A061278 a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

0, 1, 5, 20, 76, 285, 1065, 3976, 14840, 55385, 206701, 771420, 2878980, 10744501, 40099025, 149651600, 558507376, 2084377905, 7779004245, 29031639076, 108347552060, 404358569165, 1509086724601, 5631988329240, 21018866592360, 78443478040201, 292755045568445

Offset

0,3

Comments

Indices m of triangular numbers T(m) which are one-third of another triangular number: 3*T(m) = T(k); the k's are given by A001571. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002

On the previous comment: for m=0 this is actually one third of the same triangular number. - Zak Seidov, Apr 07 2011

Also numbers n such that the n-th centered 24-gonal number 12*n*(n+1)+1 is a perfect square A001834(n)^2, where A001834(n) is defined by the recursion: a(0) = 1, a(1) = 5, a(n) = 4*a(n-1) - a(n-2) + 1. - Alexander Adamchuk, Apr 21 2007

Also numbers n such that RootMeanSquare(5,...,6*n-1) is an integer. - Ctibor O. Zizka, Dec 17 2008 (Corrected by Robert K. Moniot, Jul 22 2020)

Also numbers n such that n*(n+1) = Sum_{i=1..x} n+i for some x. (This does not apply to the first term.). - Gil Broussard, Dec 23 2008

From John P. McSorley, May 26 2020: (Start)

Consecutive terms (a(n-1), a(n)) = (u,v) give all points on the hyperbola u^2 - u + v^2 - v - 4*u*v = 0 in quadrant I with both coordinates an integer.

Also related to the block sizes of small multi-set designs. (End)

If a(n) white balls and a(n+1) black balls are mixed in a bag, and a pair of balls is drawn without replacement, the probability that one ball of each color is drawn is exactly 1/3. These are the only integers for which the probability is 1/3. For example, if there are 20 white balls and 76 black balls, the probability of drawing one of each is (20/96)*(76/95) + (76/96)*(20/95) = 1/3. - Elliott Line, May 13 2022

References

R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., 58:2 (2020), 140-142.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Niccolò Castronuovo, On the number of fixed points of the map gamma, arXiv:2102.02739 [math.NT], 2021. Mentions this sequence.

Brian Lawrence and Will Sawin, The Shafarevich conjecture for hypersurfaces in abelian varieties, arXiv:2004.09046 [math.NT], 2020.

Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.

L. A. Medina and A. Straub, On multiple and infinite log-concavity, 2013.

S. Morier-Genoud, V. Ovsienko and S. Tabachnikov, 2-frieze patterns and the cluster structure of the space of polygons, Annales de l'institut Fourier, 62 no. 3 (2012), 937-987. - From N. J. A. Sloane, Dec 26 2012

Robert Phillips, A triangular number result, 2009.

Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.

Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.

Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.

Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.

Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.

Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.

Eric Weisstein, Centered Polygonal Number.

Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Formula

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

a(n) = A001075(n) - a(n-1) - 1.

a(n) = (A001835(n+1) - 1)/2 = (A001353(n+1) - A001353(n) - 1)/2.

a(n) = a(n-1) + A001353(n), i.e., partial sum of A001353.

From Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002: (Start)

a(n+2) = 4*a(n+1) - a(n) + 1 for a(0)=0, a(1)=1.

G.f.: x/((1 - x)*(1 - 4*x + x^2)).

a(n) = (1/12)*((3 - sqrt(3))*(2 - sqrt(3))^n + (3 + sqrt(3))*(2 + sqrt(3))^n-6). (End)

a(n) = (1/12)*(A003500(n) + A003500(n+1)-6). - Mario Catalani (mario.catalani(AT)unito.it), Apr 11 2003

a(n+1) = Sum_{k=0..n} U(k, 2)} = Sum_{k=0..n} S(k, 4), where U(n,x) and S(n,x) are Chebyshev polynomials. - Paul Barry, Nov 14 2003

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

a(n) = A079935(n+1) + A001571(n) for n>0, a(0)=0. - Gerry Martens, Jun 05 2015

a(n)*a(n-2) = a(n-1)*(a(n-1) - 1) for n>1. - Bruno Berselli, Nov 29 2016

From John P. McSorley, May 25 2020: (Start)

a(n)^2 - a(n) + a(n-1)^2 - a(n-1) - 4*a(n)*a(n-1) = 0.

a(n) = A001834(n-1) + a(n-2). (End)

(T(a(n)-1) + T(a(n+1)-1))/T(a(n) + a(n+1) - 1) = 2/3 where T(i) is the i-th triangular number. - Robert K. Moniot, Oct 11 2020

E.g.f.: exp(x)*(exp(x)*(3*cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)) - 3)/6. - Stefano Spezia, Feb 05 2021

a(n) = A101265(n) - 1. - Jon E. Schoenfield, Jan 01 2022

Example

a(2)=5 and T(5)=15 which is 1/3 of 45=T(9).

Maple

f:= gfun:-rectoproc({a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3), a(1)=1, a(0)=0, a(-1)=0}, a(n), remember):
map(f, [$0..50]); # Robert Israel, Jun 05 2015

Mathematica

CoefficientList[Series[x/(1 - 5*x + 5*x^2 - x^3), {x, 0, nn}], x] (* T. D. Noe, Jun 04 2012 *)
LinearRecurrence[{5, -5, 1}, {0, 1, 5}, 30] (* Harvey P. Dale, Dec 23 2012 *)

Prog

(PARI) M = [1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=1, 30, print1(([1, 0, 0]*M^i)[3], ", ")) \\ Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005
(Magma) I:=[0, 1]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2) + 1: n in [1..30]]; // Vincenzo Librandi, Dec 23 2012

Crossrefs

Cf. A001075, A001353, A001571, A001834, A001835, A079935, A101265. Also cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

Sequence in context: A275908 A290909 A270023 * A000758 A005283 A057552

Adjacent sequences: A061275 A061276 A061277 * A061279 A061280 A061281

Keyword

nonn,easy

Author

Henry Bottomley, Jun 04 2001

Extensions

More terms from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005

Status

approved