathar, Sep 06 2009

a(1) inserted and more terms added by Amiram Eldar, Apr 02 2020

A146477 Numbers k for which A146326(k) is different from A146326(j) for j < k.

Original entry on oeis.org

2, 5, 6, 17, 18, 31, 41, 43, 73, 89, 94, 106, 118, 151, 172, 193, 211, 241, 265, 268, 331, 334, 337, 379, 394, 409, 421, 433, 463, 489, 521, 526, 601, 604, 619, 634, 673, 694, 718, 721, 751, 769, 886, 919, 929, 937, 1033, 1039, 1114, 1174, 1201, 1249, 1291, 1321, 1324, 1471, 1516, 1579, 1609

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

This sequence is sorted A146343.

Original name was: a(n) = smallest numbers which continued fractions have different period.

Robert Israel, Table of n, a(n) for n = 1..700

Cf. A000290, A078370, A146326-A146345, A146348-A146360, A146363.

f:= proc(n) if issqr(n) then 0 else nops(numtheory:-cfrac((1+sqrt(n))/2,periodic,quotients)[2]) fi end proc:
S:= {0}: R:= NULL: count:= 0:
for n from 2 while count < 30 do
  v:= f(n);
  if not member(v,S) then
     count:= count+1; R:= R, n; S:= S union {v};
  fi
od:
R; # Robert Israel, May 02 2021

$MaxExtraPrecision = 300; s = 10; aa = {}; Do[k = ContinuedFraction[(1 + Sqrt[n])/2, 1000]; If[Length[k] < 190, AppendTo[aa, 0], m = 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; s = s + 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; AppendTo[aa, m]], {n, 1, 1200}]; Print[aa]; bb = {}; Do[k = 1; yes = 0; Do[If[aa[[k]] == n && yes == 0, AppendTo[bb, k]; yes = 1], {k, 1, Length[aa]}], {n, 1, 22}]; Sort[bb]

19 replaced by 18, 331 and 334 inserted by R. J. Mathar, Nov 08 2008

Name clarified by Robert Israel, May 02 2021

A146348 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 3.

Original entry on oeis.org

17, 37, 61, 101, 197, 257, 317, 401, 461, 557, 577, 677, 773, 1129, 1297, 1429, 1601, 1877, 1901, 2917, 3137, 4357, 4597, 5417, 5477, 6053, 7057, 8101, 8761, 8837, 10733, 11621, 12101, 13457, 13877, 14401, 15277, 15377, 15877, 16333, 16901, 17737, 17957, 18329, 21317, 22501, 23593, 24337, 25601, 28901, 30137, 30977, 32401, 33857, 41453, 41617, 42437, 44101

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

Primes in A146328. Finite A050952 is subset of this sequence.
From Michel Lagneau, Sep 03 2014: (Start)
The primes of the form p = n^2+1 for n>2 are in the sequence, and the continued fraction of (1+sqrt(p))/2 is [n/2; 1, 1, n-1, 1, 1, n-1, 1, 1, ...] with the period (1, 1, n-1).
We observe that the other primes {61, 317, 461, 557, 773, 1129, 1429, ...} are prime divisors of composite numbers of the form k^2+1 where k = 11, 114, 48, 118, 317, 168, 620, ... .
(End)
Another possibly infinite subset of the sequence is primes of the form 100*k^2-44*k+5, where the continued fraction is [5*k-1; 2, 2, 10*k-3, ...] with period [2, 2, 10*k-3].  This includes {61, 317, 773, 1429, 4597, 6053, ...}. - Robert Israel, Sep 03 2014

Amiram Eldar, Table of n, a(n) for n = 1..2500 (terms 1..200 from Zak Seidov)

Cf. A000290, A050952, A078370, A146326-A146345, A146348-A146360, A146363.

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146348 := proc(n) RETURN(isprime(n) and A146326(n) = 3) ; end: for n from 2 to 4000 do if isA146348(n) then printf("%d,\n",n) ; fi; od: # R. J. Mathar, Sep 06 2009

okQ[n_] := Length[ContinuedFraction[(1 + Sqrt[n])/2][[2]]] == 3; Select[Prime[Range[100]], okQ]

1019 removed; more terms added by R. J. Mathar, Sep 06 2009

More terms from Zak Seidov, Mar 09 2011

A146360 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 15: primes in A146338.

Original entry on oeis.org

193, 281, 1861, 1933, 2089, 2141, 2437, 2741, 2837, 3037, 3121, 3413, 4001, 4637, 4877, 5821, 6653, 7673, 8117, 10069, 10273, 10457, 11197, 11549, 11821, 12409, 13037, 14653, 15061, 15077, 18661, 20549, 22921, 23117, 24169, 25621, 28837, 35597, 35869, 36389, 38569

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

Amiram Eldar, Table of n, a(n) for n = 1..2000

Cf. A000290, A050950-A050969, A078370, A146326-A146345, A146348-A146360.

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146360 := proc(n) RETURN(isprime(n) and A146326(n) = 15) ; end: for n from 2 to 30000 do if isA146360(n) then printf("%d,\n",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Select[Prime[Range[1500]],Length[ContinuedFraction[(Sqrt[#]+1)/2][[2]]] == 15&] (* Harvey P. Dale, Aug 16 2014 *)

8539 removed by R. J. Mathar, Sep 06 2009

More terms from Amiram Eldar, Mar 30 2020

A005473 Primes of form k^2 + 4.

Original entry on oeis.org

5, 13, 29, 53, 173, 229, 293, 733, 1093, 1229, 1373, 2029, 2213, 3253, 4229, 4493, 5333, 7229, 7573, 9029, 9413, 10613, 13229, 13693, 15629, 18229, 18773, 21613, 24029, 26573, 27893, 31333, 33493, 37253, 41213, 42853, 46229, 47093, 54293

Offset: 1

N. J. A. Sloane

a(n) mod 24 = 5 or 13 and if a(n) mod 24 =13 then a(n) mod 72 = 13.
From Artur Jasinski, Oct 30 2008: (Start)
Primes p such that the continued fraction of (1+sqrt(p))/2 has period 1.
Primes in A078370 = primes of the form 4*k^2 + 4*k + 5 = (2*k+1)^2 + 4.
(End)
Starting at a(3) all the primes in this sequence can be expressed as the following sum: ((2*k+1)*(2*k+3)+(2*k+3)*(2*k+5)+(2*k+5)+(2*k+7)+(2*k+7)*(2*k+9))/4 for some values (not all!) of k>=0. Thus for a(5)=173 the sum is (9*11 + 11*13 + 13*15 + 15*17)/4=173. - J. M. Bergot, Nov 03 2014

			a(2)=29 since 29=5^2+4 is prime.

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..4600
D. Shanks, The simplest cubic fields, Math. Comp., 28 (1974), 1137-1152.
Eric Weisstein's World of Mathematics, Near-Square Prime

Subsequence of A185086.
a(n)-4 is contained in A016754. (a(n)-5)/8 is contained in A000217. Either (a(n)-5)/24 is contained in A001318 (if a(n) mod 24=5) or (a(n)-13)/72 is contained in A000217 (if a(n) mod 24=13). Floor[a(n)/24] is contained in A001840.
Cf. A146326, A010051, A000290, A138353, A098062.

a005473 n = a005473_list !! (n-1)
a005473_list = filter ((== 1) . a010051') $ map (+ 4) a000290_list
-- Reinhard Zumkeller, Mar 12 2012

[a: n in [0..300] | IsPrime(a) where a is n^2+4]; // Vincenzo Librandi, Nov 30 2011

select(isprime,[seq(4*k^2 + 4*k + 5, k=0..1000)]); # Robert Israel, Nov 02 2014

Intersection[Table[n^2+4,{n,0,10^2}],Prime[Range[9*10^3]]] ...or... For[i=4,i<=4,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 *)
aa = {}; Do[If[PrimeQ[4 k^2 + 4 k + 5], AppendTo[aa, 4 k^2 + 4 k + 5]], {k, 0, 200}]; aa (* Artur Jasinski, Oct 30 2008 *)
Select[Table[n^2+4,{n,0,7000}],PrimeQ] (* Vincenzo Librandi, Nov 30 2011 *)

for(n=1,1e3,if(isprime(t=n^2+4),print1(t","))) \\ Charles R Greathouse IV, Jul 05 2011

a(n) = 24*A056904(n)+m, where m=13 if A056904(n) is three times a triangular number (and n>0) and m=5 if A056904(n) is not three times a triangular number (or n=0).

For n>=2, a(n) = A098062(n-1). - Zak Seidov, Apr 12 2007

More terms and additional comments from Henry Bottomley, Jul 06 2000

A146363 a(n) = smallest prime p such that continued fraction of (1 + sqrt(p))/2 has period length n.

Original entry on oeis.org

5, 2, 17, 7, 41, 19, 89, 31, 73, 43, 541, 103, 421, 179, 193, 191, 521, 139, 241, 151, 337, 491, 433, 271, 929, 211, 409, 487, 673, 379, 937, 463, 601, 331, 769, 1439, 2297, 619, 1033, 1399, 1777, 571, 1753, 823, 1993, 739, 1249, 631, 4337, 1051, 1321, 751, 1201

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

Artur Jasinski, Table of n, a(n) for n=1..1000

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

A := proc(n) option remember ; local c; try c := numtheory[cfrac](1/2+sqrt(n)/2,'periodic,quotients') ; RETURN(nops(c[2]) ); catch: RETURN(-1) end try ; end: A146363 := proc(n) local p,i ; for i from 1 do p := ithprime(i) ; if A(p) = n then RETURN(p) ; fi; od; end: for n from 1 do printf("%d, ",A146363(n)) ; od: # R. J. Mathar, Nov 08 2008

$MaxExtraPrecision = 300; s = 10; aa = {}; Do[k = ContinuedFraction[(1 + Sqrt[n])/2, 1000]; If[Length[k] < 190, AppendTo[aa, 0], m = 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; s = s + 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; AppendTo[aa, m]], {n, 1, 1200}]; Print[aa]; bb = {}; Do[k = 1; yes = 0&&PeimeQ[k]; Do[If[aa[[k]] == n && yes == 0, AppendTo[bb, k]; yes = 1], {k, 1, Length[aa]}], {n, 1, 22}]; bb (* Artur Jasinski *)
aa = {}; Do[n = 1; While[m != Length[ContinuedFraction[(1 + Sqrt[Prime[n]])/2][[2]]], n++ ]; AppendTo[aa, Prime[n]], {m, 1, 100}]; aa (* Artur Jasinski, Feb 03 2010 *)

a(25) replaced by 929 and extended by R. J. Mathar, Nov 08 2008

A146335 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 11.

Original entry on oeis.org

265, 541, 593, 661, 701, 857, 1061, 1109, 1217, 1237, 1709, 1733, 1949, 2333, 2509, 2557, 2957, 3125, 3229, 3677, 3701, 4181, 4373, 4685, 5081, 5237, 5309, 6133, 6425, 6445, 7013, 7025, 8185, 8545, 8693, 9305, 9533, 9553, 10333, 10525, 10853, 10961, 11125, 11141

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

For primes in this sequence see A146356.

			a(4) = 661 because continued fraction of (1+sqrt(661))/2 = 13, 2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25, 2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25, 2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25, 2, 1, 4, 2, 8, 8 ... has period (2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25) length 11.

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

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

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146335 := proc(n) RETURN(A146326(n) = 11) ; end: for n from 2 to 2000 do if isA146335(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 11 &]  (* Amiram Eldar, Mar 31 2020 *)

916 removed by R. J. Mathar, Sep 06 2009

More terms from Amiram Eldar, Mar 31 2020

A146362 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 17 : primes in A146340.

Original entry on oeis.org

521, 617, 709, 1433, 1597, 2549, 2909, 3581, 3821, 4013, 4649, 5501, 5693, 5813, 6197, 7853, 8093, 8573, 9281, 9677, 10597, 10973, 11273, 13109, 13613, 15413, 15641, 15737, 16001, 16477, 17093, 20261, 22637, 24697, 26717, 32413, 35537, 38177, 43717, 46649, 47681

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

Amiram Eldar, Table of n, a(n) for n = 1..2000

Cf. A000290, A050950-A050969, A078370, A146326-A146345, A146348-A146360.

Select[Range[2*10^4], PrimeQ[#] && Length[ContinuedFraction[(1+Sqrt[#])/2][[2]]] == 17 &] (* Amiram Eldar, Mar 30 2020 *)

Period length in definition corrected, 2579, 5003 removed, 5813 inserted by R. J. Mathar, Sep 06 2009

More terms from Amiram Eldar, Mar 30 2020

A143577 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 9.

Original entry on oeis.org

73, 97, 233, 277, 349, 353, 613, 821, 877, 1073, 1181, 1189, 1277, 1285, 1313, 1385, 1613, 1637, 1693, 1745, 1865, 2357, 2581, 2777, 3233, 3557, 3989, 4157, 4469, 4517, 4553, 4709, 4889, 4925, 4933, 5245, 5261, 5305, 5597, 6113, 6205, 6253, 7213, 7585, 7837, 8885

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

For primes in this sequence see A146354.

Superset of A146354. - R. J. Mathar, Nov 05 2008

			a(1) = 73 because continued fraction of (1+sqrt(73))/2 = 4, 1, 3, 2, 1, 1, 2, 3, 1, 7, 1, 3, 2, 1, 1, 2, 3, 1, 7, 1, 3, 2, 1, 1, 2, 3, 1, 7, 1, 3, ... has period (1, 3, 2, 1, 1, 2, 3, 1, 7) length 9 .

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

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

isA143577 := proc(k) local c; try c := numtheory[cfrac](1/2+sqrt(k)/2,'periodic','quotients') ; if nops(c[2]) = 9 then RETURN(true) ; else RETURN(false) ; fi; catch: RETURN(false) ; end try; end: for k from 2 to 80000 do if isA143577(k) then printf("%d, ",k) ; fi; od: # R. J. Mathar, Nov 05 2008

Select[Range[1000], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 9 &]  (* Amiram Eldar, Mar 19 2020 *)

Extended by R. J. Mathar, Nov 05 2008

More terms from Amiram Eldar, Mar 19 2020

cp's OEIS Frontend

A146345 Indices in A146326 where records occur.

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A146344 Records in A146326.

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A146477 Numbers k for which A146326(k) is different from A146326(j) for j < k.

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A146348 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 3.

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A146360 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 15: primes in A146338.

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A005473 Primes of form k^2 + 4.

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A146363 a(n) = smallest prime p such that continued fraction of (1 + sqrt(p))/2 has period length n.

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