A129444 Numbers k such that the centered triangular number A005448(k) = 3*k*(k-1)/2 + 1 is a perfect square.

0, 1, 2, 7, 16, 65, 154, 639, 1520, 6321, 15042, 62567, 148896, 619345, 1473914, 6130879, 14590240, 60689441, 144428482, 600763527, 1429694576, 5946945825, 14152517274, 58868694719, 140095478160, 582740001361, 1386802264322

Offset: 1

Author

Alexander Adamchuk, Apr 15 2007

Keywords

• nonn
• easy

Comments

Corresponding numbers m > 0 such that m^2 is a centered triangular number are listed in A129445 = {1, 2, 8, 19, 79, 188, 782, 1861, 7741, 18422, 76628, 182359, ...}.

Examples

			G.f. = x^2 + 2*x^3 + 7*x^4 + 16*x^5 + 65*x^6 + 154*x^7 + 639*x^8 + 1520*x^9 + ...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Crossrefs

Cf. A005448 (centered triangular numbers).

Cf. A129445 (numbers k > 0 such that k^2 is a centered triangular number).

Cf. A000290, A004189, A249483.

Programs

Magma

I:=[0,1,2,7,16,65]; [n le 6 select I[n] else 11*Self(n-2) -11*Self(n-4) +Self(n-6): n in [1..40]]; // G. C. Greubel, Feb 07 2024

Mathematica

Do[ f = 3n(n-1)/2 + 1; If[ IntegerQ[ Sqrt[f] ], Print[ n ] ], {n,1,150000} ]
a[1]=0;a[2]=1;a[3]=2;a[4]=7;a[5]=16;a[6]=65;a[n_]:=a[n]=11(a[n-2]-a[n-4])+a[n-6];Table[a[n], {n, 100}] (* Zak Seidov, Apr 17 2007 *)
LinearRecurrence[{1,10,-10,-1,1},{0,1,2,7,16},30] (* Harvey P. Dale, Dec 06 2012 *)

PARI

{a(n) = my(m); m = if( n<1, 2-n, n-1); (n<1) + (-1)^(n<1) * 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 */

SageMath

def A129444_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^2*(1+x-5*x^2-x^3)/((1-x)*(1-10*x^2+x^4)) ).list()
a=A129444_list(40); a[1:] # G. C. Greubel, Feb 07 2024

Formula

a(n) = 1/2 + sqrt(1/4 + (2/3)*(A129445(n)^2 - 1)).

a(n) = 11*(a(n-2) - a(n-4)) + a(n-6); a(1)=0; a(2)=1; a(3)=2; a(4)=7; a(5)=16; a(6)=65. - Zak Seidov, Apr 17 2007

a(n) = 1 - a(-n+3) for all n in Z. - Michael Somos, Apr 05 2008

G.f.: x^2*(1 + x - 5*x^2 - x^3) / ((1 - x) * (1 - 10*x^2 + x^4)). - Michael Somos, Apr 05 2008

a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5); a(1)=0, a(2)=1, a(3)=2, a(4)=7, a(5)=16. - Harvey P. Dale, Dec 06 2012

a(n) = (1/2)*(2*[n=0] + 1 + ((1+(-1)^n)/2)*(31*b(n/2) - 3*b(n/2 + 1)) + ((1-(-1)^n)/2)*(13*b((n-1)/2) - b((n+1)/2))), where b(n)=A004189(n). - G. C. Greubel, Feb 07 2024

Extensions

More terms from Zak Seidov, Apr 17 2007