A078522 -id:A078522 - cp's OEIS frontend

A327248 Squarefree part of A078522(n+1).

6, 210, 1785, 60639, 915530, 184030, 14066106, 80753867670, 10973017315470, 372759573255306, 351745902037915, 11949006236698685, 86466986871277074, 122261486084598, 43869141307765893, 35803482505852454889891, 2162247909473892250092390, 73452778286546376583337010

Offset: 1

Tomohiro Yamada, Sep 15 2019

nonn

Also the squarefree part of (A001653(n+1)^2-1)/2 or of A002315(n)^2-1

Walsh shows that the system of simultaneous Pell equations x^2 - d*y^2 = z^2 - 2*d*y^2 = 1 has solutions in positive integers x, y, z if and only if d belongs to this sequence and, under the abc conjecture, this sequences grows exponentially.

			a(2) = 210 since A078522(3) = 840 = 210 * 2^2.

P. G. Walsh, On integer solutions to x^2 - dy^2 = 1, z^2 - 2dy^2 = 1, Acta Arithmetica 82 (1997), 69-76.

Cf. A007913, A001653, A002315, A078522.

a(n)={local(z=1+quadgen(8)); core(imag(z^(2*n+1))^2-1)}

a(n) = A007913(A078522(n+1)).

Missing a(11) inserted and more terms from Georg Fischer, Mar 02 2023

A046176 Indices of square numbers that are also hexagonal.

Original entry on oeis.org

1, 35, 1189, 40391, 1372105, 46611179, 1583407981, 53789260175, 1827251437969, 62072759630771, 2108646576008245, 71631910824649559, 2433376321462076761, 82663163018885960315, 2808114166320660573949

Offset: 1

Eric W. Weisstein

Bisection (even part) of Chebyshev sequence with Diophantine property.
(3*b(n))^2 - 2*(2*a(n+1))^2 = 1 with companion sequence b(n) = A077420(n), n >= 0.
Sequence also refers to inradius of primitive Pythagorean triangles with consecutive legs, odd followed by even. - Lekraj Beedassy, Apr 23 2003
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> oo} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12*sqrt(2). - Ant King, Nov 08 2011
Integers of the form sqrt((m+1)*(2*m+1)). The corresponding values of m form A078522. Subsequence of A284876. - Jonathan Sondow, Apr 07 2017

M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Tanya Khovanova, Recursive Sequences.
E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 4, k=1, t=3.
Serge Perrine, About the diophantine equation z^2 = 32y^2 - 16, SCIREA Journal of Mathematics (2019) Vol. 4, Issue 5, 126-139.
Maciej Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT] (2008) , v_n in Theorem 5.6.
P. H. van der Kamp, Global classification of two-component..., Found. Comput. Math. 9 (5) (2009) 559-597 near Eq. (4.7).
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A000129, A008844, A046177, A078522, A284876.

Cf. A001109, A001110 (partial sums).

List([0..20], n-> Lucas(2,-1, 4*n-2)[1]/2 ); # G. C. Greubel, Jan 13 2020

I:=[1,35]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011

seq( simplify(ChebyshevU(2*(n-1), 3)), n = 1..20); # G. C. Greubel, Jan 13 2020

LinearRecurrence[{34, -1}, {1, 35}, 15] (* Ant King, Nov 08 2011 *)
Fibonacci[4*Range[20] -2, 2]/2 (* G. C. Greubel, Jan 13 2020 *)

vector(21, n, polchebyshev(2*(n-1), 2, 3) ) \\ G. C. Greubel, Jan 13 2020

[lucas_number1(4*n-2, 2,-1)/2 for n in (1..20)] # G. C. Greubel, Jan 13 2020

a(n) = 34*a(n-1) - a(n-2); a(0)=-1, a(1)=1.
a(n+1) = S(2*n, 6) = S(n, 34) + S(n-1, 34), n >= 1, with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(n, 34) = A029547(n).
G.f.: x*(1+x)/(1-34*x+x^2).
a(n+1) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*6^(2*(n-k)), n >= 0.
a(n) = A001109(2n+1). - Lekraj Beedassy, Apr 23 2003
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(f(a(n-1),3),3). - Marcos Carreira, Dec 27 2006
From Antonio Alberto Olivares, Mar 22 2008: (Start)
a(n) = (sqrt(2)/8)*(3 + 2*sqrt(2))*(17 + 12*sqrt(2))^(n-1) - (sqrt(2)/8)*(3 - 2*sqrt(2))*(17 - 12*sqrt(2))^(n-1).
a(n) = (sqrt(2)/8)*( (17+12*sqrt(2))^(n-1/2) - (17-12*sqrt(2))^(n-1/2) ).
a(n) = (sqrt(2)/8)*( (3+2*sqrt(2))^(2n-1) - (3-2*sqrt(2))^(2n-1) ).
a(n) = (sqrt(2)/8)*( (1+sqrt(2))^(4n-2) - (1-sqrt(2))^(4n-2) ).
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3). (End)
a(n+1) = 17*a(n) + 6*sqrt(8*a(n)^2+1) for n >= 0. - Richard Choulet, May 01 2009
a(n) = b such that (-1)^(n+1) * Integral_{x=-Pi/2..Pi/2} cos((2*n-1)*x)/(3-sin(x)) dx = c + b*log(2). - Francesco Daddi, Aug 01 2011
a(n) are the nonzero integer square roots of A227970. - Richard R. Forberg, Aug 01 2013
a(n) = y/5, where y are solutions to: y^2 = 2x^2 - x - 3. - Richard R. Forberg, Nov 24 2013
a(n) = sqrt((A078522(n)+1)*(2*A078522(n)+1)). - Jonathan Sondow, Apr 07 2017
a(n) = Pell(4*n-2)/2. - G. C. Greubel, Jan 13 2020
a(n) = A001653(n)*A002315(n). - Gerry Martens, Mar 23 2024

Chebyshev comments from Wolfdieter Lang, Nov 29 2002

A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

Original entry on oeis.org

1, 25, 841, 28561, 970225, 32959081, 1119638521, 38034750625, 1292061882721, 43892069261881, 1491038293021225, 50651409893459761, 1720656898084610641, 58451683124983302025, 1985636569351347658201, 67453191674820837076801, 2291422880374557112953025

Offset: 0

N. J. A. Sloane

Numbers simultaneously square and centered square. E.g., a(1)=25 because 25 is the fourth centered square number and the fifth square number. - Steven Schlicker, Apr 24 2007
Solutions to A007913(x)=A007913(2x-1). - Benoit Cloitre, Apr 07 2002
From Ant King, Nov 09 2011: (Start)
Indices of positive hexagonal numbers that are also perfect squares.
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12 * sqrt(2).
(End)
Also indices of hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015
Also positive integers x in the solutions to 4*x^2 - 8*y^2 - 2*x + 8*y - 2 = 0, the corresponding values of y being A253826. - Colin Barker, Jan 25 2015
Squares that are sum of two consecutive squares: y^2 = (k + 1)^2 + k^2 is equivalent to x^2 - 2*y^2 = -1 with x = 2*k + 1. - Jean-Christophe Hervé, Nov 11 2015
Squares in the main diagonal of the natural number array, A000027. - Clark Kimberling, Mar 12 2023

			From _Ravi Kumar Davala_, May 26 2013: (Start)
A001333(0)=1, A001333(4)=17, A001333(8)=577, A000129(0)=0, A000129(2)=2, A000129(4)=12, A000129(8)=408 so clearly
a(n+m)=A001333(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)), with m=1,2 is true.
A002203(0)=2, A002203(4)=34, A002203(8)=1154 so clearly
a(n+m)=(1/2)*A002203(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)) is true for m=1,2
a(n+1)*a(n-1) = (a(n)+4)^2 , with n=1 is 841*1=(25+4)^2, for n=2 , 28561*25=(841+4)^2.
(End)
1 = 1 + 0, 25 = 16 + 9, 841 = 29^2 = 21^2 + 20^2 = 441 + 400.

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

Muniru A Asiru, Table of n, a(n) for n = 0..100
Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes, and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Steven C. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5, December 2011
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000290, A001844, A007913, A000384, A046177, A016754, A253826.

a := [1, 25, 841];; for i in [4..10^2] do a[i] := 35*a[i-1] - 35*a[i-2] + a[i-3]; od; a;  # Muniru A Asiru, Jan 17 2018

I:=[1,25,841]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jan 20 2018

CP := n -> 1+1/2*4*(n^2-n): N:=10: u:=3: v:=1: x:=4: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+8*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp; # Steven Schlicker, Apr 24 2007

LinearRecurrence[{35, -35, 1}, {1, 25, 841}, 15] (* Ant King, Nov 09 2011 *)
CoefficientList[Series[(1 - 10 x + x^2) / ((1 - x) (1 - 34 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Jan 20 2018 *)

a(n)=if(n<0,0,sqr(subst(poltchebi(n+1)+poltchebi(n),x,3)/4))

vector(40, n, n--; (([5, 2; 2, 1]^n)[1, 1])^2) \\ Altug Alkan, Nov 11 2015

From Benoit Cloitre, Jan 19 2003: (Start)
a(n) = A078522(n) + 1.
a(n) = ceiling(A*B^n) where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2). (End)
G.f.: (1-10x+x^2)/((1-x)(1-34x+x^2)).
a(n) = ceiling(A046176(n)/sqrt(2)). - Helge Robitzsch (hrobi(AT)math.uni-goettingen.de), Jul 28 2000
a(n+1) = 17*a(n) - 4 + 12*sqrt(2*a(n)^2 - a(n)). - Richard Choulet, Sep 14 2007
Define x(n) + y(n)*sqrt(8) = (4+sqrt(8))*(3+sqrt(8))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+4*(s(n)^2 - s(n))). - Steven Schlicker, Apr 24 2007
From Ant King, Nov 09 2011: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
a(n) = 34*a(n-1) - a(n-2) - 8.
a(n) = 1/8 * ((1 + sqrt(2))^(4*n-2) + (1 - sqrt(2))^(4*n-2) + 2).
a(n) = ceiling((1/8) * (1 + sqrt(2))^(4*n-2)). (End)
From Ravi Kumar Davala, May 26 2013: (Start)
a(n+2) = 577*a(n) - 144 + 408*sqrt(2*a(n)^2 - a(n)).
a(n+m) = A001333(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+m) = (1/2)*A002203(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+1)*a(n-1) = (a(n)+4)^2. (End)
a(n) = A001652(n)^2 + A046090(n)^2. - César Aguilera, Jan 15 2018
Limit_{n -> infinity} a(n)/a(n-1) = A156164. - César Aguilera, Jan 28 2018
sqrt(2*a(n))-1 = A002315(n). - Ezhilarasu Velayutham, Apr 05 2019
4*a(n) = 1 +3*A077420(n). - R. J. Mathar, Mar 05 2024
Product_{n>=0} (1 + 4/a(n)) = 2*sqrt(2) + 3 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Entry edited by N. J. A. Sloane, Sep 14 2007

A157014 Expansion of x(1-x)/(1 - 22x + x^2).

Original entry on oeis.org

1, 21, 461, 10121, 222201, 4878301, 107100421, 2351330961, 51622180721, 1133336644901, 24881784007101, 546265911511321, 11992968269241961, 263299036011811821, 5780585823990618101, 126909589091781786401, 2786230374195208682721, 61170158643202809233461

Offset: 1

Paul Weisenhorn, Feb 21 2009

This sequence is part of a solution of a general problem involving 2 equations, three sequences a(n), b(n), c(n) and a constant A:
    A    * c(n)+1 = a(n)^2,
   (A+1) * c(n)+1 = b(n)^2, where solutions are given by the recurrences:
a(1) = 1, a(2) = 4*A+1, a(n) = (4*A+2)*a(n-1)-a(n-2) for n>2, resulting in a(n) terms 1, 4*A+1, 16*A^2+12*A+1, 64*A^3+80*A^2+24*A+1, ...;
b(1) = 1, b(2) = 4*A+3, b(n) = (4*A+2)*b(n-1)-b(n-2) for n>2, resulting in b(n) terms 1, 4*A+3, 16*A^2+20*A+5, 64*A^3+112*A^2+56*A+7, ...;
c(1) = 0, c(2) = 16*A+8, c(3) = (16*A^2+16*A+3)*c(2), c(n) = (16*A^2+16*A+3) * (c(n-1)-c(n-2)) + c(n-3) for n>3, resulting in c(n) terms 0, 16*A+8, 256*A^3+384*A^2+176*A+24, 4096*A^5 + 10240*A^4 + 9472*A^3 + 3968*A^2 + 736*A + 48, ... .
A157014 is the a(n) sequence for A=5.
For other A values the a(n), b(n) and c(n) sequences are in the OEIS:
  A   a-sequence    b-sequence     c-sequence
  1     A001653       A002315(n-1)   A078522
  2     A072256       A054320(n-1)   A045502(n-1)
  3     A001570       A028230        A059989(n-1)
  4     A007805       A049629(n-1)   A157459
  5  -> A157014 <-    A133283        A157460
  6     A153111       A157461        A157874
  7     A157877       A157878        A157879
  8     A077420(n-1)  A046176        A157880
  9     A097315(n-1)  A097314(n-1)   A157881
Positive values of x (or y) satisfying x^2 - 22xy + y^2 + 20 = 0. - Colin Barker, Feb 19 2014
From Klaus Purath, Apr 22 2025: (Start)
Nonnegative solutions to the Diophantine equation 5*b(n)^2 - 6*a(n)^2 = -1. The corresponding b(n) are A133283(n). Note that (b(n+1)^2 - b(n)*b(n+2))/4 = 6 and (a(n)*a(n+2) - a(n+1)^2)/4 = 5.
(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A077421(n-1) = Lucas U(22,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.
a(n)=(t(i+2*n-1) + t(i))/(t(i+n) + t(i+n-1)) as long as t(i+n) + t(i+n-1) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 21*t(i+2) - 21*t(i+1) + t(i) or t(i+2) = 22*t(i+1) - t(i) without regard to initial values and including this sequence itself. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (22,-1).

Cf. similar sequences listed in A238379.

a:=[1,21];; for n in [3..20] do a[n]:=22*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2020

I:=[1,21]; [n le 2 select I[n] else 22*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2014

seq( simplify(ChebyshevU(n-1,11) - ChebyshevU(n-2,11)), n=1..20); # G. C. Greubel, Jan 14 2020

CoefficientList[Series[(1-x)/(1-22x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Feb 21 2014 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A041049 *)
a[30, 20] (* Gerry Martens, Jun 07 2015 *)
Table[ChebyshevU[n-1, 11] - ChebyshevU[n-2, 11], {n,20}] (* G. C. Greubel, Jan 14 2020 *)

Vec((1-x)/(1-22*x+x^2)+O(x^20)) \\ Charles R Greathouse IV, Sep 23 2012

[chebyshev_U(n-1,11) - chebyshev_U(n-2,11) for n in (1..20)] # G. C. Greubel, Jan 14 2020

G.f.: x*(1-x)/(1-22*x+x^2).
a(1) = 1, a(2) = 21, a(n) = 22*a(n-1) - a(n-2) for n>2.
5*A157460(n)+1 = a(n)^2 for n>=1.
6*A157460(n)+1 = A133283(n)^2 for n>=1.
a(n) = (6+sqrt(30)-(-6+sqrt(30))*(11+2*sqrt(30))^(2*n))/(12*(11+2*sqrt(30))^n). - Gerry Martens, Jun 07 2015
a(n) = ChebyshevU(n-1, 11) - ChebyshevU(n-2, 11). - G. C. Greubel, Jan 14 2020

Edited by Alois P. Heinz, Sep 09 2011

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

Original entry on oeis.org

1, 14, 209, 3121, 46606, 695969, 10392929, 155197966, 2317576561, 34608450449, 516809180174, 7717529252161, 115246129602241, 1720974414781454, 25699370092119569, 383769576967012081, 5730844284413061646, 85578894689228912609, 1277952576054020627489

Offset: 1

Paul Weisenhorn, May 23 2009

This summarizes the case C=13 of common solutions to C*k+1=A^2, (C+4)*k+1=B^2.
The 2 equations are equivalent to the Pell equation x^2-C*(C+4)*y^2=1,
with x=(C*(C+4)*k+C+2)/2; y=A*B/2 and with smallest values x(1) = (C+2)/2, y(1)=1/2.
Generic recurrences are:
A(j+2)=(C+2)*A(j+1)-A(j) with A(1)=1; A(2)=C+1.
B(j+2)=(C+2)*B(j+1)-B(j) with B(1)=1; B(2)=C+3.
k(j+3)=(C+1)*(C+3)*( k(j+2)-k(j+1) )+k(j) with k(1)=0; k(2)=C+2; k(3)=(C+1)*(C+2)*(C+3).
x(j+2)=(C^2+4*C+2)*x(j+1)-x(j) with x(1)=(C+2)/2; x(2)=(C^2+4*C+1)*(C+2)/2;
Binet-type of solutions of these 2nd order recurrences are:
R=C^2+4*C; S=C*sqrt(R); T=(C+2); U=sqrt(R); V=(C+4)*sqrt(R);
A(j)=((R+S)*(T+U)^(j-1)+(R-S)*(T-U)^(j-1))/(R*2^j);
B(j)=((R+V)*(T+U)^(j-1)+(R-V)*(T-U)^(j-1))/(R*2^j);
x(j)+sqrt(R)*y(j)=((T+U)*(C^2*4*C+2+(C+2)*sqrt(R))^(j-1))/2^j;
k(j)=(((T+U)*(R+2+T*U)^(j-1)+(T-U)*(R+2-T*U)^(j-1))/2^j-T)/R. [Paul Weisenhorn, May 24 2009]
.C -A----- -B----- -k-----
01 A001519 A002878 A058038
02 A001653 A002315 A045899/2
03 A004253 A030221 A160695
04 A001653 A002315 A078522/4
05 A049685 A033890 A161582
06 A070997 A057080 A159683/2
07 A070998 A057081 A161585
08 A072256 A054320 A045502/4
09 A078922 A097783 A161586
10 A077417 A077416 A159681/2
11 A085260 A126816 A161588
12 A001570 A028230 A059989/4
13 A160682 A161591 A161584
14 A157456 A159678 A159679/2
15 A161595 A161599 A161583
16 A007805 A049629 A157459/4
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(13)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]
Positive values of x (or y) satisfying x^2 - 15xy + y^2 + 13 = 0. - Colin Barker, Feb 11 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
J.-C. Novelli, J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Index entries for the Pell equation
Index entries for linear recurrences with constant coefficients, signature (15,-1).

Cf. similar sequences listed in A238379.

I:=[1,14]; [n le 2 select I[n] else 15*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 12 2014

LinearRecurrence[{15,-1},{1,14},20] (* Harvey P. Dale, Oct 08 2012 *)
CoefficientList[Series[(1 - x)/(1 - 15 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)

a(n) = round((2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221)) \\ Colin Barker, Jul 25 2016

a(n) = 15*a(n-1)-a(n-2).
G.f.: (1-x)*x/(1-15*x+x^2).
a(n) = (2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221). - Colin Barker, Jul 25 2016

Edited, extended by R. J. Mathar, Sep 02 2009

First formula corrected by Harvey P. Dale, Oct 08 2012

A278310 Numbers m such that T(m) + 3*T(m+1) is a square, where T = A000217.

Original entry on oeis.org

3, 143, 4899, 166463, 5654883, 192099599, 6525731523, 221682772223, 7530688524099, 255821727047183, 8690408031080163, 295218051329678399, 10028723337177985443, 340681375412721826703, 11573138040695364122499, 393146012008229658338303, 13355391270239113019379843

Offset: 1

Bruno Berselli, Nov 17 2016

Equivalently, both m+1 and 2*m+3 are squares for nonnegative m.
Corresponding triangular numbers T(m): 6, 10296, 12002550, 13855048416, 15988853699286, 18451128064030200, 21292585958400815526, ...
Square roots of T(m) + 3*T(m+1) are listed by A082405 (after 0).
Negative values of m for which T(m) + 3*T(m+1) is a square: -1, -2, -26, -842, -28562, -970226, -32959082, ...

			3 is in the sequence because T(3) + 3*T(4) = 6 + 3*10 = 6^2.
For n=5 is a(5) = 5654883, therefore floor(sqrt(5654883)) = 2377 = A182189(5) - 2 = 2379 - 2.

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Subsequence of A000466.
Cf. A000217, A001109, A029546, A046176, A082405, A156164, A182189.
Cf. A278438: numbers m such that T(m) + 2*T(m+1) is a square.
Cf. A078522: numbers m such that 3*T(m) + T(m+1) is a square.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: A084703 (k=-1), A076218 (k=3), this sequence (k=-5).

Iv:=[3,143]; [n le 2 select Iv[n] else 34*Self(n-1)-Self(n-2)+40: n in [1..20]];

P:=proc(q) local n; for n from 3 to q do if type(sqrt(2*n^2+5*n+3),integer) then print(n); fi; od; end: P(10^9); # Paolo P. Lava, Nov 18 2016

Table[((1 + Sqrt[2])^(4 n) + (1 - Sqrt[2])^(4 n))/8 - 5/4, {n, 1, 20}]
RecurrenceTable[{a[1] == 3, a[2] == 143, a[n] == 34 a[n - 1] - a[n - 2] + 40}, a, {n, 1, 20}]
LinearRecurrence[{35, -35, 1}, {3, 143, 4899}, 50] (* G. C. Greubel, Nov 20 2016 *)

Vec(x*(3 + 38*x - x^2)/((1 - x)*(1 - 34*x + x^2)) + O(x^50)) \\ G. C. Greubel, Nov 20 2016

def A278310():
    a, b = 3, 143
    yield a
    while True:
        yield b
        a, b = b, 34*b - a + 40
a = A278310(); print([next(a) for  in range(18)]) # _Peter Luschny, Nov 18 2016

O.g.f.: x*(3 + 38*x - x^2)/((1 - x)*(1 - 34*x + x^2)).
E.g.f.: (exp((1-sqrt(2))^4*x) + exp((1+sqrt(2))^4*x) - 10*exp(x))/8 + 1.
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n>3.
a(n) = 34*a(n-1) - a(n-2) + 40 for n>2.
a(n) = a(-n) = ((1 + sqrt(2))^(4*n) + (1 - sqrt(2))^(4*n))/8 - 5/4.
a(n) = 4*A001109(n)^2 - 1.
a(n) = -A029546(n) + 38*A029546(n-1) + 3*A029546(n-2) for n>1.
Lim_{n -> infinity} a(n)/a(n-1) = A156164.
Floor(sqrt(a(n))) = A182189(n) - 2.
a(n) - a(n-1) = 4*A046176(n) for n>1.

A377016 Semiperimeter of the unique primitive Pythagorean triple whose short leg is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

Original entry on oeis.org

1, 28, 861, 28680, 970921, 32963140, 1119662181, 38034888528, 1292062686481, 43892073946540, 1491038320325421, 50651410052600280, 1720656899012149561, 58451683130389395028, 1985636569382856677301, 67453191675004485098400, 2291422880375627492063521, 77840924741066359629967420

Offset: 0

Miguel-Ángel Pérez García-Ortega, Oct 13 2024

a(0) = 1 is included by convention. This corresponds to the Pythagorean triple 1^2 + 0^2 = 1^2.

			For n=2, the short leg is A002315(2) = 41 and the hypotenuse is A008844(n) = 841 so the semiperimeter is then a(2) = (41 + 840 + 841)/2 = 861.

Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2024.

Index entries for linear recurrences with constant coefficients, signature (41,-246,246,-41,1).

Cf. A002315, A078522, A008844.

s[n_]:=s[n]=Module[{a, b},a=((1+Sqrt[2])^(2n+1)-(Sqrt[2]-1)^(2n+1))/2;b=(a^2-1)/2;{(a+2b+1)/2}];semis={};Do[semis=Join[semis,FullSimplify[s[n]]],{n,0,17}];semis

Vec((1 - 13*x - 41*x^2 + 21*x^3)/((1 - 34*x + x^2)*(1 - 6*x + x^2)*(1 - x)) + O(x^20)) \\ Andrew Howroyd, Oct 14 2024

a(n) = (A002315(n) + 2*A008844(n) - 1)/2.

G.f.: (1 - 13*x - 41*x^2 + 21*x^3)/((1 - 34*x + x^2)*(1 - 6*x + x^2)*(1 - x)). - Andrew Howroyd, Oct 14 2024

A377017 Area of the unique primitive Pythagorean triple whose short leg is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

Original entry on oeis.org

0, 84, 17220, 3412920, 675761016, 133797385260, 26491207202460, 5245125232676784, 1038508304885968560, 205619399242324129860, 40711602541676078766516, 8060691683852625858745320, 1595976241800278270688414120, 315995235184771245126273789084, 62565460590342906257639745449100

Offset: 0

Miguel-Ángel Pérez García-Ortega, Oct 13 2024

nonn

a(0)=0 is included by convention. This corresponds to the Pythagorean triple 1^2 + 0^2 = 1^2.

All terms in this sequence are divisible by 84.

			For n=2, the short leg is A002315(2) = 41 and the long leg is A008844(2)-1 = 840 so the area is then a(2) = 41*840/2 = 17220.

Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2024.

Index entries for linear recurrences with constant coefficients, signature (204,-1190,204,-1).

Cf. A377016, A002315, A078522, A008844.

s[n_]:=s[n]=Module[{a, b}, a=((1+Sqrt[2])^(2n+1)-(Sqrt[2]-1)^(2n+1))/2; b=(a^2-1)/2; {(a*b)/2}]; areas={}; Do[areas=Join[areas, FullSimplify[s[n]]], {n, 0, 17}]; areas

a(n) = A002315(n)*(A008844(n)-1)/2.

G.f.: 84*x*(1 + x)/((1 - 198*x + x^2)*(1 - 6*x + x^2)). - Andrew Howroyd, Oct 14 2024

A284876 Positive integers that are square roots of products a(a+d)(a+2*d) with coprime a > 0, d >= 0.

Original entry on oeis.org

1, 35, 120, 1189, 1547, 1560, 2737, 4080, 8400, 13175, 24360, 29520, 31080, 39997, 40391, 52633, 62279, 65773, 80520, 93023, 131040, 133055, 133560, 185640, 212219, 240240, 241345, 379680, 385440, 393805, 399960, 434231, 449497, 471240, 510229, 555360, 585395

Offset: 1

Jonathan Sondow, Apr 05 2017

nonn

The main entry for this sequence is A284666, formed by the triples a, a+d, a+2*d. The pairs a, d form A284874.

sqrt((1+d)*(1+2*d)) is a member if and only if d is in A078522. The values of sqrt((1+d)*(1+2*d)) form the subsequence A046176.

			gcd(1,24)=1 and 1*(1+24)*(1+2*24) = 25*49 = (5*7)^2, so 5*7 = 35 is a member.
gcd(18,7)=1 and 18*(18+7)*(18+2*7) = 18*25*32 = 9*25*64 = (3*5*8)^2, so 3*5*8 = 120 is in the sequence.

Giovanni Resta, Table of n, a(n) for n = 1..416 (terms < 10^9)

Cf. A046176, A078522, A284666, A284874.

nn = 50000; t = {};
p[a_, b_, c_] := a b c; Do[
If[p[a, a + d, a + 2 d] <= 2 nn^2 && GCD[a, d] == 1 &&
   IntegerQ[Sqrt[p[a, a + d, a + 2 d]]],
  AppendTo[t, Sqrt[p[a, a + d, a + 2 d]]]], {a, 1, nn}, {d, 0, nn}]; Sort[t]

is(n,s)={!fordiv(n*=n,a,a^3>n && return;issquare(n\a*8+a^2,&s) && (s-=3*a)%4==0 && gcd(s\4,a)==1 && break)} \\ M. F. Hasler, Apr 06 2017

a(k+1)^2 = A284666(3*k+1)*A284666(3*k+2)*A284666(3*k+3) = A284874(2*k+1)*(A284874(2*k+1) + A284874(2*k+2))*(A284874(2*k+1) + 2*A284874(2*k+2)) for k >= 0.

a(19)-a(37) from Giovanni Resta, Apr 06 2017

A284666 List of 3-term arithmetic progressions of coprime positive integers whose product is a square.

Original entry on oeis.org

1, 1, 1, 1, 25, 49, 18, 25, 32, 1, 841, 1681, 49, 169, 289, 50, 169, 288, 49, 289, 529, 128, 289, 450, 98, 625, 1152, 289, 625, 961, 800, 841, 882, 162, 1681, 3200, 288, 1369, 2450, 529, 1369, 2209, 1, 28561, 57121, 49, 5329, 10609, 961, 1681, 2401, 289, 2809, 5329

Offset: 1

Jonathan Sondow, Mar 31 2017

This is a 3-column table read by rows a, a+d, a+2*d. Each row has product a square. The rows are ordered by the products. The square roots of the products form A284876, which contains A046176. The pairs a,d form A284874.
Goldbach proved that a product of 3 consecutive positive integers is never a square.
Euler proved that a product of 4 consecutive positive integers is never a square.
Erdos and Selfridge (1975) proved that a product of 2 or more consecutive positive integers is never a square or a higher power.
Saradha (1998) proved that 18, 25, 32 is the only arithmetic progression a, a+d, ..., a+(k-1)*d whose product is a square if a>=1, 1=3 with gcd(a,d)=1. In 1997 she showed that the product is not a square or a higher power if a>=1, 1=3 with gcd(a,d)=1.
(1, 1+d, 1+2*d) is in the table if and only if d is in A078522. - Robert Israel, Apr 05 2017 - Jonathan Sondow, Apr 06 2017

			18*(18+7)*(18+2*7) = 18*25*32 = 9*25*64 = (3*5*8)^2 and gcd(18,25,32) = 1, so 18,25,32 is in the sequence.

Giovanni Resta, Table of n, a(n) for n = 1..1248 (triples with product < 10^18)
P. Erdős and J.L. Selfridge, The product of consecutive integers is never a power, Illinois J. Math., 19 (1975), 292-301.
N. Saradha, On perfect powers in products with terms from arithmetic progressions, Acta Arith., 82 (1997), 147-172.
N. Saradha, Squares in products with terms in an arithmetic progression, Acta Arith., 86 (1998), 27-43.

Cf. A046176, A078522, A284874, A284876.

N:= 10^11: # to get all triples where the product <= N
Res:= [1,0]:
for a from 1 to floor(N^(1/3)) do
  for d from 1 while a*(a+d)*(a+2*d) <= N do
     if igcd(a,d) = 1 and issqr(a*(a+d)*(a+2*d)) then
       Res:= Res, [a,d]
     fi
  od
od:
Res:= sort([Res], (s,t) -> s[1]*(s[1]+s[2])*(s[1]+2*s[2]) <= t[1]*(t[1]+t[2])*(t[1]+2*t[2])):
map(t -> (t[1],t[1]+t[2],t[1]+2*t[2]), Res); # Robert Israel, Apr 05 2017

nn = 50000; t = {};
p[a_, b_, c_] := a *b*c; Do[
If[p[a, a + d, a + 2 d] <= 2 nn^2 && GCD[a, d] == 1 &&
   IntegerQ[Sqrt[p[a, a + d, a + 2 d]]],
  AppendTo[t, {a, a + d, a + 2 d}]], {a, 1, nn}, {d, 0, nn}];
Sort[t, p[#1[[1]], #1[[2]], #1[[3]]] <
    p[#2[[1]], #2[[2]], #2[[3]]] &] // Flatten

a(3*k+1) = A284874(2*k+1) and a(3*k+2) = A284874(2*k+1)+A284874(2*k+2) and a(3*k+3) = A284874(2*k+1)+2*A284874(2*k+2) and a(3*k+1)*a(3*k+2)*a(3*k+3) = A284876(k+1)^2 for k>=0.

cp's OEIS Frontend

A327248 Squarefree part of A078522(n+1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

Extensions

A046176 Indices of square numbers that are also hexagonal.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

PARI

Formula

Extensions

A157014 Expansion of x*(1-x)/(1 - 22*x + x^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A160682 The list of the A values in the common solutions to 13*k+1 = A^2 and 17*k+1 = B^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A278310 Numbers m such that T(m) + 3*T(m+1) is a square, where T = A000217.

Views

Author

A157014 Expansion of x(1-x)/(1 - 22x + x^2).

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

A284876 Positive integers that are square roots of products a(a+d)(a+2*d) with coprime a > 0, d >= 0.