A001570 -id:A001570 - cp's OEIS frontend

A156712 Star numbers (A003154) that are also triangular numbers (A000217).

1, 7, 91, 1261, 17557, 244531, 3405871, 47437657, 660721321, 9202660831, 128176530307, 1785268763461, 24865586158141, 346332937450507, 4823795538148951, 67186804596634801, 935791468814738257, 13033893758809700791, 181538721154521072811, 2528508202404485318557

Offset: 1

Aaron Meyerowitz, Feb 14 2009

From Colin Barker, Jan 06 2015: (Start)
Also indices of centered square numbers (A001844) which are also centered triangular numbers (A005448).
Also indices of centered octagonal numbers (A016754) which are also centered hexagonal numbers (A003215).
Also positive integers y in the solutions to 3*x^2-4*y^2-3*x+4*y = 0, the corresponding values of x being A001922.
(End)

Colin Barker, Table of n, a(n) for n = 1..875
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 30, 56.
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Wikipedia, Star Numbers
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A000217, A001570, A001844, A001922, A003154.

Cf. A003215, A005448, A007655, A016754.

[(Evaluate(ChebyshevSecond(n+1),7) - 13*Evaluate(ChebyshevU(n), 7) + 1)/2: n in [1..30]]; // G. C. Greubel, Oct 07 2022

f:= gfun[rectoproc]({a(n+3)=15*a(n+2)-15*a(n+1)+a(n),a(1)=1,a(2)=7,a(3)=91},a(n),'remember'):
seq(f(n), n=1..30); # Robert Israel, Jan 01 2015

f[n_] := (Simplify[(2 + Sqrt@3)^(2 n - 1) + (2 - Sqrt@3)^(2 n - 1)] + 4)/8; Array[f, 17] (* Robert G. Wilson v, Oct 28 2010 *)

Vec(-x*(x^2-8*x+1)/((x-1)*(x^2-14*x+1)) + O(x^100)) \\ Colin Barker, Jan 01 2015

def A156712(n): return (1 + chebyshev_U(n, 7) - 13*chebyshev_U(n-1, 7))/2
[A156712(n) for n in range(1,31)] # G. C. Greubel, Oct 07 2022

a(n+3) = 15*a(n+2) - 15*a(n+1) + a(n).
If x^2 - 3*y^2 = 1 with x even then a(y) = (y+2)/4 evidently related to A001570 by: add 1 and halve.
G.f.: x*(1 - 8*x + x^2)/((1-x)*(1 - 14*x + x^2)). - Alexander R. Povolotsky, Feb 15 2009
a(n) = (4 + (2 + sqrt(3))*(7 - 4*sqrt(3))^n + (2 - sqrt(3))*(7 + 4*sqrt(3))^n)/8. - Colin Barker, Mar 05 2016
a(n) = (1/2)*( 1 + ChebyshevU(n, 7) - 13*ChebyshevU(n-1, 7) ). - G. C. Greubel, Oct 07 2022

a(11) onwards from Robert G. Wilson v, Oct 28 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

A232765 Values of y solving x^2 = floor(y^2/3 + y).

Original entry on oeis.org

0, 1, 4, 9, 28, 73, 144, 409, 1036, 2025, 5716, 14449, 28224, 79633, 201268, 393129, 1109164, 2803321, 5475600, 15448681, 39045244, 76265289, 215172388, 543830113, 1062238464, 2996964769, 7574576356, 14795073225, 41742334396, 105500238889, 206068786704, 581395716793, 1469428768108

Offset: 1

Richard R. Forberg, Nov 29 2013

nonn

The corresponding values of x are given by A232771.
a(n) + 3 gives the values of y solving x^2 = floor(y^2/3 - y), and yields the same values for x.
a(3n+1) are squares whose square roots are given by A005320.
Let b(n) equal the second differences of a(n) where b(1) = 2. Then, for n>0, b(3n-1) = b(3n-2) = 2* A001570(n+1);  b(3n)= 2*A011943(n);  and b(3n) = (b(3n+1) + b(3n-1))/2.

Cf. A005320, A011943, A001570, A110293.

is(n)=issquare(n^2\3+n)
print1("0, 1");for(x=3,99,y=round(sqrt(3)*x-3/2);if(is(y),print1(", "y))) \\ Charles R Greathouse IV, Dec 09 2013

Empirical g.f.: -x^2*(x+1)*(x^2+x+1)^2 / ((x-1)*(x^6-14*x^3+1)). - Colin Barker, Dec 30 2014

a(23) corrected by Colin Barker, Dec 30 2014

A106257 Numbers k such that k^2 = 12*n^2 + 13.

Original entry on oeis.org

5, 11, 59, 149, 821, 2075, 11435, 28901, 159269, 402539, 2218331, 5606645, 30897365, 78090491, 430344779, 1087660229, 5993929541, 15149152715, 83484668795, 211000477781, 1162791433589, 2938857536219, 16195595401451

Offset: 1

Pierre CAMI, Apr 28 2005

			5^2=12*1^2+13
11^2=12*3^2+13
59^2=12*17^2+13
149^2=12*43^2+13

Index entries for linear recurrences with constant coefficients, signature (0,14,0,-1).

Cf. A106256.

LinearRecurrence[{0,14,0,-1},{5,11,59,149},40] (* Harvey P. Dale, Oct 21 2021 *)

Vec(-x*(x-1)*(5*x^2+16*x+5)/((x^2-4*x+1)*(x^2+4*x+1)) + O(x^100)) \\ Colin Barker, Apr 16 2014

k(1)=5, k(2)=11, k(3)=14*k(1)-k(2), k(4)=14*k(2)-k(1) then k(n)=14*k(n-2)-k(n-4).
G.f.: -x*(x-1)*(5*x^2+16*x+5) / ((x^2-4*x+1)*(x^2+4*x+1)). - Corrected by Colin Barker, Apr 16 2014
a(2n) = (9*A001570(n)+A001570(n+1))/2, a(2n+1) = 5*A001570(n)-6*A007655(n).

Edited by Ralf Stephan, Jun 01 2007

A175155 Numbers m satisfying m^2 + 1 = x^2 * y^3 for positive integers x and y.

Original entry on oeis.org

0, 682, 1268860318, 1459639851109444, 2360712083917682, 86149711981264908618, 4392100110703410665318, 8171493471761113423918890682, 15203047261220215902863544865414318, 5484296027914919579181500526692857773246, 28285239023397517753374058381589688919682, 12439333951782387734360136352377558500557329868

Offset: 1

Michel Lagneau, Feb 27 2010

nonn

This sequence is infinite. The fundamental solution of m^2 + 1 = x^2 y^3 is (m,x,y) = (682,61,5), which means the Pellian equation m^2 - 125x^2 = -1 has the solution (m,x) = (682,61) = (m(1),x(1)). This Pellian equation admits an infinity of solutions (m(2k+1),x(2k+1)), k=1,2,..., given by the following recursive relation, starting with m(1)=682, x(1)= 61: m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1).
Squares of these terms are in A060355, since both a(n)^2 and a(n)^2 + 1 are powerful (A001694). - Charles R Greathouse IV, Nov 16 2012
It appears that y = A077426. - Robert G. Wilson v, Nov 16 2012
Also m^2 + 1 is powerful. Other solutions arise from solutions x to x^2 - k^3*y^2 = -1. - Georgi Guninski, Nov 17 2012
Although it is believed that the b-file is complete for all terms m < 10^100, the search only looked for y < 100000. - Robert G. Wilson v, Nov 17 2012

			For m=682, m^2 + 1 = 465125 = 61^2 * 5^3.

Albert H. Beiler, "The Pellian" (Chap. 22), Recreations in the Theory of Numbers, 2nd ed. NY: Dover, 1966.
A. Cayley, Report of a committee appointed for the purpose of carrying on the tables connected with the Pellian equation ..., Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 430-443.
J. M. De Koninck, Ces nombres qui nous fascinent, Ellipses, 2008, p. 108.

Robert G. Wilson v, Table of n, a(n) for n = 1..292. [This list has not been proved to be complete! - _N. J. A. Sloane_, Nov 17 2012]
E. E. Whitford, The Pell equation, New York, 1912.
Wikipedia, Pell's equation

Pellian equation: A006704, A006702, A006705, A006703, A003153, A001571, A001570.

C:=array(0..20,0..20):C[1,1]=1: C[2,1]=1: n1:=682:x1:=61:for nn from 1 by 2 to 15 do:s:=0:for i from 2 to 15 do:for j from 1 to i do:C[i,j]:= C[i-1,j] + C[i-1,j-1]: od:od:for n from 1 by 2 to nn+1 do:s:=s + C[nn+1,n] * n1^(nn-n+1)*x1^(n-1)*125^((n-1)/2):od:print (s):od: # Michel Lagneau
# 2nd program R. J. Mathar, Mar 16 2016:
# print (nonsorted!) all solutions of A175155 up to search limit
with(numtheory):
# upper limit for solutions n
nsearchlim := 10^40 :
A175155y := proc(y::integer)
    local disc;
    disc := y^3 ;
    cfrac(sqrt(disc),periodic,quotients) ;
end proc:
for y from 2 do
    if issqrfree(y) then
        # find continued fraction for x^2-(y^3=disc)*y^2=-1, sqrt(disc)
        cf := A175155y(y) ;
        nlen :=  nops(op(2,cf)) ;
        if type(nlen,odd) then
            # fundamental solution
            fuso := numtheory[nthconver](cf,nlen-1) ;
            fusolx := numer(fuso) ;
            fusoly := denom(fuso) ;
            solx := fusolx ;
            soly := fusoly ;
            while solx <= nsearchlim do
                rhhs := solx^2-y^3*soly^2 ;
                if rhhs = -1 then
                    # print("n=",solx,"x=",soly,"y=",y^3) ;
                    print(solx) ;
                end if;
                # solutions from fundamental solution
                tempx := fusolx*solx+y^3*fusoly*soly ;
                tempy := fusolx*soly+fusoly*solx ;
                solx := tempx ;
                soly := tempy ;
            end do;
        end if;
    fi;
end do:

nmax = 10^50; ymax = 100; instances = 10; fi[y_] := n /. FindInstance[0 <= n <= nmax && x > 0 && n^2 + 1 == x^2*y^3, {n, x}, Integers, instances]; yy = Select[Range[1, ymax, 2], !IntegerQ[Sqrt[#]] && OddQ[ Length[ ContinuedFraction[Sqrt[#]][[2]]]]&]; Join[{0}, fi /@ yy // Flatten // Union // Most] (* Jean-François Alcover, Jul 12 2017 *)

is(n)=ispowerful(n^2+1) \\ Charles R Greathouse IV, Nov 16 2012

m(1)=682, x(1) = 61 and m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1) m(2k+1) = C(2k+1,0) * m(1)^(2k+1) + C(2k+1,2)*m(1)^(2k-1)*x(1)^2 + ...

Added condition that x and y must be positive. Added missing initial term 0. Added warning that b-file has not been proved to be correct - there could be missing entries. - N. J. A. Sloane, Nov 17 2012

A341671 Solutions y of the Diophantine equation 3*(x^2+x+1) = y^2.

Original entry on oeis.org

3, 39, 543, 7563, 105339, 1467183, 20435223, 284625939, 3964327923, 55215964983, 769059181839, 10711612580763, 149193516948843, 2077997624703039, 28942773228893703, 403120827579808803, 5614748812888429539, 78203362552858204743, 1089232326927126436863, 15171049214426911911339

Offset: 1

Bernard Schott, Feb 17 2021

nonn

Corresponding x are in A028231.
This equation belongs to the family of equations studied by Kustaa A. Inkeri, y^m = a * (x^q-1)/(x-1) with here: m=2, a=3, q=3. This equation is exhibed in A307745 by Giovanni Resta to prove that this sequence has infinitely many terms.
This Diophantine equation 3*(x^2+x+1) = y^2 has infinitely many solutions because the Pell-Fermat equation u^2 - 3*v^2 = -2 also has infinitely many solutions. The corresponding (u,v) are in (A001834, A001835) and for each pair (u,v), the corresponding solutions of 3*(x^2+x+1) = y^2 are x = (3*u*v-1)/2 and y = 3*(u^2+1)/2.
Note that if y = 3*z, this equation becomes 3*z^2 = x^2+x+1 with solutions (x, z) = (A028231, A001570).

			The first few values for (x,y) are (1,3), (22,39), (313,543), (4366,7563), (60817,105339), ...

Kustaa A. Inkeri, On the Diophantine equation a(x^n-1)/(x-1) = y^m, Acta Arithmetica, Vol. 21, No. 1 (1972), pp. 299-311.

Cf. A001834, A001835, A001570, A028231, A307745.

Subsequence of A158235, for a(n)>3.

f[x_] := Sqrt[3*(x^2 + x + 1)]; f /@ LinearRecurrence[{15, -15, 1}, {1, 22, 313}, 20] (* Amiram Eldar, Feb 17 2021 *)

a(n) = 3*A001570(n). - Hugo Pfoertner, Feb 17 2021

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).

More terms from Amiram Eldar, Feb 17 2021

A145856 Least number k>1 such that centered n-gonal number n*k(k-1)/2+1 is a perfect square, or 0 if no such k exists.

Original entry on oeis.org

3, 0, 2, 4, 3, 8, 16, 2, 17, 9, 15, 5, 6, 16, 2, 3, 6, 0, 7, 4, 3, 40, 7, 2, 22, 8, 111, 4, 16, 8, 16, 0, 3, 9, 2, 5, 990, 9, 15, 3, 46, 16, 10, 5, 6, 336, 10, 2, 30, 0, 31, 16, 11, 416, 7, 3, 11, 33, 55, 4, 78, 56, 2, 6, 3, 8, 47751, 12, 16, 24, 48, 0, 49, 25, 17, 13, 6, 9, 2640, 2, 6721

Offset: 1

Alexander Adamchuk, Oct 22 2008

nonn

Jonathan Vos Post, When Centered Polygonal Numbers are Perfect Squares, submitted to Mathematics Magazine, 4 May 2004, manuscript no. 04-1165, unpublished, available upon request. - Jonathan Vos Post, Oct 25 2008

Eric Weisstein's World of Mathematics, Centered Polygonal Numbers
Index entries for sequences related to centered polygonal numbers

Cf. A120744, A001542, A166259, A006451, A001921, A129444, A001570, A001652, A129556, A053606, A105038, A105040, A053141, A061278.

a(n) = 0 for n in A166259.

a(n) = A120744(n) + 1. - Alexander Adamchuk, Oct 10 2009

Edited by Max Alekseyev, Jan 23 2010

cp's OEIS Frontend

A156712 Star numbers (A003154) that are also triangular numbers (A000217).

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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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A232765 Values of y solving x^2 = floor(y^2/3 + y).

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A106257 Numbers k such that k^2 = 12*n^2 + 13.

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A175155 Numbers m satisfying m^2 + 1 = x^2 * y^3 for positive integers x and y.

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Maple

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A341671 Solutions y of the Diophantine equation 3*(x^2+x+1) = y^2.

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Mathematica

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A145856 Least number k>1 such that centered n-gonal number n*k(k-1)/2+1 is a perfect square, or 0 if no such k exists.

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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.