A001652 -id:A001652 - cp's OEIS frontend

A046729 Expansion of 4x/((1+x)(1-6*x+x^2)).

0, 4, 20, 120, 696, 4060, 23660, 137904, 803760, 4684660, 27304196, 159140520, 927538920, 5406093004, 31509019100, 183648021600, 1070379110496, 6238626641380, 36361380737780, 211929657785304, 1235216565974040, 7199369738058940, 41961001862379596, 244566641436218640

Offset: 0

N. J. A. Sloane

Related to Pythagorean triples: alternate terms of A001652 and A046090.
Even-valued legs of nearly isosceles right triangles: legs differ by 1. 0 is smaller leg of degenerate triangle with legs 0 and 1 and hypotenuse 1. - Charlie Marion, Nov 11 2003
The complete (nearly isosceles) primitive Pythagorean triple is given by {a(n), a(n)+(-1)^n, A001653(n)}. - Lekraj Beedassy, Feb 19 2004
Note also that A046092 is the even leg of this other class of nearly isosceles Pythagorean triangles {A005408(n), A046092(n), A001844(n)}, i.e., {2n+1, 2n(n+1), 2n(n+1)+1} where longer sides (viz. even leg and hypotenuse) are consecutive. - Lekraj Beedassy, Apr 22 2004
Union of even terms of A001652 and A046090. Sum of legs of primitive Pythagorean triangles is A002315(n) = 2*a(n) + (-1)^n. - Lekraj Beedassy, Apr 30 2004

			[1,0,1]*[1,2,2; 2,1,2; 2,2,3]^0 gives (degenerate) primitive Pythagorean triple [1, 0, 1], so a(0) = 0. [1,0,1]*[1,2,2; 2,1,2; 2,2,3]^7 gives primitive Pythagorean triple [137903, 137904, 195025] so a(7) = 137904.
G.f. = 4*x + 20*x^2 + 120*x^3 + 696*x^4 + 4060*x^5 + 23660*x^6 + ...

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
W. Sierpiński, Pythagorean triangles, Dover Publications, Inc., Mineola, NY, 2003, p. 17. MR2002669.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).
Index entries for two-way infinite sequences

Cf. A000129, A001109, A001333, A001652, A002315, A029549, A046090.

Cf. A046727, A047235, A084158, A084159, A089499, A114336.

[4*Floor(((Sqrt(2)+1)^(2*n+1)-(Sqrt(2)-1)^(2*n+1)-2*(-1)^n) / 16): n in [0..30]]; // Vincenzo Librandi, Jul 29 2019

LinearRecurrence[{5,5,-1}, {0,4,20}, 25] (* Vincenzo Librandi, Jul 29 2019 *)

a(n)=n%2+(real((1+quadgen(8))^(2*n+1))-1)/2

a(n)=if(n<0,-a(-1-n),polcoeff(4*x/(1+x)/(1-6*x+x^2)+x*O(x^n),n))

[(lucas_number2(2*n+1,2,-1) -2*(-1)^n)/4 for n in range(41)] # G. C. Greubel, Feb 11 2023

a(n) = ((1+sqrt(2))^(2n+1) + (1-sqrt(2))^(2n+1) + 2*(-1)^(n+1))/4.
a(n) = A089499(n)*A089499(n+1).
a(n) = 4*A084158(n). - Lekraj Beedassy, Jul 16 2004
a(n) = ceiling((sqrt(2)+1)^(2*n+1) - (sqrt(2)-1)^(2*n+1) - 2*(-1)^n)/4. - Lambert Klasen (Lambert.Klasen(AT)gmx.net), Nov 12 2004
a(n) is the k-th entry among the complete near-isosceles primitive Pythagorean triple A114336(n), where k = (3*(2n-1) - (-1)^n)/2, i.e., a(n) = A114336(A047235(n)), for positive n. - Lekraj Beedassy, Jun 04 2006
a(n) = A046727(n) - (-1)^n = 2*A114620(n). - Lekraj Beedassy, Aug 14 2006
From George F. Johnson, Aug 29 2012: (Start)
2*a(n)*(a(n) + (-1)^n) + 1 = (A000129(2*n+1))^2;
n > 0, 2*a(n)*(a(n) + (-1)^n) + 1 = ((a(n+1) - a(n-1))/4)^2, a perfect square.
a(n+1) = (3*a(n) + 2*(-1)^n) + 2*sqrt(2*a(n)*(a(n) + (-1)^n)+ 1).
a(n-1) = (3*a(n) + 2*(-1)^n) - 2*sqrt(2*a(n)*(a(n) + (-1)^n)+ 1).
a(n+1) = 6*a(n) - a(n-1) + 4*(-1)^n.
a(n+1) = 5*a(n) + 5*a(n-1) - a(n-2).
a(n+1) *a(n-1) = a(n)*(a(n) + 4*(-1)^n).
a(n) = (sqrt(1 + 8*A029549(n)) - (-1)^n)/2.
a(n) = A002315(n) - A084159(n) = A084159(n) - (-1)^n.
a(n)  = A001652(n) + (1 - (-1)^n)/2 = A046090(n) - (1 + (-1)^n)/2.
Limit_{n->oo} a(n)/a(n-1) =  3 + 2*sqrt(2).
Limit_{n->oo} a(n)/a(n-2) = 17 + 12*sqrt(2).
Limit_{n->oo} a(n)/a(n-r) = (3 + 2*sqrt(2))^r.
Limit_{n->oo} a(n-r)/a(n) = (3 - 2*sqrt(2))^r. (End)
From G. C. Greubel, Feb 11 2023: (Start)
a(n) = (A001333(2*n+1) - 2*(-1)^n)/4.
a(n) = (1/2)*(A001109(n+1) + A001109(n) - (-1)^n). (End)
E.g.f.: exp(-x)*(exp(4*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - 1)/2. - Stefano Spezia, Aug 03 2024

A218395 Numbers whose square is the sum of the squares of 11 consecutive integers.

Original entry on oeis.org

11, 77, 143, 1529, 2849, 30503, 56837, 608531, 1133891, 12140117, 22620983, 242193809, 451285769, 4831736063, 9003094397, 96392527451, 179610602171, 1923018812957, 3583208949023, 38363983731689, 71484568378289, 765356655820823, 1426108158616757

Offset: 0

Paul Weisenhorn, Oct 28 2012

a(n)^2 = Sum_{j=0..10} (x(n)+j)^2 = 11*(x(n)+5)^2 + 110 and b(n) = x(n)+5 give the Pell equation a(n)^2 - 11*b(n)^2 = 110 with the 2 fundamental solutions (11; 1) and (77; 23) and the solution (10; 3) for the unit form. A198949(n+1) = b(n); A106521(n) = x(n) and x(0) = -4.

General: If the sum of the squares of c neighboring numbers is a square with c = 3*k^2-1 and 1 <= k, then a(n)^2 = Sum_{j=0..c-1} (x(n)+j)^2 and b(n) = 2*x(n)+c-1 give the Pell equation a(n)^2 - c*(b(n)/2)^2 = binomial(c+1,3)/2. a(n) = 2*e1*a(n-k) - a(n-2*k); b(n) = 2*e1*b(n-k) - b(n-2*k); a(n) = e1*a(n-k) + c*e2*b(n-k); b(n) = e2*a(n-k) + e1*b(n-k) with the solution (e1; e2) for the unit form.

			For n=6, Sum_{z=17132..17142} z^2 = 3230444569;
a(6) = sqrt(3230444569) = 56837;
b(6) = sqrt((a(6)^2-110)/11) = 17137; x(6) = b(6)-5 = 17132.

V. Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014. See Table 1 p. 7.
Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).

c=2: A001653(n+1) = a(n); A002315(n) = b(n); A001652(n) = x(n).

Cf. A001032 (11 is a term of that sequence), A198947.

s:=0: n:=-1:
for j from -5 to 5 do s:=s+j^2: end do:
for z from -4 to 100000 do
  s:=s-(z-1)^2+(z+10)^2: r:=sqrt(s):
  if (r=floor(r)) then
    n:=n+1: a(n):=r: x(n):=z:
    b(n):=sqrt((s-110)/11):
    print(n,a(n),b(n),x(n)):
  end if:
end do:

LinearRecurrence[{0,20,0,-1},{11,77,143,1529},30] (* Harvey P. Dale, Aug 15 2022 *)

a(n) = 20*a(n-2) - a(n-4); b(n) = 20*b(n-2) - b(n-4);
a(n) = 10*a(n-2) + 33*b(n-2); b(n) = 3*a(n-2) + 10*b(n-2).
a(n) = a(n-1) + 20*a(n-2) - 20*a(n-3) - a(n-4) + a(n-5).
G.f.: 11 * (1-x)*(1+8*x+x^2) / (1 - 20*x^2 + x^4).
With r=sqrt(11); s=10+3*r; t=10-3*r:
a(2*n) = ((11+r)*s^n + (11-r)*t^n)/2.
a(2*n+1) = ((77+23*r) * s^n + (77-23*r)*t^n)/2.
a(n) = 11 * A198947(n+1). - Bill McEachen, Dec 01 2022

A077442 2*a(n)^2 + 7 is a square.

Original entry on oeis.org

1, 3, 9, 19, 53, 111, 309, 647, 1801, 3771, 10497, 21979, 61181, 128103, 356589, 746639, 2078353, 4351731, 12113529, 25363747, 70602821, 147830751, 411503397, 861620759, 2398417561, 5021893803, 13979001969, 29269742059, 81475594253

Offset: 0

Gregory V. Richardson, Nov 06 2002

Lim. n -> Inf. a(n)/a(n-2) = 3 + 2*Sqrt(2) = R1*R2. Lim. k -> Inf. a(2*k-1)/a(2*k) = (9 + 4*Sqrt(2))/7 = R1 (ratio #1). Lim. k -> Inf. a(2*k)/a(2*k-1) = (11 + 6*Sqrt(2))/7 = R2 (ratio #2).

a(n) gives for n >= 0 all positive y-values solving the (generalized) Pell equation x^2 - 2*y^2 = 7. A077443(n+1) gives the corresponding x-values. See, e.g., the Nagell reference on how to find all solutions. - Wolfdieter Lang, Feb 05 2015

			a(4)^2 - 2*a(3)^2 = 27^2 - 2*19^2  = +7. - _Wolfdieter Lang_, Feb 05 2015

L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Colin Barker, Table of n, a(n) for n = 0..1000
J. J. O'Connor and E. F. Robertson, History of Pell's Equation
J. P. Robertson, Solving the Generalized Pell Equation
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A077443, A038762, A038761, A101386, A253811.

CoefficientList[Series[(1+3 x+3 x^2+x^3)/ (1-6 x^2+x^4),{x,0,50}],x]  (* Harvey P. Dale, Mar 12 2011 *)
LinearRecurrence[{0, 6, 0, -1},{1,3,9,19},50] (* Sture Sjöstedt, Oct 08 2012 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^n*[1;3;9;19])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

Vec((x+1)^3/(x^2+2*x-1)/(x^2-2*x-1) + O(x^50)) \\ Colin Barker, Mar 27 2016

For n>0, a(2n) = A046090(n) + A001653(n) + A001652(n-1); a(2n+1) = A001652(n+1) - A001652(n-1) - A001653(n-1); e.g. 53=21+29+3; 111=119-3-5. - Charlie Marion, Aug 14 2003
The same recurrences hold for the odd and even indices respectively : a(n+2) = 6*a(n+1) - a(n), a(n+1) = 3*a(n) + 2*(2*a(n)^2+7)^0.5. - Richard Choulet, Oct 11 2007
G.f.: (x+1)^3/(x^2+2*x-1)/(x^2-2*x-1). a(n)= [ -A077985(n)-3*A077985(n-1)+3*A000129(n+1)+A000129(n)]/2. - R. J. Mathar, Nov 16 2007
a(n) = 6*a(n-2) - a(n-4) with a(1)=1, a(2)=3, a(3)=9, a(4)=19. - Sture Sjöstedt, Oct 08 2012
a(n) = ((-(-1 - sqrt(2))^n*(-2+sqrt(2)) - (-1+sqrt(2))^n*(2+sqrt(2)) + (1-sqrt(2))^n*(-4+3*sqrt(2)) + (1+sqrt(2))^n*(4+3*sqrt(2))))/(4*sqrt(2)). - Colin Barker, Mar 27 2016

Edited: n in Name replaced by a(n). Pell comments moved to comment section. - Wolfdieter Lang, Feb 05 2015

A084703 Squares k such that 2*k+1 is also a square.

Original entry on oeis.org

0, 4, 144, 4900, 166464, 5654884, 192099600, 6525731524, 221682772224, 7530688524100, 255821727047184, 8690408031080164, 295218051329678400, 10028723337177985444, 340681375412721826704, 11573138040695364122500, 393146012008229658338304, 13355391270239113019379844

Offset: 0

Amarnath Murthy, Jun 08 2003

With the exception of 0, a subsequence of A075114. - R. J. Mathar, Dec 15 2008
Consequently, A014105(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011
From M. F. Hasler, Jan 17 2012: (Start)
Bisection of A079291. The squares 2*k+1 are given in A055792.
A204576 is this sequence written in binary. (End)
a(n+1), n >= 0, is the perimeter squared (x(n) + y(n) + z(n))^2 of the ordered primitive Pythagorean triple (x(n), y(n) = x(n) + 1, z(n)). The first two terms are (x(0)=0, y(0)=1, z(0)=1), a(1) = 2^2, and (x(1)=3, y(1)=4, z(1)=5), a(2) = 12^2. - George F. Johnson, Nov 02 2012

D. W. Wilson, Table of n, a(n-1) for n = 1..100 (offset=1)
Emrah Kılıç, Yücel Türker Ulutaş, and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011), Article 11.5.6, table 3, k=2.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000129, A001109, A001110, A001333, A001541, A001542, A001652, A001653.
Cf. A014105, A029549, A046090, A046176, A055792, A075114, A079291, A084702.
Cf. A089928, A204576.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: this sequence (k=-1), A076218 (k=3), A278310 (k=-5).

[4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // G. C. Greubel, Aug 18 2022

b[n_]:= b[n]= If[n<2, n, 34*b[n-1] -b[n-2] +2]; (* b=A001110 *)
a[n_]:= 4*b[n]; Table[a[n], {n, 0, 30}]
4*ChebyshevU[Range[-1,30], 3]^2 (* G. C. Greubel, Aug 18 2022 *)

[4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022

a(n) = 4*A001110(n) = A001542(n)^2.
a(n+1) = A001652(n)*A001652(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2. - Charlie Marion, Jul 01 2003
a(n) = A001653(k+n)*A001653(k-n) - A001653(k)^2, for k >= n >= 0; e.g. 144 = 5741*5 - 169^2. - Charlie Marion, Jul 16 2003
G.f.: 4*x*(1+x)/((1-x)*(1-34*x+x^2)). - R. J. Mathar, Dec 15 2008
a(n) = A079291(2n). - M. F. Hasler, Jan 16 2012
From George F. Johnson, Nov 02 2012: (Start)
a(n) = ((17+12*sqrt(2))^n + (17-12*sqrt(2))^n - 2)/8.
a(n+1) = 17*a(n) + 4 + 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1) = 17*a(n) + 4 - 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1)*a(n+1) = (a(n) - 4)^2.
2*a(n) + 1 = (A001541(n))^2.
a(n+1) = 34*a(n) - a(n-1) + 8 for n>1, a(0)=0, a(1)=4.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2) for n>0, a(0)=0, a(1)=4, a(2)=144.
a(n)*a(n+1) = (4*A029549(n))^2.
a(n+1) - a(n) = 4*A046176(n).
a(n) + a(n+1) = 4*(6*A029549(n) + 1).
a(n) = (2*A001333(n)*A000129(n))^2.
Limit_{n -> infinity} a(n)/a(n-r) = (17+12*sqrt(2))^r. (End)
Empirical: a(n) = A089928(4*n-2), for n > 0. - Alex Ratushnyak, Apr 12 2013
a(n) = 4*A001109(n)^2. - G. C. Greubel, Aug 18 2022
Product_{n>=2} (1 - 4/a(n)) = sqrt(2)/3 + 1/2 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Edited and extended by Robert G. Wilson v, Jun 15 2003

A129288 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 41)^2 = y^2.

Original entry on oeis.org

0, 36, 39, 123, 319, 336, 820, 1960, 2059, 4879, 11523, 12100, 28536, 67260, 70623, 166419, 392119, 411720, 970060, 2285536, 2399779, 5654023, 13321179, 13987036, 32954160, 77641620, 81522519, 192071019, 452528623, 475148160, 1119472036

Offset: 1

Mohamed Bouhamida, May 26 2007

Also values x of Pythagorean triples (x, x+41, y).
Corresponding values y of solutions (x, y) are in A157257.
lim_{n -> infinity} a(n)/a(n-3) = 3 + 2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (7 + 2*sqrt(2))/(7 - 2*sqrt(2)) for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (3 + 2*sqrt(2))*(7 - 2*sqrt(2))^2/(7 + 2*sqrt(2))^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. A157257, A001652, A156035 (decimal expansion of 3 + 2*sqrt(2)), A157258 (decimal expansion of 7 + 2*sqrt(2)), A157259 (decimal expansion of 7 - 2*sqrt(2)), A157260 (decimal expansion of (7 + 2*sqrt(2))/(7 - 2*sqrt(2))).

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(36+3*x+84*x^2-20*x^3-x^4-20*x^5)/((1-x)*(1-6*x^3+ x^6)))); // G. C. Greubel, May 07 2018

LinearRecurrence[{1,0,6,-6,0,-1,1},{0,36,39,123,319,336,820},40] (* Harvey P. Dale, Jan 18 2015 *)

forstep(n=0, 1200000000, [3 ,1], if(issquare(2*n^2+82*n+1681), print1(n, ",")))

a(n) = 6*a(n-3) - a(n-6) + 82 for n > 6; a(1)=0, a(2)=36, a(3)=39, a(4)=123, a(5)=319, a(6)=336.
G.f.: x*(36 + 3*x + 84*x^2 - 20*x^3 - x^4 - 20*x^5)/((1-x)*(1 - 6*x^3 + x^6)).
a(3*k + 1) = 41*A001652(k) for k >= 0.

Edited and extended by Klaus Brockhaus, Feb 26 2009

A129556 Numbers k such that the k-th centered pentagonal number A005891(k) = (5k^2 + 5k + 2)/2 is a square.

Original entry on oeis.org

0, 2, 21, 95, 816, 3626, 31005, 137711, 1177392, 5229410, 44709909, 198579887, 1697799168, 7540806314, 64471658493, 286352060063, 2448225223584, 10873837476098, 92968086837717, 412919472031679, 3530339074609680, 15680066099727722, 134059916748330141

Offset: 1

Alexander Adamchuk, Apr 20 2007

Corresponding numbers m > 0 such that m^2 is a centered pentagonal number are listed in A129557 = {1, 4, 34, 151, 1291, 5734, 49024, ...}.
From Andrea Pinos, Nov 02 2022: (Start)
By definition: 5*T(a(n)) = A129557(n)^2 - 1 where triangular number T(j) = j*(j+1)/2. This implies:
Every odd prime factor of a(n) and d(n)=a(n)+1 is present in b(n)=A129557(n)+1 or in c(n)=A129557(n)-1. (End)
From the law of cosines the non-Pythagorean triple {a(n), a(n)+1=A254332(n), A129557(n+1)} forms a near-isosceles triangle whose angle between the consecutive integer sides is equal to the central angle of the regular pentachoron polytope (4-simplex) (see A140244 and A140245). This implies that the terms {a(n)} are also those numbers k such that 1 + 5*A000217(k) is a square. - Federico Provvedi, Apr 04 2023

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Centered Pentagonal Number
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. A005891 (centered pentagonal numbers), A129557 (numbers k>0 such that k^2 is a centered pentagonal number), A221874.

Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; A233450 for k=3; A001652 for k=4; this sequence for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013

A005891 := proc(n) (5*n^2+5*n+2)/2 ; end: n := 0 : while true do if issqr(A005891(n)) then print(n) ; fi ; n := n+1 ; od : # R. J. Mathar, Jun 06 2007

Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[n] ], {n,1,40000} ]
LinearRecurrence[{1,38,-38,-1,1},{0,2,21,95,816},30] (* Harvey P. Dale, Nov 09 2017 *)
Table[(((x^(n+2))+(((-1)^n*(x^(2*n+1)+1)-x)/(x^n)))/(x^2+1)-1)/2/.x->3+Sqrt[10],{n,0,50}]//Round (* Federico Provvedi, Apr 04 2023 *)

a(n)=([0,1,0,0,0; 0,0,1,0,0; 0,0,0,1,0; 0,0,0,0,1; 1,-1,-38,38,1]^(n-1)*[0;2;21;95;816])[1,1] \\ Charles R Greathouse IV, Feb 11 2019

For n >= 5, a(n) = 38*a(n-2) - a(n-4) + 18. - Max Alekseyev, May 08 2009
G.f.: x^2*(x^3+2*x^2-19*x-2) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)). - Colin Barker, Feb 21 2013
a(n) = (A221874(n) - 1) / 2. - Bruno Berselli, Feb 21 2013
From Andrea Pinos, Oct 24 2022: (Start)
The ratios of successive terms converge to two different limits:
lower: D = lim_{n->oo} a(2n)/a(2n-1) = (7+2*sqrt(10))/3;
upper: E = lim_{n->oo} a(2n+1)/a(2n) = (13+4*sqrt(10))/3.
So lim_{n->oo} a(n+2)/a(n) = D*E = 19 + 6*sqrt(10).  (End)
a(n) = (x^(2*(n+1)) + (-1)^n*(x^(2*n+1)+1) - x) / (2*x^n*(x^2 + 1)) - (1/2), with x=3+sqrt(10). - Federico Provvedi, Apr 04 2023

More terms from R. J. Mathar, Jun 06 2007
Further terms from Max Alekseyev, May 08 2009
a(22)-a(23) from Colin Barker, Feb 21 2013

A012132 Numbers z such that x(x+1) + y(y+1) = z*(z+1) is solvable in positive integers x,y.

Original entry on oeis.org

3, 6, 8, 10, 11, 13, 15, 16, 18, 20, 21, 23, 26, 27, 28, 31, 33, 36, 37, 38, 40, 41, 43, 44, 45, 46, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 66, 67, 68, 71, 73, 74, 75, 76, 77, 78, 80, 81, 83, 86, 88, 89, 91, 92, 93

Offset: 1

Sander van Rijnswou (sander(AT)win.tue.nl)

nonn

Theorem (Sierpinski, 1963): n is a term iff n^2+(n+1)^2 is a composite number. - N. J. A. Sloane, Feb 29 2020
For n > 1, A047219 is a subset of this sequence. This is because n^2 + (n+1)^2 is divisible by 5 if n is (1 or 3) mod 5 (also see A027861). - Dmitry Kamenetsky, Sep 02 2008
From Hermann Stamm-Wilbrandt, Sep 10 2014: (Start)
For n > 0, A212160 is a subset of this sequence (n^2 + (n+1)^2 is divisible by 13 if n == (2 or 10) (mod 13)).
For n >= 0, A212161 is a subset of this sequence (n^2 + (n+1)^2 is divisible by 17 if n == (6 or 10) (mod 17)).
The above are for divisibility by 5, 13, 17; notation (1,3,5), (2,10,13), (6,10,17). Divisibility by p for a and p-a-1; notation (a,p-a-1,p). These are the next tuples: (8,20,29), (15,21,37), (4,36,41), (11,41,53), ... . The corresponding sequences are a subset of this sequence (8,20,37,49,66,78,... for (8,20,29)). These sequences have no entries in the OEIS yet. For any prime of the form 4*k+1 there is exactly one of these tuples/sequences.
For n > 1, A000217 (triangular numbers) is a subset of this sequence (3,6,10,15,...); z=A000217(n), y=z-1, x=n.
For n > 0, A001652 is a subset of this sequence; z=A001652(n), x=y=A053141(n).
For n > 1, A001108(=A115598) is a subset of this sequence; z=A001108(n), x=A076708(n), y=x+1.
For n > 0, A124124(2*n+1)(=A098790(2*n)) is a subset of this sequence (6,37,218,...); z=A124124(2*n+1), x=a(n)-1, y=a(n)+1, a(m) = 6*a(m-1) - a(m-2) + 2, a(0)=0, a(1)=4.
(End)

Aviezri S. Fraenkel, Diophantine equations involving generalized triangular and tetrahedral numbers, pp. 99-114 of A. O. L. Atkin and B. J. Birch, editors, Computers in Number Theory. Academic Press, NY, 1971.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
H. Finner and K. Strassburger, Increasing sample sizes do not necessarily increase the power of UMPU-tests for 2 X 2-tables, Metrika, 54, 77-91, (2001).
Heiko Harborth, Fermat-like binomial equations, Applications of Fibonacci numbers, Proc. 2nd Int. Conf., San Jose/Ca., August 1986, 1-5 (1988).
W. Sierpinski, On triangular numbers which are sums of two smaller triangular numbers, (Polish), Wiadom. Mat. (2) 7 (1963), 27-28. See MR0182602.

Complement of A027861. - Michael Somos, Jun 08 2000
Cf. A000217, A047219, A027861, A166080.
Cf. also A212160, A212161, A001652, A001108, A115598, A124124, A098790.

Select[Range[100], !PrimeQ[#^2 + (#+1)^2]& ] (* Jean-François Alcover, Jan 17 2013, after Michael Somos *)

More terms and references from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Feb 09 2000

A094196 Indices of the start of a string of 24 consecutive squares whose sum is a square.

Original entry on oeis.org

1, 9, 20, 25, 44, 76, 121, 197, 304, 353, 540, 856, 1301, 2053, 3112, 3597, 5448, 8576, 12981, 20425, 30908, 35709, 54032, 84996, 128601, 202289, 306060, 353585, 534964, 841476, 1273121, 2002557, 3029784, 3500233, 5295700, 8329856, 12602701

Offset: 1

Lekraj Beedassy, May 25 2004

The sequence could also include -11, -8 and -4; and if N is in the sequence, then so is -23-N.
Equivalently, 24*a(n)^2 + 552*a(n) + 4324 is a square.
All sequences of this type (i.e., sequences with fixed offset k, and a discernible pattern: k=0...23 for this sequence, k=0...22 for A269447, k=0..1 for A001652) can be extended 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

Daniel Mondot, Table of n, a(n) for n = 1..106
K. S. Brown, Sum of Consecutive Nth Powers Equals an Nth Power
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,10,-10,0,0,0,0,-1,1).

Cf. A001032, A001652, A106521, A269447.

LinearRecurrence[{1,0,0,0,0,10,-10,0,0,0,0,-1,1},{1,9,20,25,44,76,121,197,304,353,540,856,1301},60] (* Harvey P. Dale, Oct 10 2011 *)

for(n=1,15000000,if(issquare(sum(j=n,n+23,j^2)),print1(n,","))) \\ Klaus Brockhaus, Jun 01 2004

Recurrence: a(n+12) = 10a(n+6) - a(n) + 92.
O.g.f.: x*(-1-8*x-11*x^2-5*x^3-19*x^4-32*x^5-35*x^6+4*x^7+3*x^8+x^9+3*x^10+4*x^11+4*x^12) / ((-1+x) * (1-10*x^6+x^12)). - R. J. Mathar, Dec 02 2007
a(0)=1, a(1)=9, a(2)=20, a(3)=25, a(4)=44, a(5)=76, a(6)=121, a(7)=197, a(8)=304, a(9)=353, a(10)=540, a(11)=856, a(12)=1301; thereafter a(n) = a(n-1)+10*a(n-6)-10*a(n-7)-a(n-12)+a(n-13). - Harvey P. Dale, Oct 10 2011
a(1)=1, a(2)=9, a(3)=20, a(4)=25, a(5)=44, a(6)=76, a(7)=121, a(8)=197, a(9)=304, a(10)=353, a(11)=540, a(12)=856; a(n)=10*a(n-6)-a(n-12) + 92 for n>12. - Daniel Mondot, Aug 05 2016

More terms from Don Reble (djr(AT)hotmail.com) and Klaus Brockhaus, Jun 01 2004

A103200 a(1)=1, a(2)=2, a(3)=11, a(4)=19; a(n) = a(n-4) + sqrt(60a(n-2)^2 + 60a(n-2) + 1) for n >= 5.

Original entry on oeis.org

1, 2, 11, 19, 90, 153, 712, 1208, 5609, 9514, 44163, 74907, 347698, 589745, 2737424, 4643056, 21551697, 36554706, 169676155, 287794595, 1335857546, 2265802057, 10517184216, 17838621864, 82801616185, 140443172858, 651895745267, 1105706761003, 5132364345954

Offset: 1

K. S. Bhanu and M. N. Deshpande, Mar 24 2005

nonn

The original version of this question was as follows: Let a(1) = 1, a(2) = 2, a(3) = 11, a(4) = 19; for n = 1..4 let b(n) = sqrt(60 a(n)^2 + 60 a(n) + 1); for n >= 5 let a(n) = a(n-4) + b(n-2), b(n) = sqrt(60 a(n)^2 + 60 a(n) +1). Bhanu and Deshpande ask for a proof that a(n) and b(n) are always integers. The b(n) sequence is A103201.

This sequence is also the interleaving of two sequences c and d that can be extended backwards: c(0) = c(1) = 0, c(n) = sqrt(60 c(n-1)^2 + 60 c(n-1) +1) + c(n-2) giving 0,0,1,11,90,712,5609,... d(0) = 1, d(1) = 0, d(n) = sqrt(60 d(n-1)^2 + 60 d(n-1) +1) + d(n-2) giving 1,0,2,19,153,1208,9514,... and interleaved: 0,1,0,0,1,2,11,19,90,153,712,1208,5609,9514,... lim_{n->infinity} a(n)/a(n-2) = 1/(4 - sqrt(15)), (1/(4-sqrt(15)))^n approaches an integer as n -> infinity. - Gerald McGarvey, Mar 29 2005

K. S. Bhanu (bhanu_105(AT)yahoo.com) and M. N. Deshpande, An interesting sequence of quadruples and related open problems, Institute of Sciences, Nagpur, India, Preprint, 2005.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,8,-8,-1,1).

Cf. A103201, A177187 (first differences).

I:=[1,2,11,19,90]; [n le 5 select I[n] else Self(n-1)+8*Self(n-2)-8*Self(n-3)-Self(n-4)+Self(n-5): n in [1..30]]; // Vincenzo Librandi, Sep 28 2011

a[1]:=1: a[2]:=2:a[3]:=11: a[4]:=19: for n from 5 to 31 do a[n]:=a[n-4]+sqrt(60*a[n-2]^2+60*a[n-2]+1) od:seq(a[n],n=1..31); # Emeric Deutsch, Apr 13 2005

RecurrenceTable[{a[1]==1,a[2]==2,a[3]==11,a[4]==19,a[n]==a[n-4]+ Sqrt[60a[n-2]^2+60a[n-2]+1]},a[n],{n,40}] (* or *) LinearRecurrence[ {1,8,-8,-1,1},{1,2,11,19,90},40] (* Harvey P. Dale, Sep 27 2011 *)
CoefficientList[Series[-x*(1 + x + x^2)/((x - 1)*(x^4 - 8*x^2 + 1)), {x, 0, 40}], x] (* T. D. Noe, Jun 04 2012 *)

Comments from Pierre CAMI and Gerald McGarvey, Apr 20 2005: (Start)
Sequence satisfies a(0)=0, a(1)=1, a(2)=2, a(3)=11; for n > 3, a(n) = 8*a(n-2) - a(n-4) + 3.
G.f.: -x*(1 + x + x^2) / ( (x - 1)*(x^4 - 8*x^2 + 1) ). Note that the 3 = the sum of the coefficients in the numerator of the g.f., 8 appears in the denominator of the g.f. and 8 = 2*3 + 2. Similar relationships hold for other series defined as nonnegative n such that m*n^2 + m*n + 1 is a square, here m=60. Cf. A001652, A001570, A049629, A105038, A105040, A104240, A077288, A105036, A105037. (End)
a(2n) = (A105426(n)-1)/2, a(2n+1) = (A001090(n+2) - 5*A001090(n+1) - 1)/2. - Ralf Stephan, May 18 2007
a(1)=1, a(2)=2, a(3)=11, a(4)=19, a(5)=90, a(n) = a(n-1) + 8*a(n-2) - 8*a(n-3) - a(n-4) + a(n-5). - Harvey P. Dale, Sep 27 2011

More terms from Pierre CAMI and Emeric Deutsch, Apr 13 2005

A105038 Nonnegative n such that 6n^2 + 6n + 1 is a square.

Original entry on oeis.org

0, 4, 44, 440, 4360, 43164, 427284, 4229680, 41869520, 414465524, 4102785724, 40613391720, 402031131480, 3979697923084, 39394948099364, 389969783070560, 3860302882606240, 38213059042991844, 378270287547312204, 3744489816430130200, 37066627876753989800

Offset: 0

Gerald McGarvey, Apr 03 2005

Harvey P. Dale, Table of n, a(n) for n = 0..1000
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 (11,-11,1).

Cf. A001652, A001570, A049629, A105040, A104240, A077288, A105036, A103200, A105037.

CoefficientList[Series[4x/(1-11x+11x^2-x^3),{x,0,30}],x] (* or *) LinearRecurrence[{11,-11,1},{0,4,44},30] (* Harvey P. Dale, Sep 29 2013 *)

for(n=0,427284,if(issquare(6*n*(n+1)+1),print1(n,",")))

Vec(4*x/(1-11*x+11*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Nov 13 2012

G.f.: 4*x/(1-11*x+11*x^2-x^3).
a(0)=0, a(1)=4, a(2)=44, a(n)=11*a(n-1)-11*a(n-2)+a(n-3). - Harvey P. Dale, Sep 29 2013
a(n) = (-6-(5-2*sqrt(6))^n*(-3+sqrt(6))+(3+sqrt(6))*(5+2*sqrt(6))^n)/12. - Colin Barker, Mar 05 2016
a(n) = (A072256(n+1) - 1)/2.

More terms from Vladeta Jovovic, Apr 05 2005

Incorrect conjectures deleted by N. J. A. Sloane, Nov 24 2010

cp's OEIS Frontend

A046729 Expansion of 4*x/((1+x)*(1-6*x+x^2)).

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A218395 Numbers whose square is the sum of the squares of 11 consecutive integers.

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A077442 2*a(n)^2 + 7 is a square.

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A084703 Squares k such that 2*k+1 is also a square.

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A129288 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 41)^2 = y^2.

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A129556 Numbers k such that the k-th centered pentagonal number A005891(k) = (5k^2 + 5k + 2)/2 is a square.

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A046729 Expansion of 4x/((1+x)(1-6*x+x^2)).

A012132 Numbers z such that x(x+1) + y(y+1) = z*(z+1) is solvable in positive integers x,y.

A103200 a(1)=1, a(2)=2, a(3)=11, a(4)=19; a(n) = a(n-4) + sqrt(60a(n-2)^2 + 60a(n-2) + 1) for n >= 5.

A105038 Nonnegative n such that 6n^2 + 6n + 1 is a square.