A078986 -id:A078986 - cp's OEIS frontend

A132593 Nonnegative integer solutions X to the equation: X(X + 1) - 10*Y^2 = 0.

0, 9, 360, 13689, 519840, 19740249, 749609640, 28465426089, 1080936581760, 41047124680809, 1558709801289000, 59189925324301209, 2247658452522156960, 85351831270517663289, 3241121929827149048040, 123077281502161146162249

Offset: 0

Mohamed Bouhamida, Nov 14 2007

Also, numbers n such that 5*A000217(n) is a square. [Bruno Berselli, Dec 16 2013]

Seiichi Manyama, Table of n, a(n) for n = 0..500
Kenneth M. Wilke, Problem 269, Crux Mathematicorum, Vol. 3, No. 7 (1977), p. 190; Solution to Problem 269 by Lindsay Reynolds, W. J. Blundon and M. S. Klamkin, ibid., Vol. 4, No. 3 (1978), pp. 79-82; Comment by the MaScoT Problems Group, ibid., Vol. 6, No. 2 (1980), pp. 44-46.
Index entries for linear recurrences with constant coefficients, signature (39,-39,1).

Cf. A007654, A078986.

Cf. A233474 (numbers n such that 5*A000217(n)-1 is a square).

LinearRecurrence[{39,-39,1},{0,9,360},30] (* Harvey P. Dale, Jun 01 2014 *)

a(0)=0, a(1)=9 and a(n) = 38*a(n-1) - a(n-2) + 18.
a(n) = (A078986(n) - 1)/2. - Max Alekseyev, Nov 13 2009
G.f.: -9*x*(x+1)/((x-1)*(x^2-38*x+1)). - Colin Barker, Oct 24 2012
From Amiram Eldar, Feb 15 2022: (Start)
sqrt(a(n)+1) - sqrt(n) = (sqrt(10)-3)^n (Wilke, 1977).
a(n) = ((Sum_{k=0..n} binomial(2*n, 2*k) * 10^(n-k) * 9*k)- 1)/2 (Klamkin, 1978).
a(n) = sinh(n*log(sqrt(10)+3))^2 (MaScoT Problems Group, 1980). (End)

More terms from Max Alekseyev, Nov 13 2009

A207832 Numbers x such that 20*x^2 + 1 is a perfect square.

Original entry on oeis.org

0, 2, 36, 646, 11592, 208010, 3732588, 66978574, 1201881744, 21566892818, 387002188980, 6944472508822, 124613502969816, 2236098580947866, 40125160954091772, 720016798592704030

Offset: 0

Gary Detlefs, Feb 20 2012

Denote as {a,b,c,d} the second-order linear recurrence a(n) = c*a(n-1) + d*a(n-2) with initial terms a, b. The following sequences and recurrence formulas are related to integer solutions of k*x^2 + 1 = y^2.
.
   k             x                        y
   -  -----------------------  -----------------------
A001542 {0,2,6,-1}       A001541 {1,3,6,-1}
A001353 {0,1,4,-1}       A001075 {1,2,4,-1}
A060645 {0,4,18,-1}      A023039 {1,9,18,-1}
A001078 {0,2,10,-1}      A001079 {1,5,10,-1}
A001080 {0,3,16,-1}      A001081 {1,8,16,-1}
A001109 {0,1,6,-1}       A001541 {1,3,6,-1}
A084070 {0,1,38,-1}      A078986 {1,19,38,-1}
A001084 {0,3,20,-1}      A001085 {1,10,20,-1}
A011944 {0,2,14,-1}      A011943 {1,7,14,-1}
A075871 {0,180,1298,-1}  A114047 {1,649,1298,-1}
A068204 {0,4,30,-1}      A069203 {1,15,30,-1}
A001090 {0,1,8,-1}       A001091 {1,4,8,-1}
A121740 {0,8,66,-1}      A099370 {1,33,66,-1}
A202299 {0,4,34,-1}      A056771 {1,17,34,-1}
A174765 {0,39,340,-1}    A114048 {1,179,340,-1}
a(n)    {0,2,18,-1}      A023039 {1,9,18,-1}
A174745 {0,12,110,-1}    A114049 {1,55,110,-1}
A174766 {0,42,394,-1}    A114050 {1,197,394,-1}
A174767 {0,5,48,-1}      A114051 {1,24,48,-1}
A004189 {0,1,10,-1}      A001079 {1,5,10,-1}
A174768 {0,10,102,-1}    A099397 {1,51,102,-1}
The sequence of the c parameter is listed in A180495.

Bruno Berselli, Table of n, a(n) for n = 0..500
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A023039, A049660, A079962.

m:=16; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(2*x/(1-18*x+x^2))); // Bruno Berselli, Jun 19 2019

readlib(issqr):for x from 1 to 720016798592704030 do if issqr(20*x^2+1) then print(x) fi od;

LinearRecurrence[{18, -1}, {0, 2}, 16] (* Bruno Berselli, Feb 21 2012 *)
Table[2 ChebyshevU[-1 + n, 9], {n, 0, 16}]  (* Herbert Kociemba, Jun 05 2022 *)

makelist(expand(((2+sqrt(5))^(2*n)-(2-sqrt(5))^(2*n))/(4*sqrt(5))), n, 0, 15); /* Bruno Berselli, Jun 19 2019 */

a(n) = 18*a(n-1) - a(n-2).
From Bruno Berselli, Feb 21 2012: (Start)
G.f.: 2*x/(1-18*x+x^2).
a(n) = -a(-n) = 2*A049660(n) = ((2 + sqrt(5))^(2*n)-(2 - sqrt(5))^(2*n))/(4*sqrt(5)). (End)
a(n) = Fibonacci(6*n)/4. - Bruno Berselli, Jun 19 2019
For n>=1, a(n) = A079962(6n-3). - Christopher Hohl, Aug 22 2021

A248834 The numerator of curvature of touching circles inscribed in a special way in the smaller segment of circle of radius 1/6 divided by a chord of length sqrt(8/75).

Original entry on oeis.org

15, 25, 245, 3025, 39605, 525625, 6997445, 93219025, 1242045605, 16549536025, 220514700245, 2938258798225, 39150987330005, 521669482807225, 6951013841444645, 92619168339300625, 1234109231890228805, 16443956730548563225, 219108411138085022645, 2919522145350504838225

Offset: 0

Kival Ngaokrajang, Oct 15 2014

Refer to comment of A240926. Consider a circle C of radius 1/6 (in some length units) with a chord of length sqrt(8/75). This has been chosen such that the smaller sagitta has length 2/15. The input, besides the circle C, is the circle C_0 with radius R_0 = 1/15, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle is C_n = 1/R_n, n >= 0, and its numerator is conjectured to be a(n). The denominator is A000244 for n > 0. If one considers the curvature of touching circles inscribed in the larger segment (sagitta length 1/5), the sequence would be A248833. See an illustration given in the link.

Kival Ngaokrajang, Illustration of initial terms

Cf. A240926, A078986, A097315, A247512, A247335, A247512, A248833.

{
r=0.4;print1(round(6/r),", ");r1=r;dn=1;
for (n=1,40,
if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
ac=sqrt(ab^2-r^2);
if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
b=acos(r/ab)-z;
r=r*(1-cos(b))/(1+cos(b));
print1(round((6/r)*dn),", ");
dn=dn*3
)
}

Conjecture: a(n) = 17*a(n-1) - 51*a(n-2) + 27*a(n-3) for n > 3. - Colin Barker, Oct 15 2014

Empirical g.f.: 5*(54*x^3-117*x^2+46*x-3) / ((3*x-1)*(9*x^2-14*x+1)). - Colin Barker, Oct 15 2014

A249458 The numerators of curvatures of touching circles inscribed in a special way in the smaller segment of unit circle divided by a chord of length sqrt(84)/5.

Original entry on oeis.org

10, 100, 1690, 36100, 835210, 19802500, 472931290, 11318832100, 271066588810, 6492762648100, 155527144782490, 3725543446072900, 89243180863948810, 2137770243127864900, 51209104645650371290, 1226685938180259902500

Offset: 0

Kival Ngaokrajang, Oct 29 2014

The denominators are conjectured to be A169634.
Refer to comments and links of A240926. Consider a unit circle with a chord of length sqrt(84)/5. This has been chosen such that the smaller sagitta has length 3/5. The input, besides the circle C, is the circle C_0 with radius R_0 = 3/10, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle is C_n = 1/R_n, n >= 0, and its numerator is conjectured to be a(n). If one considers the curvature of touching circles inscribed in the larger segment (sagitta length 7/5), the sequence would be A249457/A005032. See an illustration given in the link.
For the proof and the formula for the rational curvatures of the circles in the smaller segment see a comment under A249864. C_n = (5/(3*7))*(7*S(n, 26/7) - 13*S(n-1, 26/7) + 7), n >= 0, with Chebyshev's S polynomials (A049310). - Wolfdieter Lang, Nov 08 2014

Kival Ngaokrajang, Illustration of initial terms.
Eric Weisstein's World of Mathematics, Sagitta.
Index entries for linear recurrences with constant coefficients, signature (33,-231,343).
Index entries for sequences related to Chebyshev polynomials.

Cf. A240926, A078986, A097315, A247512, A247335, A247512, A248834, A169634, A249457, A049310, A249863, A249864.

I:=[10, 100, 1690]; [n le 3 select I[n] else 33*Self(n-1) - 231*Self(n-2) + 343*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{33, -231, 343},{10, 100, 1690},16] (* Ray Chandler, Aug 11 2015 *)
CoefficientList[Series[10*(1 - 23*x + 70*x^2)/((1 - 26*x + (7*x)^2)*(1 - 7*x)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

{
r=0.3;dn=3;print1(round(dn/r),", ");r1=r;
for (n=1,40,
     if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
     ac=sqrt(ab^2-r^2);
     if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
     b=acos(r/ab)-z;
     r=r*(1-cos(b))/(1+cos(b)); dn=dn*7;
     print1(round(dn/r),", ");
)
}

x='x+O('x^30); Vec(10*(1 - 23*x + 70*x^2)/((1 - 26*x + (7*x)^2)*(1 - 7*x))) \\ G. C. Greubel, Dec 20 2017

Empirical g.f.: -10*(70*x^2-23*x+1) / ((7*x-1)*(49*x^2-26*x+1)). - Colin Barker, Oct 29 2014
From Wolfdieter Lang, Nov 09 2014 (Start)
a(n) = 5*(A249864(n) + 7^n) = (5*7^n)*(S(n, 26/7) - (13/7)*S(n-1, 26/7) + 1), n >= 0, with Chebyshev's S polynomials (A049310). See the comments on A249864 for the proof.
O.g.f.: 5*((1 - 13*x)/(1 - 26*x + (7*x)^2) + 1/(1-7*x)) = 10*(1 - 23*x + 70*x^2)/((1 - 26*x + (7*x)^2)*(1 - 7*x)) proving the conjecture of Colin Barker above. (End)

Edited. In name and comment small changes, keyword easy and crossrefs added. - Wolfdieter Lang, Nov 08 2014

A239364 Numbers n such that (n^2-4)/10 is a square.

Original entry on oeis.org

38, 1442, 54758, 2079362, 78960998, 2998438562, 113861704358, 4323746327042, 164188498723238, 6234839205156002, 236759701297204838, 8990633810088627842, 341407325082070653158, 12964487719308596192162, 492309126008644584648998, 18694782300609185620469762

Offset: 1

Colin Barker, Mar 17 2014

Values of x satisfying the Pellian equation x^2 - 10*y^2 = 4.

			1442 is in the sequence because (1442^2-4)/10 = 207936 = 456^2.

Colin Barker, Table of n, a(n) for n = 1..600
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Index entries for linear recurrences with constant coefficients, signature (38,-1).

Cf. A078986, A239365.

LinearRecurrence[{38,-1},{38,1442},30] (* Harvey P. Dale, Dec 19 2014 *)

Vec(-2*x*(x-19)/(x^2-38*x+1) + O(x^100))

a(n) = 2*A078986(n).
a(n) = (19+6*sqrt(10))^(-n)+(19+6*sqrt(10))^n.
a(n) = 38*a(n-1)-a(n-2).
G.f.: -2*x*(x-19) / (x^2-38*x+1).

A248833 The curvature of touching circles inscribed in a special way in the larger segment of circle of radius 1/6 divided by a chord of length sqrt(8/75).

Original entry on oeis.org

10, 25, 160, 1225, 9610, 75625, 595360, 4687225, 36902410, 290532025, 2287353760, 18008298025, 141779030410, 1116223945225, 8788012531360, 69187876305625, 544714997913610, 4288532107003225, 33763541858112160, 265819802757894025, 2092794880205040010, 16476539238882426025

Offset: 0

Kival Ngaokrajang, Oct 15 2014

Refer to comment of A240926. Consider a circle C of radius 1/6 (in some length units) with a chord of length sqrt(8/75). This has been chosen such that the larger sagitta has length 1/5. The input, besides the circle C, is the circle C_0 with radius R_0 = 1/10, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle, C_n = 1/R_n, n >= 0, is conjectured to be a(n). If one considers the curvature of touching circles inscribed in the smaller segment (sagitta length 2/15), the sequence would be A248834. See an illustration given in the link.

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Kival Ngaokrajang, Illustration of initial terms.
Eric Weisstein's World of Mathematics, Sagitta.
Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A240926, A078986, A097315, A247512, A247335, A247512, A248834.

I:=[10,25,160]; [n le 3 select I[n] else 9*Self(n-1)-9*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Oct 29 2014

CoefficientList[Series[- 5 (5 x^2 - 13 x + 2)/((x - 1) (x^2 - 8 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 29 2014 *)
LinearRecurrence[{9,-9,1}, {10,25,160}, 30] (* G. C. Greubel, Dec 20 2017 *)

{
r=0.6;print1(round(6/r),", ");r1=r;
for (n=1,40,
     if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
     ac=sqrt(ab^2-r^2);
     if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
     b=acos(r/ab)-z;
     r=r*(1-cos(b))/(1+cos(b));
     print1(round(6/r),", ");
)
}

Vec(-5*(5*x^2-13*x+2)/((x-1)*(x^2-8*x+1)) + O(x^100)) \\ Colin Barker, Oct 15 2014

From Colin Barker, Oct 15 2014: (Start)
a(n) = 9*a(n-1) - 9*a(n-2) + a(n-3).
G.f.: -5*(5*x^2-13*x+2) / ((x-1)*(x^2-8*x+1)). (End)
a(n) = 5*(2+(4-sqrt(15))^n+(4+sqrt(15))^n)/2. - Colin Barker, Mar 03 2016
E.g.f.: 5*exp(x)*(1 + exp(3*x)*cosh(sqrt(15)*x)). - Stefano Spezia, Aug 27 2025

A343034 Positive numbers m such that m^2 with last digit z deleted is still a perfect square k^2, and z divides m-k.

Original entry on oeis.org

1, 13, 19, 487, 721, 18493, 27379, 702247, 1039681, 26666893, 39480499, 1012639687, 1499219281, 38453641213, 56930852179, 1460225726407, 2161873163521, 55450123962253, 82094249361619, 2105644484839207, 3117419602578001, 79959040299927613, 118379850648602419, 3036337886912410087

Offset: 1

Bernard Schott, Apr 03 2021

This sequence is the answer to the problem A1872 proposed on French mathematical site Diophante (see link).
Equivalent to the two Diophantine equations: m^2 = 10*k^2 + z and m-k = q*z for some q >= 1.
There exist solutions iff z = 1 or z = 9.
When (m, k, z) is a solution, then (19m+60k, 6m+19k, z) is another solution.
There is only one solution such that z = m-k: (13, 4, 9), see 1st example.
There exist two distinct families of solutions corresponding to z = 1 and z = 9, odd indices correspond to z = 1 and even indices to z = 9.
-> For z = 1, all solutions m of Pell equation m^2 - 10*k^2 = 1 are terms because z = 1 divides every m-k.
   First few solutions (m, k) are (1, 0), (19, 6), (721, 228), (27379, 86568), ... with m = A078986(q) and corresponding k = 6*A078987(q).
-> For z = 9, solutions m must satisfy m^2 - 10*k^2 = 9 with 9 divides m-k. Among the 3 fundamental solutions (3, 0), (7, 2), (13, 4) of Pell equation m^2 -10*k^2 = 9, only (13, 4) gives solutions where 9 divides m-k.
   First few solutions (m, k) are (13, 4), (487, 154), (18493, 5848), ... with m = A228209(3q).

			For m = 13, 13^2 = 169, 4^2 = 16, 13^2 - 10*4^2 = 9 and 9 = 13-4 divides 13-4.
For m = 19, 19^2 = 361, 6^2 = 36, 19^2 - 10*6^2 = 1 and 1 divides 19-6 = 13.
For m = 487, 487^2 = 237169, 154^2 = 23716, 487^2 - 10*154^2 = 9 and 9 divides 487-154 = 333 = 9*37.

Diophante, A1872, Carrément magiques (in French).
Index entries for linear recurrences with constant coefficients, signature (0,38,0,-1).

Subsequence of A031149.
A078986 is a subsequence.
Cf. A078987, A228209.

LinearRecurrence[{0, 38, 0, -1}, {1, 13, 19, 487}, 24] (* Amiram Eldar, Apr 03 2021 *)

a(2n+1) = A078986(n) for n >= 0.
a(2n) = A228209(3n) for n >= 1.
a(n+4) = 38*a(n+2) - a(n), a(1) = 1, a(2) = 13, a(3)= 19, a(4) = 487.
G.f.: x*(1 + 13*x - 19*x^2 - 7*x^3)/(1 - 38*x^2 + x^4). - Stefano Spezia, Apr 03 2021

cp's OEIS Frontend

A132593 Nonnegative integer solutions X to the equation: X(X + 1) - 10*Y^2 = 0.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A207832 Numbers x such that 20*x^2 + 1 is a perfect square.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Maxima

Formula

A248834 The numerator of curvature of touching circles inscribed in a special way in the smaller segment of circle of radius 1/6 divided by a chord of length sqrt(8/75).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Formula

A249458 The numerators of curvatures of touching circles inscribed in a special way in the smaller segment of unit circle divided by a chord of length sqrt(84)/5.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

Extensions

A239364 Numbers n such that (n^2-4)/10 is a square.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A248833 The curvature of touching circles inscribed in a special way in the larger segment of circle of radius 1/6 divided by a chord of length sqrt(8/75).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

A343034 Positive numbers m such that m^2 with last digit z deleted is still a perfect square k^2, and z divides m-k.

Views

Author

Keywords

Comments

Examples