A006451 -id:A006451 - cp's OEIS frontend

A266505 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = -1, a(1) = 1, a(2) = 3, a(3) = 5.

-1, 1, 3, 5, 5, 11, 13, 27, 31, 65, 75, 157, 181, 379, 437, 915, 1055, 2209, 2547, 5333, 6149, 12875, 14845, 31083, 35839, 75041, 86523, 181165, 208885, 437371, 504293, 1055907, 1217471, 2549185, 2939235, 6154277, 7095941, 14857739, 17131117, 35869755, 41358175, 86597249, 99847467

Offset: 0

Raphie Frank, Dec 30 2015

a(n)/A266504(n) converges to sqrt(2).
Alternatively, bisection of A266506.
Alternatively, A135532(n) and A048655(n) interlaced.
Alternatively, A255236(n-1), A054490(n), A038762(n) and A101386(n) interlaced.
Let b(n) = (a(n) - (a(n) mod 2))/2, that is b(n) = {-1, 0, 1, 2, 2, 5, 6, 13, 15, 32, 37, 78, 90, ...}. Then:
A006451(n) = {b(4n+0) U b(4n+1)} gives n in N such that triangular(n) + 1 is square;
A216134(n) = {b(4n+2) U b(4n+3)} gives n in N such that triangular(n) follows form n^2 + n + 1 (twice a triangular number + 1).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).

Cf. A000129, A001333, A002203, A002965, A006451, A006452, A002965, A038761, A038762, A048654, A048655, A054490, A078343, A098586, A098790, A100525, A101386, A135532, A216134, A216162, A253811, A255236, A266504, A266505, A266507.

I:=[-1,1,3,5]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015

a:=proc(n) option remember; if n=0 then -1 elif n=1 then 1 elif n=2 then 3 elif n=3 then 5 else 2*a(n-2)+a(n-4); fi; end:  seq(a(n), n=0..50); # Wesley Ivan Hurt, Jan 01 2016

LinearRecurrence[{0, 2, 0, 1}, {-1, 1, 3, 5}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[SeriesCoefficient[(-1 + 3 x) (1 + x)^2/(1 - 2 x^2 - x^4), {x, 0, n}], {n, 0, 42}] (* Michael De Vlieger, Dec 31 2015 *)

my(x='x+O('x^40)); Vec((-1+3*x)*(1+x)^2/(1-2*x^2-x^4)) \\ G. C. Greubel, Jul 26 2018

G.f.: (-1 + 3*x)*(1 + x)^2/(1 - 2*x^2 - x^4).
a(n) = (-(1+sqrt(2))^floor(n/2)*(-1)^n - sqrt(8)*(1-sqrt(2))^floor(n/2) - (1-sqrt(2))^floor(n/2)*(-1)^n + sqrt(8)*(1+sqrt(2))^floor(n/2))/2.
a(n) = 3*(((1+sqrt(2))^floor(n/2)-(1-sqrt(2))^floor(n/2))/sqrt(8)) - (-1)^n*(((1+sqrt(2))^(floor(n/2)-(-1)^n)-(1-sqrt(2))^(floor(n/2)-(-1)^n))/sqrt(8)).
a(n) = (3*A000129(floor(n/2)) - A000129(n-(-1)^n)), where A000129 gives the Pell numbers.
a(n) = sqrt(2*A266504(n)^2 - 7*(-1)^A266504(n))*sgn(2*n-1), where A266504 gives all x in N such that 2*x^2 - 7*(-1)^x = y^2. This sequence gives associated y values.
a(2n) = (-(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n - (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n) = A135532(n).
a(2n) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) - (((1+sqrt(2))^(n-1)-(1-sqrt(2))^(n-1))/sqrt(8)) = A135532(n).
a(2n+1) = (+(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n + (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n + 1) = A048655(n).
a(2n+1) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) + (((1+sqrt(2))^(n+1)-(1-sqrt(2))^(n+1))/sqrt(8)) = A048655(n).
a(4n + 0) = 6*a(4n - 4) - a(4n - 8) = A255236(n-1).
a(4n + 1) = 6*a(4n - 3) - a(4n - 7) = A054490(n).
a(4n + 2) = 6*a(4n - 2) - a(4n - 6) = A038762(n).
a(4n + 3) = 6*a(4n - 1) - a(4n - 5) = A101386(n).
(sqrt(2*(a(2n + 1) )^2 + 14*(-1)^floor(n/2)))/2 = A266504(n).
(a(2n + 1) + a(2n))/8 = A000129(n), where A000129 gives the Pell numbers.
a(2n + 1) - a(2n) = A002203(n), where A002203 gives the companion Pell numbers.
(a(2n + 2) + a(2n + 1))/2 = A000129(n+2).
(a(2n + 2) - a(2n + 1))/2 = A000129(n-1).

A154152 Indices k such that 26 plus the k-th triangular number is a perfect square.

Original entry on oeis.org

4, 10, 37, 67, 220, 394, 1285, 2299, 7492, 13402, 43669, 78115, 254524, 455290, 1483477, 2653627, 8646340, 15466474, 50394565, 90145219, 293721052, 525404842, 1711931749, 3062283835, 9977869444, 17848298170, 58155284917, 104027505187, 338953840060

Offset: 1

R. J. Mathar, Oct 18 2009

			4*(4+1)/2+26 = 6^2. 10*(10+1)/2+26 = 9^2. 37*(37+1)/2+26 = 27^2. 67*(67+1)/2+26 = 48^2.

Colin Barker, Table of n, a(n) for n = 1..500
F. T. Adams-Watters, SeqFan Discussion, Oct 2009
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A000217, A000290, A006451.

Join[{4, 10}, Select[Range[0,10^5], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 26 &]] (* or *) LinearRecurrence[{1,6,-6,-1,1}, {4,10,37,67,220}, 25] (* G. C. Greubel, Sep 03 2016 *)

Vec(x*(-4-6*x-3*x^2+6*x^3+5*x^4)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)) + O(x^40)) \\ Colin Barker, Jul 11 2015

{k: 26+k*(k+1)/2 in A000290}.
a(n) = +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5).
G.f.: x*(-4-6*x-3*x^2+6*x^3+5*x^4)/((x-1) * (x^2-2*x-1) * (x^2+2*x-1)).
G.f.: ( 10 + (-3-6*x)/(x^2+2*x-1) + 1/(x-1) + (12+27*x)/(x^2-2*x-1) )/2.

Extended by D. S. McNeil, Dec 05 2010

A166259 Positive integers n such that a centered polygonal number nk(k+1)/2+1 is not a square for any k > 0.

Original entry on oeis.org

2, 18, 32, 50, 72, 98, 128, 162, 200, 242, 338, 392, 450, 512, 578, 648, 722, 882, 968, 1058, 1152, 1250, 1352, 1458, 1682, 1800, 1922, 2048, 2178, 2312, 2401, 2450, 2662, 2738, 2809, 2888, 3042, 3174, 3200, 3362, 3528, 3698, 3750, 4050, 4225, 4232, 4418, 4489, 4608, 4802

Offset: 1

Alexander Adamchuk, Oct 10 2009

nonn

Positive integers n such that A120744(n) = -1.

Eric Weisstein's World of Mathematics, Centered Polygonal Number.

Cf. A120744, A001542, A006451, A001921, A129444, A001570, A001652, A129556, A053606, A105038, A105040, A053141, A061278.

Edited and extended by Max Alekseyev, Jan 20 2010

A175032 a(n) is the difference between the n-th triangular number and the next perfect square.

Original entry on oeis.org

0, 0, 1, 3, 6, 1, 4, 8, 0, 4, 9, 15, 3, 9, 16, 1, 8, 16, 25, 6, 15, 25, 3, 13, 24, 36, 10, 22, 35, 6, 19, 33, 1, 15, 30, 46, 10, 26, 43, 4, 21, 39, 58, 15, 34, 54, 8, 28, 49, 0, 21, 43, 66, 13, 36, 60, 4, 28, 53, 79, 19, 45, 72, 9, 36, 64, 93, 26, 55, 85, 15, 45, 76, 3, 34, 66, 99, 22

Offset: 0

Ctibor O. Zizka, Nov 09 2009

All terms are from {0} U A175035. No terms are from A175034.

The sequence consists of ascending runs of length 3 or 4. The first run starts at n = 1 and thereafter the k-th run starts at n = A214858(k - 1). - John Tyler Rascoe, Nov 05 2022

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A001109, A214858.
Cf. A000217, A080819.
Cf. A175034, A175035.
Cf. sequences where a(m)=k: A001108 (0), A006451 (1), A154138 (3), A154139 (4), A154140 (6), A154141 (8), A154142 (9), A154143 (10), A154144 (13), A154145 (15), A154146 (16), A154147 (19), A154148 (21), A154149 (22), A154150(24), A154151 (25), A154151 (26), A154153(28), A154154 (30).

Ceiling[Sqrt[#]]^2-#&/@Accumulate[Range[0,80]] (* Harvey P. Dale, Aug 25 2013 *)

a(n) = my(t=n*(n+1)/2); if (issquare(t), 0, (sqrtint(t)+1)^2 - t); \\ Michel Marcus, Nov 06 2022

a(n) = (ceiling(sqrt(n*(n+1)/2)))^2 - n*(n+1)/2. - Ctibor O. Zizka, Nov 09 2009

a(n) = A080819(n) - A000217(n). - R. J. Mathar, Aug 24 2010

Erroneous formula variant deleted and offset set to zero by R. J. Mathar, Aug 24 2010

A175034 Offsets i such that i + n*(n+1)/2 is never a perfect square for any n>0.

Original entry on oeis.org

2, 5, 7, 11, 12, 14, 17, 18, 20, 23, 27, 29, 31, 32, 37, 38, 40, 41, 42, 44, 47, 50, 51, 52, 56, 57, 59, 62, 65, 67, 68, 69, 70, 73, 74, 77, 82, 83, 84, 86, 87, 88, 92, 95, 96, 98, 101, 102, 104, 107, 109, 110, 112, 113, 117, 119, 122, 125, 126, 127, 128, 131, 132, 135, 137, 139

Offset: 1

Ctibor O. Zizka, Nov 10 2009

Complement of A175035.

Cf. A001108, A006451, A154138, A154139, A154140, A154141

Extended by R. J. Mathar, Nov 26 2009

A229083 Numbers k such that the distance between the k-th triangular number and the nearest square is at most 1.

Original entry on oeis.org

1, 2, 4, 5, 8, 15, 25, 32, 49, 90, 148, 189, 288, 527, 865, 1104, 1681, 3074, 5044, 6437, 9800, 17919, 29401, 37520, 57121, 104442, 171364, 218685, 332928, 608735, 998785, 1274592, 1940449, 3547970, 5821348, 7428869, 11309768, 20679087, 33929305, 43298624, 65918161

Offset: 1

Ralf Stephan, Sep 13 2013

nonn

The k-th triangular number (A000217) is a square, or a square plus or minus one.

Union of A006451 (k-th triangular number is a square minus one), A072221 (k-th triangular number is a square plus one), and A001108 (k-th triangular number is square). Also, union of A229131 and A001108.

			A000217(4) = 10 and 10 - 3^2 = 1 so 4 is in the sequence.
A000217(5) = 15 and 4^2 - 15 = 1 so 5 is in the sequence.
A000217(8) = 36 = 6^2 so 8 is in sequence.

Cf. A000217, A001108, A006451, A072221, A229118, A229131.

for(n=1,10^8,for(i=-1,1,f=0;if(issquare(n*(n+1)/2+i),f=1;break));if(f,print1(n,",")))

G.f.: (x^7 - 2*x^6 + x^5 - 3*x^4 + x^3 + 2*x^2 + x + 1)/((1-2*x^2+x^4)*(1-2*x^2-x^4)*(1-x)) (conjectured).

A160970 Indices of square numbers that are also 18-gonal numbers.

Original entry on oeis.org

0, 1, 10, 44, 341, 1495, 11584, 50786, 393515, 1725229, 13367926, 58607000, 454115969, 1990912771, 15426575020, 67632427214, 524049434711, 2297511612505, 17802254205154, 78047762397956, 604752593540525, 2651326409917999, 20543785926172696, 90067050174814010

Offset: 1

Sture Sjöstedt, Jun 01 2009, Jul 02 2009

Solving the Diophantine equation A051870(m) = m*(8*m-7) = k^2 leads to the entries.

k in the sequence and a list of associated m = 0, 1, 4, 16, 121, 529, 4096, 17956, 139129, 609961...

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,34,0,-1).

Cf. A006451, A006452.

Join[{0},LinearRecurrence[{0,34,0,-1},{1,10,44,340},23]] (* Ray Chandler, Aug 01 2015 *)

is(n)=ispolygonal(n^2,18) \\ Charles R Greathouse IV, Feb 14 2013

concat(0, Vec(x^2*(x+1)*(x^2+9*x+1)/((x^2-6*x+1)*(x^2+6*x+1)) + O(x^50))) \\ Colin Barker, Jun 24 2015

a(n) = 34*a(n-2) - a(n-4), n>5. - R. J. Mathar, Oct 04 2009
G.f.: x^2*(x+1)*(x^2 + 9*x + 1)/((x^2 - 6*x + 1)*(x^2 + 6*x + 1)). - Colin Barker, Oct 07 2012
For all values excepting the leading 0, a(n) = sqrt(8*A006452(n)^2 - 7)*A006452(n) = sqrt(A006451(n-1)*(A006451(n-1) + 1)/2 + 1)*(2*A006451(n-1) + 1). - Raphie Frank, Feb 11 2013

0 added in front and extended by R. J. Mathar, Oct 04 2009

A173202 Solutions y of the Mordell equation y^2 = x^3 - 3a^2 + 1 for a = 0,1,2, ... (solutions x are given by the sequence A000466).

Original entry on oeis.org

0, 5, 58, 207, 500, 985, 1710, 2723, 4072, 5805, 7970, 10615, 13788, 17537, 21910, 26955, 32720, 39253, 46602, 54815, 63940, 74025, 85118, 97267, 110520, 124925, 140530, 157383, 175532, 195025, 215910, 238235, 262048, 287397, 314330, 342895

Offset: 1

Michel Lagneau, Feb 12 2010

For many values of k for the equation y^2 = x^3 + k, all the solutions are known. For example, we have solutions for k=-2: (x,y) = (3,-5) and (3,5). A complete resolution for all integers k is unknown. Theorem: Let k be < -1, free of square factors, with k == 2 or 3 (mod 4). Suppose that the number of classes h(Q(sqrt(k))) is not divisible by 3. Then the equation y^2 = x^3 + k admits integer solutions if and only if k = 1 - 3a^2 or 1 - 3a^2 where a is an integer. In this case, the solutions are x = a^2 - k, y = a(a^2 + 3k) or -a(a^2 + 3k) (the first reference gives the proof of this theorem). With k = -1 - 3a^2, we obtain the solutions x = 4a^2 + 1, y = a(8a^2 + 3) or -a(8a^2 + 3). For the case k = 1 - 3a^2, we obtain the solution x = 4a^2 - 1 given by the sequence A000466.

			With a=3, x = 35 and y = 207, and then 207^2 = 35^2 - 26.

T. Apostol, Introduction to Analytic Number Theory, Springer, 1976
D. Duverney, Theorie des nombres (2e edition), Dunod, 2007, p.151

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
W. J. Ellison, F. Ellison, J. Pesek, C. E. Stall & D. S. Stall, The diophantine equation y^2 + k = x^3, J. Number Theory 4 (1972), 107-117.
Helmut Richter, Solutions of Mordell's equation y^2 = x^3 + k (solutions for 0
School of Mathematics and Statistics, University of St Andrews, Louis Joel Mordell.
Eric Weisstein's World of Mathematics, Mordell Curve.
D. J. Wright, Mordell's Equation.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Diophantine equations: see also Pellian equation: (A081233, A081234), (A081231, A082394), (A081232, A082393); Mordell equation: A053755, A173200; Diophantine equations: A006452, A006451, A006454.

I:=[0, 5, 58, 207]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jul 02 2012

for a from 0 to 100 do : z := evalf(a*(8*a^2 - 3)) : print (z) :od :

CoefficientList[Series[x*(5+38*x+5*x^2)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jul 02 2012 *)
CoefficientList[Series[E^x (5 x + 24 x^2 + 8 x^3), {x, 0, 40}], x]*Table[n!, {n, 0, 40}] (* Stefano Spezia, Dec 04 2018 *)

y = a*(8*a^2 - 3).
a(n) = sqrt(A000466(n)^3 - A080663(n)). - Artur Jasinski, Nov 26 2011
From Colin Barker, Apr 26 2012: (Start)
a(n) = 8*n^3 - 24*n^2 + 21*n - 5.
G.f.: x^2*(5 + 38*x + 5*x^2)/(1 - x)^4. (End)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jul 02 2012
E.g.f.: exp(x)*(5*x + 24*x^2 + 8*x^3). - Stefano Spezia, Dec 04 2018

A175492 Numbers m >= 3 such that binomial(m,3) + 1 is a square.

Original entry on oeis.org

7, 10, 24, 26, 65, 13777

Offset: 1

Ctibor O. Zizka, May 29 2010

Related sequences:
Numbers m such that binomial(m,2) is a square: A055997;
Numbers m such that binomial(m,2) + 1 is a square: A006451 + 1;
Numbers m such that binomial(m,2) - 1 is a square: A072221 + 1;
Numbers m >= 3 such that binomial(m,3) is a square: {3, 4, 50} (Proved by A. J. Meyl in 1878);
Numbers m >= 4 such that binomial(m,4) + 1 is a square: {6, 7, 45, 55, ...};
Numbers m >= 7 such that binomial(m,7) + 1 is a square: {8, 10, 21, 143, ...}.
No additional terms up to 10 million. - Harvey P. Dale, Apr 04 2017
No additional terms up to 10 billion. - Jon E. Schoenfield, Mar 18 2022
No additional terms up to 1 trillion. The sequence is finite by Siegel's theorem on integral points. - David Radcliffe, Jan 01 2024

Wikipedia, Tetrahedral number

Cf. A006451, A007318, A055997, A072221.

Cf. A216268 (values of binomial(m, 3)) and A216269 (square roots of binomial(m, 3) + 1).

lst = {}; k = 3; While[k < 10^6, If[ IntegerQ@ Sqrt[ Binomial[k, 3] + 1], AppendTo[lst, k]]; k++ ]; lst (* Robert G. Wilson v, Jun 11 2010 *)
Select[Range[3,14000],IntegerQ[Sqrt[Binomial[#,3]+1]]&] (* Harvey P. Dale, Apr 04 2017 *)

isok(m) = (m>=3) && issquare(binomial(m,3)+1); \\ Michel Marcus, Mar 15 2022

from sympy import binomial
from sympy.ntheory.primetest import is_square
for m in range(3, 10**6):
    if is_square(binomial(m,3)+1):
        print(m) # Mohammed Yaseen, Mar 18 2022

A229081 Numbers n such that there exists a square m^2 with 3n^2 - n <= m^2 <= 3n^2 + n.

Original entry on oeis.org

1, 3, 4, 7, 8, 11, 12, 14, 15, 16, 18, 19, 22, 23, 26, 27, 29, 30, 33, 34, 37, 38, 40, 41, 42, 44, 45, 48, 49, 52, 53, 55, 56, 57, 59, 60, 63, 64, 67, 68, 70, 71, 74, 75, 78, 79, 82, 83, 85, 86, 89, 90, 93, 94, 96, 97, 98, 100, 101, 104, 105, 108, 109, 111, 112, 113, 115, 116, 119, 120, 123, 124, 126

Offset: 1

Ralf Stephan, Sep 13 2013

nonn

			There is a square between 3*4^2-4 and 3*4^2+4 (44<=49<=52) but not between 3*5^2-5=70 and 3*5^2+5=80, so 4 is in sequence but not 5.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A006451, A072221.

[n: n in [1..130] | exists{3*n^2+i: i in [-n..n] | IsSquare(3*n^2+i)}]; // Bruno Berselli, Sep 13 2013

filter:= n -> ceil(sqrt(3*n^2-n))<=floor(sqrt(3*n^2+n)):
select(filter, [$1..200]); # Robert Israel, Jan 05 2020

for(n=1,200,for(i=-n,n,f=0;if(issquare(3*n*n+i),f=1;break));if(f,print1(n,",")))

cp's OEIS Frontend

A266505 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = -1, a(1) = 1, a(2) = 3, a(3) = 5.

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A154152 Indices k such that 26 plus the k-th triangular number is a perfect square.

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A166259 Positive integers n such that a centered polygonal number n*k*(k+1)/2+1 is not a square for any k > 0.

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A175032 a(n) is the difference between the n-th triangular number and the next perfect square.

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A175034 Offsets i such that i + n*(n+1)/2 is never a perfect square for any n>0.

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Extensions

A229083 Numbers k such that the distance between the k-th triangular number and the nearest square is at most 1.

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PARI

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A160970 Indices of square numbers that are also 18-gonal numbers.

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Mathematica

PARI

PARI

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Extensions

A173202 Solutions y of the Mordell equation y^2 = x^3 - 3a^2 + 1 for a = 0,1,2, ... (solutions x are given by the sequence A000466).

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A166259 Positive integers n such that a centered polygonal number nk(k+1)/2+1 is not a square for any k > 0.