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A122769 Numbers k such that k^2 is of the form 3*m^2 + 2*m + 1 (A056109). 1
1, 11, 153, 2131, 29681, 413403, 5757961, 80198051, 1117014753, 15558008491, 216695104121, 3018173449203, 42037733184721, 585510091136891, 8155103542731753, 113585939507107651, 1582048049556775361 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

All terms are odd. Sequence is infinite. Corresponding m's are 0, 6, 88, 1230, 17136, 238678, 3324360, 46302366, 644908768, 8982420390, 125108976696, 1742543253358, 24270496570320. s^2 are squares in A056109.

The Diophantine equation A000290(x) = A000326(y) + A000326(y-1) has the solutions x = a(n) and y = (4^n+(1+sqrt(3))^(4*n-3)+(1-sqrt(3))^(4*n-3))/(3*2^(2*n-1)). - Bruno Berselli, Mar 04 2013

LINKS

Table of n, a(n) for n=1..17.

Tanya Khovanova, Recursive Sequences

Valcho Milchev and Tsvetelina Karamfilova, Domino tiling in grid - new dependence, arXiv:1707.09741 [math.HO], 2017.

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

FORMULA

Alternatively, with a different offset:

a(0) = 1, a(1) = 11, a(n) = 14*a(n-1) - a(n-2), and

a(n) = ((3 - b)*(7 - 4*b)^n + (3 + b)*(7 + 4*b)^n)/6, b=sqrt(3).

a(n) = -(1/6)*sqrt(3)*(7-4*sqrt(3))^n+(1/6)*sqrt(3)*(7+4*sqrt(3))^n+(1/2)*(7+4*sqrt(3))^n+(1/2)*(7-4*sqrt(3))^n. [Paolo P. Lava, Aug 06 2008]

G.f.: x*(1-3*x)/(1-14*x+x^2). [Philippe Deléham, Nov 17 2008]

CROSSREFS

Cf. A056109.

Sequence in context: A176365 A077577 A157186 * A051608 A191369 A223713

Adjacent sequences:  A122766 A122767 A122768 * A122770 A122771 A122772

KEYWORD

nonn,easy,changed

AUTHOR

Zak Seidov, Oct 21 2006

EXTENSIONS

Edited by N. J. A. Sloane, Oct 28 2006

STATUS

approved

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Last modified December 17 14:26 EST 2017. Contains 296104 sequences.