A129444 -id:A129444 - cp's OEIS frontend

A129445 Numbers k > 0 such that k^2 is a centered triangular number.

1, 2, 8, 19, 79, 188, 782, 1861, 7741, 18422, 76628, 182359, 758539, 1805168, 7508762, 17869321, 74329081, 176888042, 735782048, 1751011099, 7283491399, 17333222948, 72099131942, 171581218381, 713707828021, 1698478960862, 7064979148268, 16813208390239

Offset: 1

Alexander Adamchuk, Apr 15 2007, Apr 26 2007

Corresponding numbers n such that centered triangular number A005448(n) is a perfect square are listed in A129444(n).
Consider Diophantine equation 3*x*(x-1) + 2 - 2*y^2 = 0. Sequence gives solutions for y. - Zak Seidov, Jun 11 2013
Positive values of x (or y) satisfying x^2 - 10xy + y^2 + 15 = 0. - Colin Barker, Feb 09 2014
Nonnegative values of x of solutions (x, y) to the Diophantine equation 8*x^2 - 3*y^2 = 5. - Jon E. Schoenfield, Feb 02 2021

Alexander Adamchuk, Table of n, a(n) for n = 1..100
Tom Beldon and Tony Gardiner, Triangular Numbers and Perfect Squares, The Mathematical Gazette, Vol. 86, No. 507, (2002), pp. 423-431. - _Ant King_, Dec 07 2010
Index entries for linear recurrences with constant coefficients, signature (0, 10, 0, -1).

Cf. A005448, A129444.
Prime terms are listed in A129446.
Cf. A125602 (prime CTN), A184481 (semiprime CTN), A125603.
Cf. A000290, A249483.

Do[f = 3n(n-1)/2 + 1; If[IntegerQ[Sqrt[f]], Print[Sqrt[f]]], {n, 150000}]
LinearRecurrence[{0, 10, 0, -1}, {1, 2, 8, 19}, 30] (* T. D. Noe, Jun 13 2013 *)

a(n) = sqrt(3*A129444(n)*(A129444(n) - 1)/2 + 1).
G.f.: x*(1-x)*(1+3*x+x^2)/(1-10*x^2+x^4). - Colin Barker, Apr 11 2012
a(n) = 10*a(n-2) - a(n-4), a(1..4) =  1, 2, 8, 19. - Zak Seidov, Jun 11 2013

More terms from Alexander Adamchuk, Apr 26 2007

A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

Original entry on oeis.org

0, 1, 6, 15, 64, 153, 638, 1519, 6320, 15041, 62566, 148895, 619344, 1473913, 6130878, 14590239, 60689440, 144428481, 600763526, 1429694575, 5946945824, 14152517273, 58868694718, 140095478159, 582740001360, 1386802264321, 5768531318886, 13727927165055

Offset: 1

Bruno Berselli, Dec 10 2013

For n>1, partial sums of A080872 starting from A080872(1).

			153 is in the sequence because 3*153*154/2+1 = 188^2.

Bruno Berselli, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Sequence A129444 gives n+1.
Cf. A000217, A080872, A129445 (square roots of 3*A000217(a(n))+1), A132596 (numbers m such that 3*A000217(m) is a square).
Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; this sequence for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6.

LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 6, 15, 64}, 30]

G.f.: x^2*(1 + 5*x - x^2 - x^3) / ((1 - x)*(1 - 10*x^2 + x^4)).
a(n) = a(n-1) +10*a(n-2) -10*a(n-3) -a(n-4) +a(n-5) for n>5, a(1)=0, a(2)=1, a(3)=6, a(4)=15, a(5)=64.
a(n) = -1/2 + ( (-3*(-1)^n + 2*sqrt(6))*(5 + 2*sqrt(6))^floor(n/2) - (3*(-1)^n + 2*sqrt(6))*(5 - 2*sqrt(6))^floor(n/2) )/12.

A120744 Least k>0 such that a centered polygonal number nk(k+1)/2+1 is a perfect square; or -1 if no such number exists.

Original entry on oeis.org

2, -1, 1, 3, 2, 7, 15, 1, 16, 8, 14, 4, 5, 15, 1, 2, 5, -1, 6, 3, 2, 39, 6, 1, 21, 7, 110, 3, 15, 7, 15, -1, 2, 8, 1, 4, 989, 8, 14, 2, 45, 15, 9, 4, 5, 335, 9, 1, 29, -1, 30, 15, 10, 415, 6, 2, 10, 32, 54, 3, 77, 55, 1, 5, 2, 7, 47750, 11, 15, 23, 47, -1, 48, 24, 16, 12, 5, 8, 2639, 1, 6720, 704, 38, 4, 2, 39, 505, 3, 13, 56, 9, 20, 13, 1631, 41

Offset: 1

Alexander Adamchuk, Apr 26 2007

sign

			a(5) = 2 because A129556(2) = 2>1 and A129556(1) = 0<1.

Max Alekseyev, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Centered Polygonal Number

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

a(n) = -1 for n in A166259.

a(n) = 1 for n = k^2-1.

Edited and b-file provided by Max Alekseyev, Jan 20 2010

A249483 Squares (A000290) which are also centered triangular numbers (A005448).

Original entry on oeis.org

1, 4, 64, 361, 6241, 35344, 611524, 3463321, 59923081, 339370084, 5871850384, 33254804881, 575381414521, 3258631508224, 56381506772644, 319312633001041, 5524812282304561, 31289379402593764, 541375222159074304, 3066039868821187801, 53049246959306977201

Offset: 1

Colin Barker, Jan 13 2015

Apart from the first term the same as A165516. - R. J. Mathar, Jan 20 2015

			64 is in the sequence because the 8th square is 64, which is also the 7th centered triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).

Cf. A000290, A005448, A129444, A129445.

I:=[1,4,64,361,6241]; [n le 5 select I[n] else Self(n-1)+98*Self(n-2)-98*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Jan 20 2015

CoefficientList[Series[(x^2 - 5 x + 1) (x^2 + 8 x + 1) / ((1 - x) (x^2 - 10 x + 1) (x^2 + 10 x + 1)), {x, 0, 70}], x] (* Vincenzo Librandi, Jan 20 2015 *)

Vec(-x*(x^2-5*x+1)*(x^2+8*x+1)/((x-1)*(x^2-10*x+1)*(x^2+10*x+1)) + O(x^100))

a(n) = a(n-1)+98*a(n-2)-98*a(n-3)-a(n-4)+a(n-5).

G.f.: -x*(x^2-5*x+1)*(x^2+8*x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)).

A129446 Primes p such that p^2 is a centered triangular number; or primes in A129445.

Original entry on oeis.org

2, 19, 79, 1861, 7741, 1647522841025041

Offset: 1

Alexander Adamchuk, Apr 15 2007, Apr 26 2007

nonn

The next terms are too large to include, see the b-file.

Alexander Adamchuk, Apr 26 2007, Table of n, a(n) for n = 1..9

Cf. A005448 (centered triangular numbers).
Cf. A129444 (numbers k such that centered triangular number A005448(k) = 3*k*(k-1)/2 + 1 is a perfect square).
Cf. A129445 (numbers k > 0 such that k^2 is a centered triangular number).

Do[ f = 3n(n-1)/2 + 1; If[ PrimeQ[ Sqrt[f] ], Print[ Sqrt[f] ] ], {n,1,150000} ]

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

A257274 Numbers whose square can be written as sum of at least 3 consecutive triangular numbers (A000217).

Original entry on oeis.org

8, 10, 19, 26, 44, 58, 79, 84, 91, 105, 111, 121, 135, 140, 154, 172, 184, 188, 195, 203, 208, 212, 217, 222, 230, 240, 246, 265, 276, 286, 292, 316, 322, 329, 338, 364, 390, 426, 429, 462, 490, 498, 506, 548, 550, 605, 704, 714, 715, 763, 770, 780, 782, 807

Offset: 1

M. F. Hasler, May 02 2015

nonn

Any square can trivially be written as sum of two consecutive triangular numbers T = A000217, since T(n-1) + T(n) = n(n-1)/2 + n(n+1)/2 = n*2n/2 = n^2. But it seems nontrivial to determine the squares that can be written as sum of more than 2 consecutive triangular numbers.
Some of these have two different decompositions of this form, e.g., 286^2 = T(13)+...+T(78) = T(75)+...+T(96), 826^2 = T(13)+...+T(159) = T(43)+...+T(160). What is the sequence of these numbers?
The terms > 2 of A129445, numbers k > 0 such that k^2 is a centered triangular number, form a subsequence: they correspond to k^2 = T(n-2) + T(n-1) + T(n), with n in sequence A129444: numbers n such that centered triangular number A005448(n) = 3n(n-1)/2 + 1 is a perfect square.
Terms > 2 of sequence A075870 also form a subsequence, namely the numbers whose square is the sum of four triangular numbers T(n-3)+...+T(n), with n given by twice the terms > 1 of A046090 or A182435.

			8^2 = T(5)+T(6)+T(7), 10^2 = T(5)+T(6)+T(7)+T(8), 19^2 = T(14)+T(15)+T(16), 26^2 = T(3)+...+T(15), 44^2 = T(13)+...+T(23), ...

Chai Wah Wu, Table of n, a(n) for n = 1..10000

{a=[];(S(n)=binomial(n+2,3)); for(n=1,999,for(k=1,n-3,issquare(S(n)-S(k))&&a=concat(a,sqrtint(S(n)-S(k))))); Set(a)[1..50]}

a(14), a(43)-a(54) from Chai Wah Wu, Jan 20 2016

A136331 The discriminant of the characteristic polynomial of the O+ and O- submatrix for spin 3 of the nuclear electric quadrupole Hamiltonian is a perfect square for these values.

Original entry on oeis.org

0, 3, 6, 21, 48, 195, 462, 1917, 4560, 18963, 45126, 187701, 446688, 1858035, 4421742, 18392637, 43770720, 182068323, 433285446, 1802290581, 4289083728, 17840837475, 42457551822, 176606084157, 420286434480, 1748220004083

Offset: 0

Lorenz H. Menke, Jr., Mar 27 2008

Perfect square values for discriminants are used to classify the Galois group of a polynomial. The O+ discriminant component is sqrt(6*(x^2-3*x+6)) (used to generate these values) and for the O- discriminant sqrt(6*(x^2+3*x+6)).
This sequence is the negative of the O+ sequence. Also, note that if 3*a(n) represents the positive terms, the negative terms are generated from 3 - 3*a(n).
For the O- sequence, reverse the O+ sequence and change all of the signs to generate ..., -446688, -45126, -18963, -4560, -1917, -462, -195, -48, -21, -6, -3, 0, 3, 18, 45, 192, 459, 1914, 4557, 18960, 45123, 187698, 446685.
Note that the difference equation a(n) generates the above sequence divided by 3 or ..., -148895, -62566, -15041, -6320, -1519, -638, -153, -64, -15, -6, -1, 0, 1, 2, 7, 16, 65, 154, 639, 1520, 6321, 15042, 148896, ...
This sequence, its reverse, and the division-by-3 form all appear to be new.

			G.f. = 3*x + 6*x^2 + 21*x^3 + 48*x^4 + 195*x^5 + 462*x^6 + 1917*x^7 + ...

Mohammad K. Azarian, On the Hyperfactorial Function, Hypertriangular Function, and the Discriminants of Certain Polynomials, International Journal of Pure and Applied Mathematics, Vol. 36, No. 2, 2007, pp. 251-257. Mathematical Reviews, MR2312537. Zentralblatt MATH, Zbl 1133.11012.
The physics reference is G. W. King, "The Asymmetric Rotor I. Calculation and Symmetry Classification of Energy Levels", Journal of Chemical Physics, Jan 1943, Volume 11, pp. 27-42.

Cf. A129444, A138976.

Do[If[IntegerQ[Sqrt[6 (6 - 3 x + x^2)]], Print[{x, Sqrt[6 (6 - 3 x + x^2)]}]], {x, -1000, 1000}]; Do[If[IntegerQ[Sqrt[6 (6 + 3 x + x^2)]], Print[{x, Sqrt[6 (6 + 3 x + x^2)]}]], {x, -1000, 1000}];

{a(n) = my(m); m = if( n<0, m = 1-n, n); 3*(n<0) + 3*(-1)^(n<0) * polcoeff( (x + x^2 - 5*x^3 - x^4) / ((1 - x) * (1 - 10*x^2 + x^4)) + x*O(x^m), m)}; /* Michael Somos, Apr 05 2008 */

The difference equation is a(n) = 11*(a(n-2) - a(n-4)) + a(n-6) with a(0)=0, a(1)=1, a(2)=2, a(3)=7, a(4)=16, a(5)=65. The solution is a(n) = 1/2 - (1/12)*(3+2*sqrt(6))*(5-2*sqrt(6))^(n/2) + (1/12)*(-3+2*sqrt(6))*(5+2*sqrt(6))^(n/2) for even n, a(n) = 1/2 - (1/12)*(3*sqrt(2) + sqrt(3))*(5-2*sqrt(6))^(n/2) + (1/12)*(3*sqrt(2) - sqrt(3))*(5+2*sqrt(6))^(n/2) for odd n. Multiply the resultant sequence by 3 to generate the present sequence.
G.f.: 3 * (x + x^2 - 5*x^3 - x^4) / (1 - x - 10*x^2 + 10*x^3 + x^4 - x^5). - Michael Somos, Apr 05 2008
a(n) = A138976(-n) for all n in Z. a(n) = 3 * A129444(n+1).

I rather feel that this should be broken up into two sequences, one each for the positive and negative terms, both starting at 0. - N. J. A. Sloane, Apr 04 2008

More terms from Michael Somos, Apr 05 2008

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

A129445 Numbers k > 0 such that k^2 is a centered triangular number.

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A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

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A120744 Least k>0 such that a centered polygonal number nk(k+1)/2+1 is a perfect square; or -1 if no such number exists.

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A249483 Squares (A000290) which are also centered triangular numbers (A005448).

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A129446 Primes p such that p^2 is a centered triangular number; or primes in A129445.

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Mathematica

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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A257274 Numbers whose square can be written as sum of at least 3 consecutive triangular numbers (A000217).

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PARI

Extensions

A136331 The discriminant of the characteristic polynomial of the O+ and O- submatrix for spin 3 of the nuclear electric quadrupole Hamiltonian is a perfect square for these values.

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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.