A046090 -id:A046090 - cp's OEIS frontend

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

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

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

A006062 Star-hex numbers.

Original entry on oeis.org

1, 37, 1261, 42841, 1455337, 49438621, 1679457781, 57052125937, 1938092824081, 65838103892821, 2236557439531837, 75977114840189641, 2580985347126915961, 87677524687474953037, 2978454854027021487301, 101179787512231255615201, 3437134320561835669429537

Offset: 1

N. J. A. Sloane, Jul 11 1991

Martin Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 22.
Harvey J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Paolo Xausa, Table of n, a(n) for n = 1..650
Harvey J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A001109, A001652, A001653, A046090.

A006062:=-(z+1)**2/(z-1)/(z**2-34*z+1); [Simon Plouffe in his 1992 dissertation for offset zero.]

LinearRecurrence[{35, -35, 1}, {1, 37, 1261}, 20] (* Paolo Xausa, Sep 02 2024 *)

a(n) = 34*a(n-1) - a(n-2) + 4.
G.f.: -x*(x + 1)^2/(x - 1)/(x^2 - 34*x + 1). [from Simon Plouffe, see Maple code].
From Charlie Marion, Aug 03 2005: (Start)
a(n) = A001109(n-1)^2 + A001109(n)^2; e.g., 1261 = 6^2 + 35^2.
a(n) = A001652(n-1)*A046090(n-1) + A001653(n-1)^2; e.g., 1261 = 20*21 + 29^2. (End)

A076295 Consider all Pythagorean triples (Y-7,Y,Z); sequence gives Y values.

Original entry on oeis.org

4, 7, 12, 15, 28, 55, 72, 147, 304, 403, 840, 1755, 2332, 4879, 10212, 13575, 28420, 59503, 79104, 165627, 346792, 461035, 965328, 2021235, 2687092, 5626327, 11780604, 15661503, 32792620, 68662375, 91281912, 191129379, 400193632, 532029955, 1113983640

Offset: 0

Henry Bottomley, Oct 05 2002

nonn

First two terms included for consistency with A076293.

			15 is in the sequence as the longer leg of the (8,15,17) triangle.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A046090, A076293, A076294, A076296.

LinearRecurrence[{1,0,6,-6,0,-1,1},{4,7,12,15,28,55,72},40] (* Harvey P. Dale, Feb 02 2012 *)

a(n) =6a(n-3)-a(n-6)-14 =(A076293(n)+7)/2 =sqrt(A076294(n)^2-A076296(n)^2) =A076296(n)+7.
a(3n+1) = 7*A046090(n).
a(0)=4, a(1)=7, a(2)=12, a(3)=15, a(4)=28, a(5)=55, a(6)=72, a(n)= a(n-1)+ 6*a(n-3)-6*a(n-4)-a(n-6)+a (n-7). - Harvey P. Dale, Feb 02 2012
G.f.: -(3*x^6-3*x^5-5*x^4-21*x^3+5*x^2+3*x+4) / ((x-1)*(x^6-6*x^3+1)). - Colin Barker, Sep 14 2014

A115598 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z-(X+1) values.

Original entry on oeis.org

1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161, 384199200, 2239277041, 13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200, 87784138523761, 511643454094368, 2982076586042449, 17380816062160328, 101302819786919521

Offset: 1

N. J. A. Sloane, Mar 14 2006

Old name was A001653(n) - A046090(n).

Vincenzo Librandi, Table of n, a(n) for n = 1..200
L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Essentially a duplicate of A001108.

LinearRecurrence[{7,-7,1},{1,8,49},30] (* Harvey P. Dale, Oct 27 2013 *)
CoefficientList[Series[-(x + 1)/((x - 1) (x^2 - 6 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Jul 28 2014 *)

a(n) = (-2+(3-2*sqrt(2))^n+(3+2*sqrt(2))^n)/4.
a(n) = 7*a(n-1)-7*a(n-2)+a(n-3).
G.f.: -x*(x+1) / ((x-1)*(x^2-6*x+1)).
a(n)^2 + (a(n)+1)^2 = A001542(n)^2 + 1^2. - Hermann Stamm-Wilbrandt, Jul 27 2014

Corrected and edited by Colin Barker, Jul 31 2013

A182435 a(n) = 6*a(n-1) - a(n-2) - 2 with n>1, a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 4, 21, 120, 697, 4060, 23661, 137904, 803761, 4684660, 27304197, 159140520, 927538921, 5406093004, 31509019101, 183648021600, 1070379110497, 6238626641380, 36361380737781, 211929657785304, 1235216565974041, 7199369738058940, 41961001862379597

Offset: 0

Kenneth J Ramsey, Apr 28 2012

It appears that for n>0, A143608(n) divides a(n).
The sequence a(n)/A143608(n) appears to generate A001541 interleaved with A001653. - R. J. Mathar, Jul 04 2012
It also appears that if p equals a prime of the form 8*r +/- 3 then a(p + 1) == 0 (mod p); and that if p is a prime in the form of 8*r +/- 1 then a(p - 1) == 0 (mod p), inherited from A143608.

Bruno Berselli, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A001108, A143608, A001541 (first differences).

Essentially a duplicate of A046090.

[n le 2 select n-1 else 6*Self(n-1)-Self(n-2)-2: n in [1..24]]; // Bruno Berselli, May 15 2012

m = -20;
n = -3;
c = 0;
list3 = Reap[While[c < 20,t = 6 n - m - 2;Sow[t];m = n;n = t; c++]][[2,1]]
LinearRecurrence[{7,-7,1},{0,1,4},30] (* Harvey P. Dale, May 11 2018 *)

concat(0,Vec((1-3*x)/(1-x)/(1-6*x+x^2)+O(x^98))) \\ Charles R Greathouse IV, Jun 11 2013

a(n) = A046090(n-1), for n>=1.
G.f.: x*(1-3*x)/((1-x)*(1-6*x+x^2)). - Bruno Berselli, May 15 2012
a(n) = A001652(n-1)+1 with A001652(-1)=-1. - Bruno Berselli, May 16 2012
2*a(n)*(a(n)-1)+1 = A001653(n)^2 for n>0. - Bruno Berselli, Oct 23 2012

A301383 Expansion of (1 + 3x - 2x^2)/(1 - 7x + 7x^2 - x^3).

Original entry on oeis.org

1, 10, 61, 358, 2089, 12178, 70981, 413710, 2411281, 14053978, 81912589, 477421558, 2782616761, 16218279010, 94527057301, 550944064798, 3211137331489, 18715879924138, 109084142213341, 635788973355910, 3705649697922121, 21598109214176818, 125883005587138789, 733699924308655918

Offset: 0

Bruno Berselli, Mar 20 2018

y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y+2). The corresponding x values are listed in A075841.
y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y+1) are in A002315, and A075870 gives the x values.
y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y) are in A046090, and A001653 gives the x values.
Also, indices y for which 4*A000217(y) + 5 is a square. The next integers k such that k*A000217(y) + 5 is a square for infinitely many y values are 11, 20, 22, 29, 31, ...
First differences are in A106329.

Robert Israel, Table of n, a(n) for n = 0..1304
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Subsequence of A301451.

Cf. A000217, A000290, A001652, A002315, A033539, A046090, A053142, A075841, A077444, A106329, A241976.

using Nemo
function A301383List(len)
    R, x = PowerSeriesRing(ZZ, len+2, "x")
    f = divexact(1+3*x-2*x^2, 1-7*x+7*x^2-x^3)
    [coeff(f, k) for k in 0:len]
end
A301383List(23) |> println # Peter Luschny, Mar 21 2018

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3)));

f:= gfun:-rectoproc({a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3), a(0)=1,a(1)=10,a(2)=61},a(n),remember):
map(f, [$0..50]); # Robert Israel, Mar 21 2018

CoefficientList[Series[(1 + 3 x - 2 x^2)/(1 - 7 x + 7 x^2 - x^3), {x, 0, 30}], x]

makelist(coeff(taylor((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3), x, 0, n), x, n), n, 0, 30);

Vec((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3)+O(x^30))

m=30; L. = PowerSeriesRing(ZZ, m); f=(1+3*x-2*x^2)/(1-7*x+7*x^2-x^3); print(f.coefficients())

O.g.f.: (1 + 3*x - 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) = 6*a(n-1) - a(n-2) + 2.
a(n) = (3/4)*((1 + sqrt(2))^(2*n + 1) + (1 - sqrt(2))^(2*n + 1)) - 1/2.
a(n) = A033539(2*n+2) = A241976(n+1) + 1 = 3*A001652(n) + 1 = 3*A046090(n) - 2.
a(n) = A053142(n+1) + 3*A053142(n) - 2*A053142(n-1), n>0.
2*a(n) = 3*A002315(n)   - 1.
4*a(n) = 3*A077444(n+1) - 2.
E.g.f.: (3*exp(3*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - cosh(x) - sinh(x))/2. - Stefano Spezia, Mar 06 2020
Let T(n) be the n-th triangular number, A000217(n).  Then T(a(n)-3) + 2*T(a(n)-2) + 3*T(a(n)-1) + 4*T(a(n)) + 3*T(a(n)+1) + 2*T(a(n)+2) + T(a(n)+3) = (A001653(n) + A001653(n+2))^2. - Charlie Marion, Mar 16 2021

A056869 Prime hypotenuses of Pythagorean triangles with consecutive integer sides.

Original entry on oeis.org

5, 29, 5741, 33461, 44560482149, 1746860020068409, 68480406462161287469, 13558774610046711780701, 4125636888562548868221559797461449, 4760981394323203445293052612223893281

Offset: 1

Harvey P. Dale, Sep 02 2000

nonn

These primes belong to A001653.
From Jianing Song, Jan 02 2019: (Start)
Essentially the same sequence as A086383.
If p is a term then it is a unique-period prime in base sqrt(2*p^2 - 1). (End)

			29 is included because it is prime and it is the hypotenuse of the 20, 21, 29 Pythagorean triangle.

Robert Israel, Table of n, a(n) for n = 1..22

Cf. A000129, A001653, A001652, A046090, A086383.

f:=[1,5];; for n in [3..60] do f[n]:=6*f[n-1]-f[n-2]; od; a:=Filtered(f,IsPrime);; Print(a); # Muniru A Asiru, Jan 03 2019

f:= gfun:-rectoproc({a(n)=6*a(n-1)-a(n-2),a(1)=1,a(2)=5},a(n),remember):
select(isprime, [seq(f(n),n=1..1000)]); # Robert Israel, Oct 13 2015

Select[Sqrt[#^2+(#+1)^2]&/@With[{p=3+2Sqrt[2]},NestList[Floor[p #]+3&,3,120]],PrimeQ] (* Harvey P. Dale, May 02 2018 *)

t(n) = if(n<3, 5^(n-1), 6*t(n-1)-t(n-2));
for(n=1, 50, if(isprime(t(n)), print1(t(n)", "))) \\ Altug Alkan, Oct 13 2015

a(n) = A086383(n+1). - Jianing Song, Jan 02 2019

Incorrect link to index entries for linear recurrences with constant coefficients removed by Colin Barker, Oct 13 2015

Offset changed to 1 by Colin Barker, Oct 13 2015

A111647 a(n) = A001541(n)A001653(n+1)A002315(n).

Original entry on oeis.org

1, 105, 20213, 3998709, 791704585, 156753394977, 31036379835581, 6145046450172525, 1216688160731724433, 240898110778299543129, 47696609245941810082565, 9443687732585695622131557

Offset: 0

Charlie Marion, Aug 24 2005

			a(1) = 105 = 3*5*7.

G. C. Greubel, Table of n, a(n) for n = 0..434
Index entries for linear recurrences with constant coefficients, signature (204,-1190,204,-1).

Cf. A001541, A001653, A002315, A111648, A111649.

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)))); // G. C. Greubel, Jul 15 2018

CoefficientList[Series[(1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)), {x, 0, 30}], x] (* G. C. Greubel, Jul 15 2018 *)

x='x+O('x^30); Vec((1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1))) \\ G. C. Greubel, Jul 15 2018

2*a(n) = A001109(3*n+1) + A001109(n+1).
a(n) = sqrt(A011900(2*n)*A046090(2*n)*A001109(2*n+1)).
a(n) = A001541(3*n) + 2*A001109(n)*A001541(n-1)*A001541(n).
For n>0, a(n) = A001652(3*n) - A053141(2*n)*A002315(n-1) - A001652(n-1).
G.f.: (1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
2*a(n) = A001109(n+1) + A097731(n) + 6*A097731(n-1). - R. J. Mathar, Jan 31 2024

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

Extensions

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

Extensions

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A006062 Star-hex numbers.

Views

Author

Keywords

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A076295 Consider all Pythagorean triples (Y-7,Y,Z); sequence gives Y values.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A115598 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z-(X+1) values.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A182435 a(n) = 6*a(n-1) - a(n-2) - 2 with n>1, a(0)=0, a(1)=1.

Views

Author

Keywords

Comments

A301383 Expansion of (1 + 3x - 2x^2)/(1 - 7x + 7x^2 - x^3).

A111647 a(n) = A001541(n)A001653(n+1)A002315(n).