A002350 -id:A002350 - cp's OEIS frontend

A081231 Let p = n-th prime of the form 4k+3, take smallest solution (x,y) to the Pellian equation x^2 - p*y^2 = 1 with x and y >= 1; sequence gives value of x.

2, 8, 10, 170, 24, 1520, 3482, 48, 530, 48842, 3480, 80, 82, 227528, 962, 4730624, 10610, 77563250, 1728148040, 64080026, 168, 4190210, 8994000, 16266196520, 278354373650, 224, 226, 6195120, 3674890, 139128, 115974983600, 138274082

Offset: 1

N. J. A. Sloane, Apr 18 2003

			For n=3, p = 11, x=10, y=3 since we have 10^2 = 11*3^2 + 1, so a(3) = 10.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Values of y are in A082394. Equals A002350(p). Cf. A082393.

PellSolve[(m_Integer)?Positive] := Module[{cf, n, s}, cf = ContinuedFraction[ Sqrt[m]]; n = Length[ Last[cf]]; If[ OddQ[n], n = 2*n]; s = FromContinuedFraction[ ContinuedFraction[ Sqrt[m], n]]; {Numerator[s], Denominator[s]}]; Transpose[ PellSolve /@ Select[ Prime[ Range[72]], Mod[ #, 4] == 3 &]][[1]] (* Robert G. Wilson v, Sep 02 2004 *)

More terms from Robert G. Wilson v, Sep 02 2004

A081232 Let p = n-th prime of the form 4k+1, take smallest solution (x,y) to the Pellian equation x^2 - p*y^2 = 1 with x and y >= 1; sequence gives value of x.

Original entry on oeis.org

9, 649, 33, 9801, 73, 2049, 66249, 1766319049, 2281249, 500001, 62809633, 201, 158070671986249, 1204353, 6083073, 25801741449, 46698728731849, 2499849, 2469645423824185801, 6224323426849, 393, 5848201, 1072400673

Offset: 1

N. J. A. Sloane, Apr 18 2003

			For n = 1, p = 5, x=9, y=4 since 9^2 = 5*4^2 + 1, so a(1) = 9.

Vincenzo Librandi, Table of n, a(n) for n = 1..2000
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)

Values of y are in A082393. Cf. A082394, A081233. Equals A002350(p).

PellSolve[(m_Integer)?Positive] := Module[{cof, n, s}, cof = ContinuedFraction[Sqrt[m]]; n = Length[Last[cof]]; If[ OddQ[n], n = 2*n]; s = FromContinuedFraction[ContinuedFraction[Sqrt[m], n]]; {Numerator[s], Denominator[s]}]; First /@ PellSolve /@ Select[Prime@Range@54, Mod[ #, 4] == 1 &] (* Robert G. Wilson v *)

More terms from Robert G. Wilson v, Feb 28 2006

A298211 Smallest n such that A001353(a(n)) == 0 (mod n), i.e., x=A001075(a(n)) and y=A001353(a(n)) is the fundamental solution of the Pell equation x^2 - 3(ny)^2 = 1.

Original entry on oeis.org

1, 2, 3, 2, 3, 6, 4, 4, 9, 6, 5, 6, 6, 4, 3, 8, 9, 18, 5, 6, 12, 10, 11, 12, 15, 6, 27, 4, 15, 6, 16, 16, 15, 18, 12, 18, 18, 10, 6, 12, 7, 12, 11, 10, 9, 22, 23, 24, 28, 30, 9, 6, 9, 54, 15, 4, 15, 30, 29, 6, 30, 16, 36, 32, 6, 30, 17, 18, 33, 12, 7, 36, 18, 18, 15

Offset: 1

A.H.M. Smeets, Jan 15 2018

nonn

The fundamental solution of the Pell equation x^2 - 3*(n*y)^2 = 1 is the smallest solution of x^2 - 3*y^2 = 1 satisfying y == 0 (mod n).

For primes p > 2, 2^p-1 is a Mersenne prime if and only if a(2^p-1) = 2^(p-1). For example, a(7) = 4, a(31) = 16, a(127) = 64, but a(2047) = 495 < 1024. - Jianing Song, Jun 02 2022

Michael J. Jacobson, Jr. and Hugh C. Williams, Solving the Pell Equation, Springer, 2009, pages 1-17.

A.H.M. Smeets, Table of n, a(n) for n = 1..20000
H. W. Lenstra Jr., Solving the Pell Equation, Notices of the AMS, Vol.49, No.2, Feb. 2002, p.182-192.

Cf. A091338, A298210, A298212.

With[{s = Array[ChebyshevU[-1 + #, 2] &, 75]}, Table[FirstPosition[s, k_ /; Divisible[k, n]][[1]], {n, Length@ s}]] (* Michael De Vlieger, Jan 15 2018, after Eric W. Weisstein at A001353 *)

xf, yf = 2, 1
x, n = 2*xf, 0
while n < 20000:
    n = n+1
    y1, y0, i = 0, yf, 1
    while y0%n != 0:
        y1, y0, i = y0, x*y0-y1, i+1
    print(n, i)

a(n) <= n.
a(A038754(n)) = A038754(n).
A001075(a(n)) = A002350(3*n^2).
A001353(a(n)) = A002349(3*n^2).
if n | m then a(n) | a(m).
a(3^m) = 3^m  and a(2*3^m) = 2*3^m for m>=0.
In general: if p is prime and p == 3 (mod 4) then: a(n) = n iff n = p^m or n = 2*p^m, for m>=0.
a(k*A005385(n)) = a(k)*A005384(n) for n>2 and k > 0 (conjectured).
a(p) | (p-A091338(p)) for p is an odd prime. - A.H.M. Smeets, Aug 02 2018
From Jianing Song, Jun 02 2022: (Start)
a(p) | (p-A091338(p))/2 for p is an odd prime > 3.
a(p^e) = a(p)*p^(e-r) for e >= r, where r is the largest number such that a(p^r) = a(p). r can be greater than 1, for p = 2, 103, 2297860813 (Cf. A238490).
If gcd(m,n) = 1, then a(m*n) = lcm(a(m),a(n)). (End)

A379815 a(n) is the smallest integer k > n such that sqrt(1/n + 1/k) is a rational number; or 0 if no such k exists.

Original entry on oeis.org

0, 16, 9, 0, 20, 12, 441, 64, 16, 90, 1089, 36, 4212, 98, 225, 0, 272, 144, 549081, 25, 567, 2156, 13225, 48, 144, 650, 81, 98, 142100, 150, 71622369, 256, 363, 578, 1225, 64, 1332, 684, 468, 360, 41984, 252, 521345889, 198, 180, 559682, 108241, 144, 63, 400, 127449, 117, 1755572, 108, 2420, 392, 4275, 568458

Offset: 1

Felix Huber, Feb 07 2025

nonn

a(1) = a(4) = a(16) = 0. Proof: See Huber link.

k > n exists for n > 16.

			a(3) = 9 because sqrt(1/3 + 1/4) = sqrt(7/12) is irrational, sqrt(1/3 + 1/5) = sqrt(8/15) is irrational, sqrt(1/3 + 1/6) = sqrt(1/2) is irrational, sqrt(1/3 + 1/7) = sqrt(10/21) is irrational, sqrt(1/3 + 1/8) = sqrt(11/24) is irrational, but sqrt(1/3 + 1/9) = sqrt(4/9) = 2/3 is rational.

Felix Huber, Proof that a(1) = a(4) = a(16) = 0

Cf. A002350, A024352, A076600, A378501, A379816.

Cf. A357372, A257522.

A379815:=proc(n)
    local k;
    if n=1 or n=4 or n=16 then
        return 0
    else
        for k from n+1 do
            if type(sqrt(1/n+1/k),rational) then
                return k
            fi
        od
    fi;
end proc;
seq(A379815(n),n=1..58);

a(n) = if ((n==1) || (n==4) || (n==16), return(0)); my(k=n+1); while (!issquare(1/n + 1/k), k++); k; \\ Michel Marcus, Feb 08 2025

a(n) <= n*A002350(n)^2 - n if n is not a square; a(m^2) <= A076600(m)^2. - Jinyuan Wang, Feb 11 2025

A379816 a(n) is the smallest integer k > n such that sqrt(1/n - 1/k) is a rational number; or 0 if no such k exists.

Original entry on oeis.org

0, 4, 12, 0, 25, 18, 448, 16, 12, 100, 1100, 18, 4225, 112, 240, 18, 289, 36, 549100, 25, 588, 2178, 13248, 72, 45, 676, 108, 126, 142129, 180, 71622400, 64, 396, 612, 1260, 48, 1369, 722, 507, 400, 42025, 294, 521345932, 242, 225, 559728, 108288, 72, 112, 100, 127500, 169, 1755625, 162, 2475, 448, 4332, 568516, 16573100, 150

Offset: 1

Felix Huber, Feb 07 2025

nonn

a(1) = a(4) = 0. Proof: See Huber link.
k > n exists for n > 4.
Continues a(61) <= 53872731025, a(62) = 1984, a(63) = 112, a(64) = 72, a(65) = 4225, a(66) = 2178, a(67) <= 159831244588, a(68) = 1156, a(69) = 268272, a(70) = 8820, a(71) <= 859838400, a(72) = 144, a(73) <= 83265625, a(74) = 136900, a(75) = 300, a(76) = 6498, a(77) = 13552, a(78) = 2106, a(79) = 505600, a(80) = 100, a(81) = 108, a(82) = 6724, a(83) = 558092, a(84) = 147, a(85) = 12145225, a(86) = 447458, a(87) = 68208, a(88) = 8712, a(89) = 22250089, a(90) = 100, a(91) = 225450316, a(92)=1150, a(93) = 565068, a(97) <= 3046267249, a(101) = 10201, a(103) <= 5332206050752, a(107) = 99022508. - R. J. Mathar, Feb 21 2025
For nonsquare n, solutions k (not necessarily the smallest ones) are given by k= n*A002350(n)^2 such that 1/n-1/k= (A002349(n)/A002350(n))^2. - R. J. Mathar, Feb 25 2025
For n=p^2, squared odd primes, solutions k (not necessarily the smallest) are constructed by k=p*(p+1)^2/4 such that 1/n -1/k = 1/p^2-1/k =  [(p-1)/(p*(p+1))]^2 is a rational square. For other perfect squares n, start from such a solution for a prime factor p|n, n=(p*f)^2, and multiply the 3 terms of both sides of that solution with 1/f^2 to find a solution for n. - R. J. Mathar, Feb 25 2025

			a(16) = 18 because sqrt(1/16 - 1/17) = sqrt(1/272) is irrational but sqrt(1/16 - 1/18) = sqrt(1/144) = 1/12 is rational.

Felix Huber, Proof that a(1) = a(4) = 0

Cf. A024352, A378501, A379815, A002349, A002350.

A379816:=proc(n)
    local k;
    if n=1 or n=4 then
        return 0
    else
        for k from n+1 do
            if type(sqrt(1/n-1/k),rational) then
                return k
            fi
        od
    fi;
end proc;
seq(A379816(n),n=1..58);

a(n) = if ((n==1) || (n==4), return(0)); my(k=n+1); while (!issquare(1/n - 1/k), k++); k; \\ Michel Marcus, Feb 08 2025

Terms a(59-60) from R. J. Mathar, Feb 12 2025

A033319 Incrementally largest values of minimal y satisfying Pell equation x^2-Dy^2=1.

Original entry on oeis.org

0, 2, 4, 6, 180, 1820, 3588, 9100, 226153980, 15140424455100, 183567298683461940, 9562401173878027020, 42094239791738433660, 1238789998647218582160, 189073995951839020880499780706260

Offset: 1

Eric W. Weisstein

nonn

Records in A033317 (or A002349).

Ray Chandler, Table of n, a(n) for n = 1..63
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)
Eric Weisstein's World of Mathematics, Pell Equation.

Cf. A000037, A033317, A033318, A002349, A002350, A033316.

PellSolve[(m_Integer)?Positive] := Module[{cf, n, s}, cf = ContinuedFraction[Sqrt[m]]; n = Length[Last[cf]]; If[n == 0, Return[{}]]; If[OddQ[n], n = 2 n]; s = FromContinuedFraction[ ContinuedFraction[ Sqrt[m], n]]; {Numerator[s], Denominator[s]}];
yy = DeleteCases[PellSolve /@ Range[10^5], {}][[All, 2]];
Reap[Module[{y, record = 0}, Sow[0]; For[i = 1, i <= Length@yy, i++, y = yy[[i]]; If[y > record, record = y; Sow[y]]]]][[2, 1]] (* Jean-François Alcover, Nov 21 2020, after N. J. A. Sloane in A002350 *)

A298212 Smallest n such that A060645(a(n)) = 0 (mod n), i.e., x=A023039(a(n)) and y=A060645(a(n)) is the fundamental solution of the Pell equation x^2 - 5(ny)^2 = 1.

Original entry on oeis.org

1, 1, 2, 1, 5, 2, 4, 2, 2, 5, 5, 2, 7, 4, 10, 4, 3, 2, 3, 5, 4, 5, 4, 2, 25, 7, 6, 4, 7, 10, 5, 8, 10, 3, 20, 2, 19, 3, 14, 10, 10, 4, 22, 5, 10, 4, 8, 4, 28, 25, 6, 7, 9, 6, 5, 4, 6, 7, 29, 10, 5, 5, 4, 16, 35, 10, 34, 3, 4, 20, 35, 2, 37, 19, 50

Offset: 1

A.H.M. Smeets, Jan 15 2018

nonn

The fundamental solution of the Pell equation x^2 - 5*(n*y)^2 = 1, is the smallest solution of x^2 - 5*y^2 = 1 satisfying y = 0 (mod n).

Michael J. Jacobson, Jr. and Hugh C. Williams, Solving the Pell Equation, Springer, 2009, pages 1-17.

A.H.M. Smeets, Table of n, a(n) for n = 1..20000
H. W. Lenstra Jr., Solving the Pell Equation, Notices of the AMS, Vol.49, No.2, Feb. 2002, pp. 182-192.

Cf. A298210, A298211.

b[n_] := b[n] = Switch[n, 0, 0, 1, 4, _, 18 b[n - 1] - b[n - 2]];
a[n_] := For[k = 1, True, k++, If[Mod[b[k], n] == 0, Return[k]]];
a /@ Range[100] (* Jean-François Alcover, Nov 16 2019 *)

xf, yf = 9, 4
x, n = 2*xf, 0
while n < 20000:
    n = n+1
    y1, y0, i = 0, yf, 1
    while y0%n != 0:
        y1, y0, i = y0, x*y0-y1, i+1
    print(n, i)

a(n) <= n.
a(A000351(n)) = A000351(n).
A023039(a(n)) = A002350(5*n^2).
A060645(a(n)) = A002349(5*n^2).
if n | m then a(n) | a(m).
a(5^m) = 5^m for m>=0.
In general: if p is prime and p = 1 (mod 4) then: a(n) = n iff n = p^m, for m>=0.

A074076 One-sixth of the area of some primitive Heronian triangles with a distance of 2n+1 between the median and altitude points on the longest side.

Original entry on oeis.org

60, 4620, 2024, 5984, 11480, 22960, 41580, 8096, 45920, 521640, 226884, 392920, 438944, 803320, 6725544, 207900, 37966500, 1544620, 6846840, 2295200, 2785484, 9009000, 4016600, 188375200, 3383500, 149240, 5738000, 875124, 12013456, 8848840

Offset: 1

Lekraj Beedassy, Aug 28 2002

nonn

With N=2n+1, such a triangle has sides N*u +/- M, 2*M*u (the latter being cut into M*u +/- N by the corresponding altitude) and inradius M*(N - M)*v. The first entry, in particular, is associated with sequence A023039.

Ray Chandler, Table of n, a(n) for n = 1..499
Eric Weisstein's World of Mathematics, Heronian Triangle.
Wikipedia, Heronian Triangle.

Cf. A074074, A074075, A023039, A002349, A002350.

A033313 := proc(Dcap) local c,i,fr,nu,de ; if issqr(Dcap) then -1; else c := numtheory[cfrac](sqrt(Dcap)) ; for i from 1 do try fr := numtheory[nthconver](c,i) ; nu := numer(fr) ; de := denom(fr) ; if nu^2-Dcap*de^2=1 then RETURN(nu) ; fi; catch: RETURN(-1) ; end try; od: fi: end:
A074076 := proc(n) local Dmin,xmin,Dcap ; Dmin := -1; xmin := -1; mmin := -1; ymin := -1; for m from 1 to n do Dcap := (2*n+1+2*m)*(2*n+1-2*m) ; x := A033313(Dcap) ; if xmin = -1 or (x >0 and xA074076(n),n=1..80) ; # R. J. Mathar, Sep 21 2009

a(n) = M(n)*D(n)*u(n)*v(n)/6, where (u, v) is the fundamental solution to x^2 - D*y^2 = 1, with M = 2*A074075(n); D = A074074(n) = N^2 - M^2.

Removed assertion that these are the minimum areas - R. J. Mathar, Sep 21 2009

A128972 n^3 - 1 divided by its largest cube divisor.

Original entry on oeis.org

7, 26, 63, 124, 215, 342, 511, 91, 37, 1330, 1727, 2196, 2743, 3374, 4095, 614, 17, 254, 7999, 9260, 10647, 12166, 13823, 1953, 17575, 19682, 813, 24388, 26999, 29790, 32767, 4492, 39303, 42874, 46655, 1876, 54871, 59318, 63999, 8615, 74087, 79506

Offset: 2

Jonathan Vos Post, Apr 28 2007

In other words, cubefree part of n^3-1, or cubefree kernel of n^3-1. Cube analog of A068310.

			a(9) = (9^3-1)/8 = (2^3 * 7 * 13)/(2^3) = 728/8 = 91.
a(10) = (10^3-1)/27 = (3^3 * 37)/(3^3) = 999/27 = 37.
a(18) = (18^3-1)/343 = (7^3 * 17)/(7^3) = 5831/343 = 17.

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A002350, A004709, A007948, A062378, A067872, A033314, A068310.

a:= n -> mul(f[1]^(f[2] mod 3), f = ifactors(n^3-1)[2]):
seq(a(n),n=2..100); # Robert Israel, Sep 24 2014

a(n) = A062378(A068601(n)) = A062378(n^3-1).

More terms from Carl R. White, Nov 09 2010

A094680 a(n+1) = 4a(n)^3 - 3a(n), with a(0) = 2.

Original entry on oeis.org

2, 26, 70226, 1385331749802026, 10634604778476758291777057017318241822792488226

Offset: 0

Jose Eduardo Blazek, Jun 07 2004

Smallest positive integer x satisfying the Pell equation x^2 - 3^(2*n-3) * y^2 = 1. - A.H.M. Smeets, Sep 29 2017

Term a(5) has 139 decimal digits and a(6) has 417 decimal digits. - Andrew Howroyd, Feb 25 2018

Andrew Howroyd, Table of n, a(n) for n = 0..6

Cf. A001075, A002812.

NestList[4 #^3 - 3 # &, 2, 5] (* Michael De Vlieger, Oct 02 2017 *)

a(n) = if (n==0, 2, 4*a(n-1)^3 - 3*a(n-1)); \\ Michel Marcus, Oct 03 2017

a(n) = polchebyshev(3^n, 1, 2); \\ Michel Marcus, Oct 03 2017

a(n) = cosh(3^n*arccosh(2)).
a(n) = ChebyshevT(3^n, 2). - Vladeta Jovovic, Jun 11 2004
From A.H.M. Smeets, Oct 02 2017: (Start)
a(n) = A001075(3^(n-2))
a(n) = A002350(3^(2n-3)). (End)

More terms from Vladeta Jovovic, Jun 11 2004

Offset corrected by Michel Marcus, Oct 03 2017

cp's OEIS Frontend

A081231 Let p = n-th prime of the form 4k+3, take smallest solution (x,y) to the Pellian equation x^2 - p*y^2 = 1 with x and y >= 1; sequence gives value of x.

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A081232 Let p = n-th prime of the form 4k+1, take smallest solution (x,y) to the Pellian equation x^2 - p*y^2 = 1 with x and y >= 1; sequence gives value of x.

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A298211 Smallest n such that A001353(a(n)) == 0 (mod n), i.e., x=A001075(a(n)) and y=A001353(a(n)) is the fundamental solution of the Pell equation x^2 - 3*(n*y)^2 = 1.

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A379815 a(n) is the smallest integer k > n such that sqrt(1/n + 1/k) is a rational number; or 0 if no such k exists.

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A379816 a(n) is the smallest integer k > n such that sqrt(1/n - 1/k) is a rational number; or 0 if no such k exists.

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A033319 Incrementally largest values of minimal y satisfying Pell equation x^2-Dy^2=1.

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Mathematica

A298212 Smallest n such that A060645(a(n)) = 0 (mod n), i.e., x=A023039(a(n)) and y=A060645(a(n)) is the fundamental solution of the Pell equation x^2 - 5*(n*y)^2 = 1.

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Author

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Comments

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Programs

Mathematica

Python

Formula

A074076 One-sixth of the area of some primitive Heronian triangles with a distance of 2n+1 between the median and altitude points on the longest side.

Views

A298211 Smallest n such that A001353(a(n)) == 0 (mod n), i.e., x=A001075(a(n)) and y=A001353(a(n)) is the fundamental solution of the Pell equation x^2 - 3(ny)^2 = 1.

A298212 Smallest n such that A060645(a(n)) = 0 (mod n), i.e., x=A023039(a(n)) and y=A060645(a(n)) is the fundamental solution of the Pell equation x^2 - 5(ny)^2 = 1.

A094680 a(n+1) = 4a(n)^3 - 3a(n), with a(0) = 2.