A006452 -id:A006452 - cp's OEIS frontend

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

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

A306561 Square numbers that are also central polygonal numbers (i.e., square numbers found in the Lazy Caterer's sequence).

Original entry on oeis.org

1, 4, 16, 121, 529, 4096, 17956, 139129, 609961, 4726276, 20720704, 160554241, 703893961, 5454117904, 23911673956, 185279454481, 812293020529, 6294047334436, 27594051024016, 213812329916329, 937385441796001, 7263325169820736, 31843510970040004

Offset: 1

Moshe Monk, Feb 23 2019

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000
J.S. Seneschal, Relation To Square, Oblong, & Triangular Numbers On Multiplication Table
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A000124, A000290, A006452.

Cf. A000217, A002378.

R:=PowerSeriesRing(Integers(), 25); Coefficients(R!( x*(1+3*x-22*x^2+3*x^3+x^4)/((1-x)*(1-34*x^2 +x^4)) )); // G. C. Greubel, Apr 10 2019

LinearRecurrence[{1,34,-34,-1,1}, {1,4,16,121,529}, 25] (* G. C. Greubel, Apr 10 2019 *)

my(x='x+O('x^25)); Vec(x*(1+3*x-22*x^2+3*x^3+x^4)/((1-x)*(1-34*x^2 +x^4))) \\ G. C. Greubel, Apr 10 2019

lista(nn) = {for (n=0, nn, if (issquare(cpn = (n^2 + n) / 2 + 1), print1(cpn, ", ")););} \\ Michel Marcus, Apr 11 2019

a=(x*(1+3*x-22*x^2+3*x^3+x^4)/((1-x)*(1-34*x^2 +x^4))).series(x, 25).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Apr 10 2019

From Alois P. Heinz, Feb 23 2019: (Start)
G.f.: x*(1+3*x-22*x^2+3*x^3+x^4)/((1-x)*(1+6*x+x^2)*(1-6*x+x^2)).
a(n) = A006452(n)^2 for n >= 1.
{ A000124 } intersect { A000290 }. (End)

More terms from Alois P. Heinz, Feb 23 2019

A358682 Numbers k such that 8k^2 + 8k - 7 is a square.

Original entry on oeis.org

1, 7, 43, 253, 1477, 8611, 50191, 292537, 1705033, 9937663, 57920947, 337588021, 1967607181, 11468055067, 66840723223, 389576284273, 2270616982417, 13234125610231, 77134136678971, 449570694463597, 2620290030102613, 15272169486152083, 89012726886809887, 518804191834707241

Offset: 1

Stefano Spezia, Nov 26 2022

a(n) is the n-th almost cobalancing number of second type (see Tekcan and Erdem).

			a(2) = 7 is a term since 8*7^2 + 8*7 - 7 = 441 = 21^2.

Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A002315, A006452, A077443, A077446, A106328, A106329, A216134.

Mathematica
```
LinearRecurrence[{7,-7,1},{1,7,43},24]
```

a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) for n > 3.
a(n) = (3*(3 - 2*sqrt(2))^n*(2 + sqrt(2)) + 3*(2 - sqrt(2))*(3 + 2*sqrt(2))^n - 4)/8.
O.g.f.: x*(1 + x^2)/((1 - x)*(1 - 6*x + x^2)).
E.g.f.: (3*(2 + sqrt(2))*(cosh(3*x - 2*sqrt(2)*x) + sinh(3*x - 2*sqrt(2)*x)) + 3*(2 - sqrt(2))*(cosh(3*x + 2*sqrt(2)*x) + sinh(3*x + 2*sqrt(2)*x)) - 4*(cosh(x) + sinh(x)) - 8)/8.
a(n) = 3*A011900(n) - 2 = 6*A053142(n) + 1. - Hugo Pfoertner, Nov 26 2022

A364319 a(n) = (A077446(n) + 1)/2 for n >= 0.

Original entry on oeis.org

0, 1, 3, 6, 16, 33, 91, 190, 528, 1105, 3075, 6438, 17920, 37521, 104443, 218686, 608736, 1274593, 3547971, 7428870, 20679088, 43298625, 120526555, 252362878, 702480240, 1470878641, 4094354883, 8572908966, 23863649056, 49966575153, 139087539451

Offset: 0

Wolfdieter Lang, Aug 15 2023

a(n) and b(n) = A006452(n+1), for n >= 0, give the nonnegative solution of the equation binomial(a(n), 2) = b(n)^2 - 1.

This shows that the number of independent elements of an antisymmetric a(n) X a(n) matrix coincides with the number of independent elements of a traceless b(n) X b(n) matrix. The n = 0 case is trivial: 0 = 0. (The question about this coincidence was posed to W. L. by Martin Bordemann, Mar 03 1991.)

			The solutions (a(n), b(n)) begin:
  n: 0 1 2 3  4  5  6   7   8    9   10   11    12    13     14     15 ...
  ------------------------------------------------------------------------
  a: 0 1 3 6 16 33 91 190 528 1105 3075 6438 17920 37521 104443 218686 ...
  b: 1 1 2 4 11 23 64 134 373  781 2174 4552 12671 26531  73852 154634 ...

Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A001109, A001541, A006452, A049310, A077446.

LinearRecurrence[{1, 6, -6, -1, 1}, {0, 1, 3, 6, 16}, 31] (* Robert P. P. McKone, Aug 29 2023 *)

a(n) = 6*a(n-2) - a(n-4) - 2, for n >= 0, with a(-4) = -32, a(-3) = -15, a(-2) = -5, a(-1) = -2.
O.g.f.: G(x) = x*(1 + 2*x - 3*x^2 - 2*x^3)/((1 - x)*(1 - 6*x^2 + x^4)) = x*(1 + 2*x - 3*x^2 - 2*x^3)/((1 - x)*(1 - 2*x - x^2)*(1 + 2*x - x^2)).
Bisection: a(2*k) = (5*S(k-1, 6) + S(k-2, 6) + 1)/2 and a(2*k+1) = (S(k, 6) + 5*S(k-1, 6) + 1)/2, for k >= 0, with the Chebyshev polynomials S(n, x) (A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n+1).
Bisection: a(2*k) = (1 + 8*q(k) - p(k))/2 and a(2*k+1) = (1 + 8*q(k) + p(k))/2, for k >= 0, with p(k) = A001541(k) = S(k, 6) - 3*S(k-1, 6) and q(k) = A001109(k) = S(k-1, 6).
E.g.f.: (cosh(x) - cosh(sqrt(2)*x)*(cosh(x) - 3*sinh(x)) + sinh(x) - sqrt(2)*(cosh(x) - 2*sinh(x))*sinh(sqrt(2)*x))/2. - Stefano Spezia, Aug 29 2023

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

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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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A306561 Square numbers that are also central polygonal numbers (i.e., square numbers found in the Lazy Caterer's sequence).

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A358682 Numbers k such that 8*k^2 + 8*k - 7 is a square.

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A364319 a(n) = (A077446(n) + 1)/2 for n >= 0.

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A358682 Numbers k such that 8k^2 + 8k - 7 is a square.