A078370 -id:A078370 - cp's OEIS frontend

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

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

A146334 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 10.

Original entry on oeis.org

43, 67, 116, 129, 134, 161, 162, 184, 218, 242, 243, 246, 270, 274, 297, 301, 314, 338, 339, 345, 354, 356, 407, 411, 451, 452, 459, 465, 475, 498, 515, 517, 532, 534, 561, 563, 590, 591, 595, 597, 603, 611, 638, 648, 657, 665, 669, 671, 690, 705, 715

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

For primes in this sequence see A146355.

			a(1) = 43 because continued fraction of (1+Sqrt[43])/2 = 3, 1, 3, 1, 1, 12, 1, 1, 3, 1, 5, 1, 3, 1, 1, 12, 1, 1, 3, 1, 5, 1, 3, 1, 1, 12, 1, 1, 3, 1, ... has period (1, 3, 1, 1, 12, 1, 1, 3, 1, 5) length 10.

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: isA146334 := proc(n) RETURN(A146326(n) = 10) ; end: for n from 2 to 715 do if isA146334(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

cf10Q[n_]:=Module[{s=(1+Sqrt[n])/2,x},x=If[IntegerQ[s],1,Length[ ContinuedFraction[ s][[2]]]];x==10]; Select[Range[750],cf10Q] (* Harvey P. Dale, Sep 22 2015 *)

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

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

A088307 Triangle, read by rows, T(n,k) = n^2 + k^2 if gcd(n,k)=1, otherwise 0.

Original entry on oeis.org

2, 5, 0, 10, 13, 0, 17, 0, 25, 0, 26, 29, 34, 41, 0, 37, 0, 0, 0, 61, 0, 50, 53, 58, 65, 74, 85, 0, 65, 0, 73, 0, 89, 0, 113, 0, 82, 85, 0, 97, 106, 0, 130, 145, 0, 101, 0, 109, 0, 0, 0, 149, 0, 181, 0, 122, 125, 130, 137, 146, 157, 170, 185, 202, 221, 0

Offset: 1

Reinhard Zumkeller, Nov 05 2003

(n^2-k^2, 2*k*n, T(n,k)) is a primitive Pythagorean triple iff T(n,k) > 0.

			Triangle begins:
   2;
   5,  0;
  10, 13,  0;
  17,  0, 25,  0;
  26, 29, 34, 41,  0;
  37,  0,  0,  0, 61, 0;
  ...

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
Eric Weisstein's World of Mathematics, Pythagorean Triple

Cf. A000007, A000010, A001844, A002522, A008527, A053818.

Cf. A057427, A070216, A078370, A079273, A190816.

function A088307(n,k)
  if GCD(k,n) eq 1 then return n^2+k^2;
  else return 0;
  end if; return A088307;
end function;
[A088307(n,k): k in [1..n], n in [1..13]]; // G. C. Greubel, Dec 16 2022

Table[If[CoprimeQ[n,k],n^2+k^2,0],{n,20},{k,n}]//Flatten (* Harvey P. Dale, Jul 13 2018 *)

def A088307(n,k):
    if (gcd(n,k)==1): return n^2 + k^2
    else: return 0
flatten([[A088307(n,k) for k in range(1,n+1)] for n in range(1,14)]) # G. C. Greubel, Dec 16 2022

T(n, n) = 2*A000007(n-1).
T(n, 1) = A002522(n).
T(2*n+1, 2) = A078370(n).
Sum_{k=1..n} A057427(T(n,m)) = A000010(n).
From G. C. Greubel, Dec 15 2022: (Start)
T(n, n-1) = A001844(n).
T(n, n-2) = ((1-(-1)^n)/2) * A008527((n+1)/2).
T(2*n, n) = 5*A000007(n-1).
T(2*n+1, n) = A079273(n+1).
T(2*n-1, n) = A190816(n).
Sum_{k=1..floor((n+1)/2)} T(n-k+1, k) = A053818(n+1) + [n=1]. (End)

A146364 a(n) = smallest primes whose continued fraction have different period.

Original entry on oeis.org

2, 5, 7, 17, 19, 31, 41, 43, 73, 89, 103, 139, 151, 179, 191, 193, 211, 241, 271, 331, 337, 379, 409, 421, 433, 463, 487, 491, 521, 541, 571, 601, 619, 631, 673, 739, 751, 769, 823, 919, 929, 937, 1033, 1039, 1051, 1201, 1249, 1291, 1321, 1399, 1439, 1471, 1531, 1579, 1609, 1699, 1747, 1753, 1759

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

This sequence is sorted A146363.

Robert Israel, Table of n, a(n) for n = 1..1500

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

g:= proc(n) local c;
      c:= NumberTheory:-ContinuedFraction((1+sqrt(n))/2);
      nops(Term(c,periodic)[2]);
end proc:
R:= NULL: S:= {}: count:= 0:
p:= 1:
while count < 100 do
  p:= nextprime(p);
  v:= g(p);
  if not member(v,S) then
    R:= R,p; count:= count+1; S:= S union {v};
    if count mod 20 = 0 then printf("%d %d\n",count,p) fi
  fi
od:
R; # Robert Israel, Jun 14 2024

$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}]; Sort[bb] (*Artur Jasinski*)

More terms from Robert Israel, Jun 14 2024

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

A146478 a(n) = length of period of continued fraction (1 + sqrt(prime(n)))/2.

Original entry on oeis.org

2, 2, 1, 4, 2, 1, 3, 6, 4, 1, 8, 3, 5, 10, 4, 1, 6, 3, 10, 8, 9, 4, 2, 7, 9, 3, 12, 6, 7, 7, 12, 6, 7, 18, 5, 20, 5, 18, 4, 1, 14, 5, 16, 15, 3, 20, 26, 4, 2, 1, 9, 12, 19, 14, 3, 12, 5, 24, 9, 15, 18, 1, 14, 16, 19, 3, 34, 21, 14, 9, 9, 4, 20, 7, 30, 8, 7, 5, 3, 27, 18, 13, 16, 23, 4, 2, 19, 23, 3

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

Subsequence of A146326 (length of period continued fraction of (1 + sqrt(n))/2).

Zak Seidov, Table of n, a(n) for n = 1..1000

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

A146478 := proc(p) local c; c := numtheory[cfrac](1/2+sqrt(p)/2,'periodic','quotients') ; nops(c[2]) ; end: for n from 1 to 100 do printf("%d,",A146478(ithprime(n))) ; od: # R. J. Mathar, Nov 05 2008

Table[Length[ContinuedFraction[(1+Sqrt[Prime[n]])/2][[2]]],{n,100}] (* Zak Seidov, Mar 22 2011 *)

a(59) changed from 7 to 9 by R. J. Mathar, Nov 05 2008

A308464 Squarefree numbers of the form m^2 + 4.

Original entry on oeis.org

5, 13, 29, 53, 85, 173, 229, 293, 365, 445, 533, 629, 733, 965, 1093, 1229, 1373, 1685, 1853, 2029, 2213, 2405, 2605, 2813, 3029, 3253, 3485, 3973, 4229, 4493, 4765, 5045, 5333, 5629, 5933, 6245, 6565, 6893, 7229, 7573, 8285, 8653, 9029, 9413, 9805

Offset: 1

Alonso del Arte, May 29 2019

Yokoi's conjecture posits that, except for most of the values less than 365, the ring of algebraic integers of Q(sqrt(a(n))) has class number greater than 1. Only one counterexample to this conjecture may exist, and it would also be a counterexample to the generalized Riemann hypothesis, according to Mollin (1996).

All terms == 5 (mod 8). - Robert Israel, Jun 05 2019

Richard A. Mollin, Quadratics. Boca Raton, Florida: CRC Press (1996): 176 - 177.

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

Cf. A078370, A087475 (supersequences).

select(numtheory:-issqrfree,[seq(m^2+4,m=1..1000,2)]); # Robert Israel, Jun 05 2019

Select[(2Range[50] - 1)^2 + 4, MoebiusMu[#] != 0 &]
Select[Table[i^2 + 4, {i, 1, 100}], SquareFreeQ] (* Navvye Anand, Jun 20 2024 *)

is(n) = issquarefree(n) && issquare(n-4) \\ Felix Fröhlich, May 29 2019

A319749 a(n) is the numerator of the Heron sequence with h(0)=3.

Original entry on oeis.org

3, 11, 119, 14159, 200477279, 40191139395243839, 1615327685887921300502934267457919, 2609283532796026943395592527806764363779539144932833602430435810559

Offset: 0

Paul Weisenhorn, Sep 27 2018

The denominator of the Heron sequence is in A319750.
The following relationship holds between the numerator of the Heron sequence and the numerator of the continued fraction A041018(n)/A041019(n) convergent to sqrt(13).
n even: a(n)=A041018((5*2^n-5)/3).
n  odd: a(n)=A041018((5*2^n-1)/3).
More generally, all numbers c(n)=A078370(n)=(2n+1)^2+4 have the same relationship between the numerator of the Heron sequence and the numerator of the continued fraction convergent to 2n+1.
sqrt(c(n)) has the continued fraction 2n+1; n,1,1,n,4n+2.
hn(n)^2-c(n)*hd(n)^2=4 for n>1.
From Peter Bala, Mar 29 2022: (Start)
Applying Heron's method (sometimes called the Babylonian method) to approximate the square root of the function x^2 + 4, starting with a guess equal to x, produces the sequence of rational functions [x, 2*T(1,(x^2+2)/2)/x, 2*T(2,(x^2+2)/2)/( 2*x*T(1,(x^2+2)/2) ), 2*T(4,(x^2+2)/2)/( 4*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2) ), 2*T(8,(x^2+2)/2)/( 8*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2)*T(4,(x^2+2)/2) ), ...], where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. The present sequence is the case x = 3. Cf. A001566 and A058635 (case x = 1), A081459 and A081460 (essentially the case x = 4). (End)

			A078370(2)=29.
hn(0)=A041046(0)=5; hn(1)=A041046(3)=27; hn(2)=A041046(5)=727;
hn(3)=A041046(13)=528527.

P. Liardet and P. Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Wikipedia, Engel Expansion
Wikipedia, Methods of computing square roots
Index entries for sequences of form a(n+1)=a(n)^2 + ...

2*T(2^n,x/2) modulo differences of offset: A001566 (x = 3 and x = 7), A003010 (x = 4), A003487 (x = 5), A003423 (x = 6), A346625 (x = 8), A135927 (x = 10), A228933 (x = 18).

Cf. A041018, A041019, A057076, A078370, A081459, A010122, A319750, A041046.

hn[0]:=3:  hd[0]:=1:
for n from 1 to 6 do
hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
hd[n]:=hn[n-1]*hd[n-1]:
   printf("%5d%40d%40d\n", n, hn[n], hd[n]):
end do:
#alternative program
a := n -> if n = 0 then 3 else simplify( 2*ChebyshevT(2^(n-1), 11/2) ) end if:
seq(a(n), n = 0..7); # Peter Bala, Mar 16 2022

def aupton(nn):
    hn, hd, alst = 3, 1, [3]
    for n in range(nn):
        hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
        alst.append(hn)
    return alst
print(aupton(7)) # Michael S. Branicky, Mar 16 2022

h(n) = hn(n)/hd(n); hn(0)=3; hd(0)=1.
hn(n+1) = (hn(n)^2+13*hd(n)^2)/2.
hd(n+1) = hn(n)*hd(n).
A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
From Peter Bala, Mar 16 2022: (Start)
a(n) = 2*T(2^(n-1),11/2) for n >= 1, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = 2*T(2^n, 3*sqrt(-1)/2) for n >= 2.
a(n) = ((11 + 3*sqrt(13))/2)^(2^(n-1)) + ((11 - 3*sqrt(13))/2)^(2^(n-1)) for n >= 1.
a(n+1) = a(n)^2 - 2 for n >= 1.
a(n) = A057076(2^(n-1)) for n >= 1.
Engel expansion of (1/6)*(13 - 3*sqrt(13)); that is, (1/6)*(13 - 3*sqrt(13)) = 1/3 + 1/(3*11) + 1/(3*11*119) + .... (Define L(n) = (1/2)*(n - sqrt(n^2 - 4)) for n >= 2 and show L(n) = 1/n + L(n^2-2)/n. Iterate this relation with n = 11. See also Liardet and Stambul, Section 4.)
sqrt(13) = 6*Product_{n >= 0} (1 - 1/a(n)).
sqrt(13) = (9/5)*Product_{n >= 0} (1 + 2/a(n)). See A001566. (End)

a(6) and a(7) added by Peter Bala, Mar 16 2022

A319750 a(n) is the denominator of the Heron sequence with h(0) = 3.

Original entry on oeis.org

1, 3, 33, 3927, 55602393, 11147016454528647, 448011292165037607943004375755833, 723685043824607606355691108666081531638582859833105061571146291527

Offset: 0

Paul Weisenhorn, Sep 27 2018

The numerators of the Heron sequence are in A319749.
There is the following relationship between the denominator of the Heron sequence and the denominator of the continued fraction A041018(n)/ A041019(n) convergent to sqrt(13).
n even: a(n) = A041019((5*2^n-5)/3).
n  odd: a(n) = A041019((5*2^n-1)/3).
General: all numbers c(n) = A078370(n) = (2*n+1)^2 + 4 have the same relationship between the denominator of the Heron sequence and the denominator of the continued fraction convergent to 2*n+1.
sqrt(c(n)) has the continued fraction [2*n+1; n, 1, 1, n, 4*n+2].
hn(n)^2 - c(n)*hd(n)^2 = 4 for n > 1.

			A078370(2) = 29.
hd(0) = A041047(0) = 1, hd(1) = A041047(3) = 5,
hd(2) = A041047(5) = 135, hd(3) = A041047(13) = 38145.

Cf. A041018, A041019, A078370, A010122, A041047, A319749.

hn[0]:=3: hd[0]:=1:
for n from 1 to 6 do
  hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
  hd[n]:=hn[n-1]*hd[n-1]:
  printf("%5d%40d%40d\n", n, hn[n], hd[n]):
end do:

def aupton(nn):
    hn, hd, alst = 3, 1, [1]
    for n in range(nn):
        hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
        alst.append(hd)
    return alst
print(aupton(7)) # Michael S. Branicky, Mar 15 2022

h(n) = hn(n)/hd(n), hn(0) = 3, hd(0) = 1.
hn(n+1) = (hn(n)^2 + 13*hd(n)^2)/2.
hd(n+1) = hn(n)*hd(n).
A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
a(0) = 1, a(1) = 3 and a(n) = 2*T(2^(n-2), 11/2)*a(n-1) for n >= 2, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Mar 16 2022

a(5) corrected and terms a(6) and a(7) added by Peter Bala, Mar 15 2022

cp's OEIS Frontend

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

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A146334 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 10.

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A146344 Records in A146326.

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A088307 Triangle, read by rows, T(n,k) = n^2 + k^2 if gcd(n,k)=1, otherwise 0.

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A146364 a(n) = smallest primes whose continued fraction have different period.

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

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A146478 a(n) = length of period of continued fraction (1 + sqrt(prime(n)))/2.

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