A001652 -id:A001652 - cp's OEIS frontend

A106521 Numbers m such that Sum_{k=0..10} (m+k)^2 is a square.

18, 38, 456, 854, 9192, 17132, 183474, 341876, 3660378, 6820478, 73024176, 136067774, 1456823232, 2714535092, 29063440554, 54154634156, 579811987938, 1080378148118, 11567176318296, 21553408328294

Offset: 1

Ralf Stephan, May 30 2005

Equivalently, 11*a(n)^2 + 110*a(n) + 385 is a square.
11*((m+5)^2+10) is a square iff the second factor is divisible by 11 and the quotient is a square, i.e., iff m = 11*k - 4 or m = 11*k - 6 and 11*k^2 +- 2 k + 1 is a square. Thus a(n) == (7,5,5,7,7,5,5,7,...) (mod 11), repeating with period 4 and the values are obtained by solving these Pell-type equations (cf. link to Dario Alpern's quadratic solver). The corresponding recurrence equations (see PARI code) should make it possible to prove the conjectured g.f. - M. F. Hasler, Jan 27 2008
All sequences of this type (i.e., sequences with fixed offset k, and a discernible pattern: k=0...10 for this sequence, k=0..1 for A001652) can be continued using a formula such as x(n) = a*x(n-p) - x(n-2p) + b, where a and b are various constants, and p is the period of the series. Alternatively 'p' can be considered the number of concurrent series. - Daniel Mondot, Aug 05 2016

			Since 18^2 + 19^2 + ... + 28^2 = 5929 = 77^2, 18 is in the sequence. - _Michael B. Porter_, Aug 07 2016

Vincenzo Librandi, Table of n, a(n) for n = 1..500
Dario Alpern, Quadratic two integer variable equation solver
Index entries for linear recurrences with constant coefficients, signature (1,20,-20,-1,1).

Cf. A001032, A094196.

LinearRecurrence[{1,20,-20,-1,1},{18,38,456,854,9192},30] (* Harvey P. Dale, May 07 2011 *)

A106521(n)={local(xy=[ -4-2*(n%2);11],PQRS=[10,3;33,10],KL=[45;165]);until(0>=n-=2,xy=PQRS*xy+KL);xy[1]} \\ M. F. Hasler, Jan 27 2008

G.f.: 2*x*(9+10*x+29*x^2-x^3-2*x^4)/(1-x)/(1-20*x^2+x^4). - Vladeta Jovovic, May 31 2005; adapted to the offset by Bruno Berselli, May 16 2011
a(1)=18, a(2)=38, a(3)=456, a(4)=854, a(5)=9192; thereafter a(n)=a(n-1)+20*a(n-2)- 20*a(n-3)-a(n-4)+a(n-5). - Harvey P. Dale, May 07 2011
a(n) = A198949(n+1)-5. - Bruno Berselli, Feb 12 2012
a(1)=18, a(2)=38, a(3)=456, a(4)=854; thereafter a(n) = 20*a(n-2) - a(n-4) + 90. - Daniel Mondot, Aug 05 2016

Edited and extended by M. F. Hasler, Jan 27 2008

A118676 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+79)^2 = y^2.

Original entry on oeis.org

0, 20, 161, 237, 341, 1140, 1580, 2184, 6837, 9401, 12921, 40040, 54984, 75500, 233561, 320661, 440237, 1361484, 1869140, 2566080, 7935501, 10894337, 14956401, 46251680, 63497040, 87172484, 269574737, 370088061, 508078661, 1571196900, 2157031484, 2961299640

Offset: 1

Mohamed Bouhamida, May 19 2006

Also values x of Pythagorean triples (x, x+79, y).
Corresponding values y of solutions (x, y) are in A159758.
For the generic case x^2+(x+p)^2 = y^2 with p = m^2-2 a (prime) number > 7 in A028871, see A118337.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (83+18*sqrt(2))/79 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (10659+6110*sqrt(2))/79^2 for n mod 3 = 0.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A159758, A028871, A118337, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A159759 (decimal expansion of (83+18*sqrt(2))/79), A159760 (decimal expansion of (10659+6110*sqrt(2))/79^2).

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(20+141*x+76*x^2-16*x^3-47*x^4-16*x^5)/((1-x)*(1- 6*x^3+x^6)))); // G. C. Greubel, May 07 2018

LinearRecurrence[{1,0,6,-6,0,-1,1},{0,20,161,237,341,1140,1580},75] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2012 *)

forstep(n=0, 100000000, [1, 3], if(issquare(2*n^2+158*n+6241), print1(n, ",")))

a(n) = 6*a(n-3) -a(n-6) +158 for n > 6; a(1)=0, a(2)=20, a(3)=161, a(4)=237, a(5)=341, a(6)=1140.
G.f.: x*(20+141*x+76*x^2-16*x^3-47*x^4-16*x^5)/((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 79*A001652(k) for k >= 0.

Edited by Klaus Brockhaus, Apr 30 2009

A129298 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+89)^2 = y^2.

Original entry on oeis.org

0, 51, 120, 267, 540, 931, 1780, 3367, 5644, 10591, 19840, 33111, 61944, 115851, 193200, 361251, 675444, 1126267, 2105740, 3936991, 6564580, 12273367, 22946680, 38261391, 71534640, 133743267, 223003944, 416934651, 779513100, 1299762451

Offset: 1

Mohamed Bouhamida, May 26 2007

Also values x of Pythagorean triples (x, x+89, y).
Corresponding values y of solutions (x, y) are in A160055.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (107+42*sqrt(2))/89 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (8979+2990*sqrt(2))/89^2 for n mod 3 = 0.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A160055, A129289, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A160056 (decimal expansion of (107+42*sqrt(2))/89), A160057 (decimal expansion of (8979+2990*sqrt(2))/89^2).

I:=[0,51,120,267,540,931,1780]; [n le 7 select I[n] else Self(n-1) +6*Self(n-3) -6*Self(n-4) - Self(n-6) + Self(n-7): n in [1..30]]; // G. C. Greubel, Apr 19 2018

LinearRecurrence[{1,0,6,-6,0,-1,1},{0,51,120,267,540,931,1780},30] (* Harvey P. Dale, Sep 21 2013 *)

{forstep(n=0, 10000000, [3, 1], if(issquare(2*n^2+178*n+7921), print1(n, ",")))};

x='x+O('x^30); concat(0, Vec(x*(51+69*x+147*x^2-33*x^3-23*x^4-33*x^5)/((1-x)*(1-6*x^3+x^6)))) \\ G. C. Greubel, Apr 19 2018

a(n) = 6*a(n-3) - a(n-6) + 178 with for n > 6; a(1)=0, a(2)=51, a(3)=120, a(4)=267, a(5)=540, a(6)=931.
G.f.: x*(51+69*x+147*x^2-33*x^3-23*x^4-33*x^5) / ((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 89*A001652(k), k >= 0. (Zak Seidov, May 28 2007)
a(1)=0, a(2)=51, a(3)=120, a(4)=267, a(5)=540, a(6)=931, a(7)=1780, a(n) = a(n-1) + 6*a(n-3) - 6*a(n-4) - a(n-6) + a(n-7). - Harvey P. Dale, Sep 21 2013

Edited and three terms added by Klaus Brockhaus, May 04 2009

A076218 Numbers n such that 2n^2 - 3n + 1 is a square.

Original entry on oeis.org

0, 1, 5, 145, 4901, 166465, 5654885, 192099601, 6525731525, 221682772225, 7530688524101, 255821727047185, 8690408031080165, 295218051329678401, 10028723337177985445, 340681375412721826705, 11573138040695364122501, 393146012008229658338305

Offset: 1

Gregory V. Richardson, Nov 03 2002

Limit_{n -> infinity} a(n)/a(n-1) = 33.970562748477140585620264690516... = 17 + 12*sqrt(2).
Conjecture: a nonzero number occurs twice in A055524 if and only if it's in this sequence. - J. Lowell, Jul 23 2016
Equivalently, n=0 or both n-1 and 2*n-1 are perfect squares. - Sture Sjöstedt, Feb 22 2017

			5 is in the sequence since 2*5^2 - 3*5 + 1 = 50 - 15 + 1 = 36 is a square. - _Michael B. Porter_, Jul 24 2016

Colin Barker, Table of n, a(n) for n = 1..650
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: A084703 (k=-1), this sequence (k=3), A278310 (k=-5).

Join[{0},LinearRecurrence[{35,-35,1},{1,5,145},20]] (* Harvey P. Dale, Nov 27 2012 *)

a(n)=if(n>1,([0,1,0;0,0,1;1,-35,35]^n*[145;5;1])[1,1],0) \\ Charles R Greathouse IV, Jul 24 2016

concat(0, Vec(x^2*(1-30*x+5*x^2) / ((1-x)*(1-34*x+x^2)) + O(x^30))) \\ Colin Barker, Nov 21 2016

From Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 04 2002: (Start)
a(n) = ( (3+(17+12*sqrt(2))^(n-1)) + (3+(17-12*sqrt(2))^(n-1)) )/8 for n>=1.
a(n) = 35 * a(n-1) - 35 * a(n-2) + a(n-3).
G.f.: (x-30*x^2+5*x^3)/(1-35*x+35*x^2-x^3). (End)
Product of adjacent odd-indexed Pell numbers (A000129). - Gary W. Adamson, Jun 07 2003
sqrt(2) - 1 = 0.414213562... = 2/5 + 2/145 + 2/4901 + 2/166465 + ... = Sum_{n>=2} 2/a(n). - Gary W. Adamson, Jun 07 2003
For n > 0, one more than square of adjacent even-indexed Pell numbers (A000129). - Charlie Marion, Mar 09 2005
a(n) = A001652(n-1) + 2*A001652(n-1)*A001652(n-2) + A001652(n-2) + 2. - Charlie Marion, Nov 24 2018

A118675 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+47)^2 = y^2.

Original entry on oeis.org

0, 16, 85, 141, 225, 616, 940, 1428, 3705, 5593, 8437, 21708, 32712, 49288, 126637, 190773, 287385, 738208, 1112020, 1675116, 4302705, 6481441, 9763405, 25078116, 37776720, 56905408, 146166085, 220178973, 331669137, 851918488, 1283297212, 1933109508

Offset: 1

Mohamed Bouhamida, May 19 2006

nonn

Also values x of Pythagorean triples (x, x+47, y).
Corresponding values y of solutions (x, y) are in A159750.
For the generic case x^2+(x+p)^2 = y^2 with p = m^2-2 a (prime) number > 7 in A028871, see A118337.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (51+14*sqrt(2))/47 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (3267+1702*sqrt(2))/47^2 for n mod 3 = 0.

Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A159750, A028871, A118337, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A159751 (decimal expansion of (51+14*sqrt(2))/47), A159752 (decimal expansion of (3267+1702*sqrt(2))/47^2).

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(16+69*x+56*x^2-12*x^3-23*x^4-12*x^5)/((1-x)*(1-6*x^3 +x^6)))); // G. C. Greubel, May 07 2018

Select[Range[0,100000],IntegerQ[Sqrt[#^2+(#+47)^2]]&] (* or *) LinearRecurrence[{1,0,6,-6,0,-1,1},{0,16,85,141,225,616,940},50] (* Vladimir Joseph Stephan Orlovsky, Feb 02 2012 *)

{forstep(n=0, 100000000, [1, 3], if(issquare(2*n^2+94+2209), print1(n, ",")))}

a(n) = 6*a(n-3) -a(n-6) +94 for n > 6; a(1)=0, a(2)=16, a(3)=85, a(4)=141, a(5)=225, a(6)=616.
G.f.: x*(16+69*x+56*x^2-12*x^3-23*x^4-12*x^5)/((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 47*A001652(k) for k >= 0.

Edited by Klaus Brockhaus, Apr 30 2009

A124124 Nonnegative integers n such that 2n^2 + 2n - 3 is square.

Original entry on oeis.org

1, 2, 6, 13, 37, 78, 218, 457, 1273, 2666, 7422, 15541, 43261, 90582, 252146, 527953, 1469617, 3077138, 8565558, 17934877, 49923733, 104532126, 290976842, 609257881, 1695937321, 3551015162, 9884647086, 20696833093, 57611945197, 120629983398, 335787024098

Offset: 1

John W. Layman, Nov 29 2006

First differences are apparently in A143608. [R. J. Mathar, Jul 17 2009]

Alternative definition: T_n and (T_n - 1)/2 are triangular numbers. - Raphie Frank, Sep 06 2012

Michael De Vlieger, Table of n, a(n) for n = 1..2613
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A001108, A001652, A001653, A006452, A008844, A046090, A046172, A077442, A098790, A216134.

A124124 := proc(n)
coeftayl(x*(1+x-2*x^2+x^3+x^4)/((1-x)*(x^2-2*x-1)*(x^2+2*x-1)), x=0, n);
end proc:
seq(A124124(n), n=1..20); # Wesley Ivan Hurt, Aug 04 2014
# Alternative:
a[1]:= 1: a[2]:= 2: a[3]:= 6:
for n from 4 to 1000 do
a[n]:= (3 + 2*(n mod 2))*(a[n-1]-a[n-2])+a[n-3]
od:
seq(a[n],n=1..100); # Robert Israel, Aug 13 2014

LinearRecurrence[{1,6,-6,-1,1},{1,2,6,13,37},40] (* Harvey P. Dale, Nov 05 2011 *)
CoefficientList[Series[(1 + x - 2*x^2 + x^3 + x^4)/((1 - x)*(x^2 - 2*x - 1)*(x^2 + 2*x - 1)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 04 2014 *)

for(n=1,10^10,if(issquare(2*n^2+2*n-3),print1(n,", "))) \\ Derek Orr, Aug 13 2014

It appears that a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) if n is even, a(n) = 5*a(n-1)-5*a(n-2)+a(n-3) if n is odd. Can anyone confirm this?
Corrected and confirmed (using the g.f.) by Robert Israel, Aug 27 2014
2*a(n) = sqrt(7+2*A077442(n-1)^2)-1. - R. J. Mathar, Dec 03 2006
a(n) = a(n-1)+6*a(n-2)-6*a(n-3)-a(n-4)+a(n-5). G.f.: -x*(1+x-2*x^2+x^3+x^4)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)). [R. J. Mathar, Jul 17 2009]
For n>0, a(2n-1) = 2*A001653(n) - A046090(n-1) and a(2n) = 2*A001653(n) + A001652(n-1). - Charlie Marion, Jan 03 2012
From Raphie Frank, Sep 06 2012: (Start)
If y = A006452(n), then a(n) = 2y + ((sqrt(8y^2 - 7) - 1)/2 - (1 - sgn(n))).
Also see A216134 [a(n) = y + ((sqrt(8y^2 - 7) - 1)/2 - (1 - sgn(n)))].
(End)
From Hermann Stamm-Wilbrandt, Aug 27 2014: (Start)
a(2*n+2) = A098586(2*n).
a(2*n+1) = A098790(2*n).
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>6. (End)
a(2*n+1)^2 + (a(2*n+1)+1)^2 = A038761(n)^2 + 2^2. - Hermann Stamm-Wilbrandt, Aug 31 2014

More terms from Harvey P. Dale, Feb 07 2011

More terms from Wesley Ivan Hurt, Aug 04 2014

A129289 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+73)^2 = y^2.

Original entry on oeis.org

0, 44, 95, 219, 455, 744, 1460, 2832, 4515, 8687, 16683, 26492, 50808, 97412, 154583, 296307, 567935, 901152, 1727180, 3310344, 5252475, 10066919, 19294275, 30613844, 58674480, 112455452, 178430735, 341980107, 655438583, 1039970712, 1993206308, 3820176192

Offset: 1

Mohamed Bouhamida, May 26 2007

nonn

Also values x of Pythagorean triples (x, x+73, y).
Corresponding values y of solutions (x, y) are in A160041.
lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (89+36*sqrt(2))/73 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (5907+1802*sqrt(2))/73^2 for n mod 3 = 0.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A160041, A129288, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A160042 (decimal expansion of (89+36*sqrt(2))/73), A160043 (decimal expansion of (5907+1802*sqrt(2))/73^2).

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(44+51*x+124*x^2-28*x^3-17*x^4-28*x^5)/((1-x)*(1-6*x^3+x^6)))); // G. C. Greubel, May 07 2018

Select[Range[0,100000],IntegerQ[Sqrt[#^2+(#+73)^2]]&] (* or *) LinearRecurrence[{1,0,6,-6,0,-1,1},{0,44,95,219,455,744,1460},70] (* Vladimir Joseph Stephan Orlovsky, Feb 02 2012 *)

{forstep(n=0, 100000000, [3 ,1], if(issquare(2*n^2+146*n+5329), print1(n, ",")))}

a(n) = 6*a(n-3) -a(n-6) +146 for n > 6; a(1)=0, a(2)=44, a(3)=95, a(4)=219, a(5)=455, a(6)=744.
G.f.: x*(44+51*x+124*x^2-28*x^3-17*x^4-28*x^5)/((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 73*A001652(k) for k >= 0.

Edited and two terms added by Klaus Brockhaus, May 04 2009

A157096 Consider all consecutive integer Pythagorean 11-tuples (X, X+1, X+2, X+3, X+4, X+5, Z-4, Z-3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.

Original entry on oeis.org

0, 55, 1260, 27715, 608520, 13359775, 293306580, 6439385035, 141373164240, 3103770228295, 68141571858300, 1496010810654355, 32844096262537560, 721074106965172015, 15830786256971246820, 347556223546402258075, 7630406131763878430880, 167521378675258923221335

Offset: 0

Charlie Marion, Mar 12 2009

In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0, first(1) = k*(2k+1) and, for n > 1, first(n) = (4k+2)*first(n-1) - first(n-2) + 2*k^2; e.g., if k=6, then first(2) = 2100 = 26*78 - 0 + 72.
In general, the first and last terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n > 0, let first(n) = (2k+1)*first(n-1) + 2k*last(n-1) + k and last(n) = (2k+2)*first(n-1) + (2k+1)*last(n-1) + 2k; e.g., if k=6 and n=2, then first(2) = 2100 = 13*78 + 12*90 + 6 and last(2) = 2274 = 14*78 + 13*90 + 12.
In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(n) = (k^(n+1)((1+sqrt((k+1)/k))^(2n+1) + (1-sqrt((k+1)/k))^(2n+1)) - 2*k)/4; e.g., if k=6 and n=2, then first(2) = 2100 = (6^3((1+sqrt((7/6))^5 + (1-sqrt(7/6))^5) - 2*6)/4.
In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction convergent to sqrt((k+1)/k), then the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(2n+1) = k*u(2n)*u(2n+1) and, for n > 0, first(2n) = (k+1)*e(2n-1)*e(2n); e.g., a(1) = 55 = 5*1*11 and a(2) = 1260 = 6*10*21.
In general, if first(n) is the first term of the n-th consecutive integer Pythagorean 2k+1-tuple, then lim_{n->inf} first(n+1)/first(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

			a(2)=55 since 55^2 + 56^2 + 57^2 + 58^2 + 59^2 + 60^2 = 61^2 + 62^2 + 63^2 + 64^2 + 65^2.

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.
W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.

G. C. Greubel, Table of n, a(n) for n = 0..740 (terms 0..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (23,-23,1).

Cf. A001652, A157084, A157088, A157092.

I:=[0, 55, 1260]; [n le 3 select I[n] else 23*Self(n-1) - 23*Self(n-2) + Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jun 09 2012

CoefficientList[Series[5*x*(x-11)/((x-1)*(x^2-22*x+1)),{x,0,20}],x] (* Vincenzo Librandi, Jun 09 2012 *)

x='x+O('x^50); concat([0], Vec(5*x*(x-11)/((x-1)*(x^2-22*x+1)))) \\ G. C. Greubel, Nov 04 2017

For n > 1, a(n) = 22*a(n-1) - a(n-2) + 50.
For n > 0, a(n) = 11*a(n-1) + 10*A157097(n-1) + 5.
a(n) = (5^(n+1)*((1+sqrt(6/5))^(2n+1) + (1-sqrt(6/5))^(2n+1)) - 2*5)/4.
Lim_{n->inf} a(n+1)/a(n) = 5(1+sqrt(6/5))^2 = 11+2*sqrt(30).
G.f.: 5*x*(x-11)/((x-1)*(x^2-22*x+1)). - Colin Barker, Jun 08 2012
a(n) = 23*a(n-1) - 23*a(n-2) + a(n-3). Vincenzo Librandi, Jun 09 2012
a(n) = 5*(-1/2+1/20*(11+2*sqrt(30))^(-n)*(5-sqrt(30)+(5+sqrt(30))*(11+2*sqrt(30))^(2*n))). - Colin Barker, Mar 03 2016

Terms a(15) onward added by G. C. Greubel, Nov 06 2017

A200993 Triangular numbers, T(m), that are two-thirds of another triangular number; T(m) such that 3T(m) = 2T(k) for some k.

Original entry on oeis.org

0, 10, 990, 97020, 9506980, 931587030, 91286021970, 8945098566040, 876528373449960, 85890835499530050, 8416425350580494950, 824723793521388975060, 80814515339745539060940, 7918997779501541438997070, 775980967875811315482651930, 76038215854050007375860892080

Offset: 0

Charlie Marion, Dec 15 2011

For n>1, a(n) = 98*a(n-1) - a (n-2) + 10. In general, for m>0, let b(n) be those triangular numbers such that for some triangular number c(n), (m+1)*b(n) = m*c(n). Then b(0) = 0, b(1)= A014105(m) and for n>1, b(n) = 2*A069129(m+1)*b(n-1) - b(n-2) + A014105(m).

Further, c(0) = 0, c(1) = A000384(m+1) and for n>1, c(n) = 2*A069129(m+1)*c(n-1) - c(n-2) + A000384(m+1).

			3*0 = 2*0.
3*10 = 2*15.
3*990 = 2*1485.
3*97020 = 2*145530.

Colin Barker, Table of n, a(n) for n = 0..500
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A001652, A029549, A053141, A075528, A200994-A201008.

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(10*x/((1-x)*(x^2-98*x+1)))); // G. C. Greubel, Jul 15 2018

LinearRecurrence[{99,-99,1},{0,10,990},20] (* Harvey P. Dale, Feb 25 2018 *)

concat(0, Vec(10*x/((1-x)*(1-98*x+x^2)) + O(x^40))) \\ Colin Barker, Mar 02 2016

G.f. 10*x / ((1-x)*(x^2-98*x+1)). - R. J. Mathar, Dec 20 2011
a(n) = 99*a(n-1)-99*a(n-2)+a(n-3) for n>2. - Colin Barker, Mar 02 2016
a(n) = (-10+(5-2*sqrt(6))*(49+20*sqrt(6))^(-n)+(5+2*sqrt(6))*(49+20*sqrt(6))^n)/96. - Colin Barker, Mar 07 2016

A129993 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x+199)^2 = y^2.

Original entry on oeis.org

0, 21, 504, 597, 704, 3441, 3980, 4601, 20540, 23681, 27300, 120197, 138504, 159597, 701040, 807741, 930680, 4086441, 4708340, 5424881, 23818004, 27442697, 31619004, 138821981, 159948240, 184289541, 809114280, 932247141, 1074118640

Offset: 1

Mohamed Bouhamida, Jun 14 2007

Also values x of Pythagorean triples (x, x+199, y).
Corresponding values y of solutions (x, y) are in A159548.
For the generic case x^2+(x+p)^2 = y^2 with p = 2*m^2-1 a (prime) number in A066436 see A118673 or A129836.
Lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
Lim_{n -> infinity} a(n)/a(n-1) = (201+20*sqrt(2))/199 for n mod 3 = {1, 2}.
Lim_{n -> infinity} a(n)/a(n-1) = (91443+58282*sqrt(2))/199^2 for n mod 3 = 0.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A159548, A066436, A118673, A118674, A129836, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A159549 (decimal expansion of (201+20*sqrt(2))/199), A159550 (decimal expansion of (91443+58282*sqrt(2))/199^2).

I:=[0,21,504,597,704,3441,3980]; [n le 7 select I[n] else Self(n-1) + 6*Self(n-3) - 6*Self(n-4) - Self(n-6) + Self(n-7): n in [1..50]]; // G. C. Greubel, Mar 31 2018

LinearRecurrence[{1,0,6,-6,0,-1,1},{0,21,504,597,704,3441,3980},30] (* Harvey P. Dale, Jun 03 2012 *)

{forstep(n=0, 500000000, [1, 3], if(issquare(2*n^2+398*n+39601), print1(n, ",")))};

a(n) = 6*a(n-3) - a(n-6) + 398 for n > 6; a(1)=0, a(2)=21, a(3)=504, a(4)=597, a(5)=704, a(6)=3441.
G.f.: x*(21+483*x+93*x^2-19*x^3-161*x^4-19*x^5) / ((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 199*A001652(k) for k >= 0.
a(1)=0, a(2)=21, a(3)=504, a(4)=597, a(5)=704, a(6)=3441, a(7)=3980, a(n)=a(n-1)+6*a(n-3)-6*a(n-4)-a(n-6)+a(n-7). - Harvey P. Dale, Jun 03 2012

Edited and two terms added by Klaus Brockhaus, Apr 14 2009

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A106521 Numbers m such that Sum_{k=0..10} (m+k)^2 is a square.

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A118676 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+79)^2 = y^2.

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A129298 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+89)^2 = y^2.

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A076218 Numbers n such that 2*n^2 - 3*n + 1 is a square.

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A118675 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+47)^2 = y^2.

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A124124 Nonnegative integers n such that 2n^2 + 2n - 3 is square.

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A076218 Numbers n such that 2n^2 - 3n + 1 is a square.

A200993 Triangular numbers, T(m), that are two-thirds of another triangular number; T(m) such that 3T(m) = 2T(k) for some k.