A136392 a(n) = 6*n^2 - 10*n + 5.

1, 9, 29, 61, 105, 161, 229, 309, 401, 505, 621, 749, 889, 1041, 1205, 1381, 1569, 1769, 1981, 2205, 2441, 2689, 2949, 3221, 3505, 3801, 4109, 4429, 4761, 5105, 5461, 5829, 6209, 6601, 7005, 7421, 7849, 8289, 8741, 9205, 9681, 10169

Offset

1,2

Comments

Binomial transform of [1, 8, 12, 0, 0, 0, ...].

Numbers k such that 6*k-5 is the square of a number of the form 6*k-5, contained in A199859. - _Eleonora Echeverri-Toro_, Nov 29 2011

Central terms of the triangle A033292. - _Reinhard Zumkeller_, Feb 06 2012

Sequence found by reading the line from 1, in the direction 1, 9, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - _Omar E. Pol_, Jul 18 2012

Links

Table of n, a(n) for n=1..42.

John Elias, Illustration: Centered nesting cube frames

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

a(n) = n*(3n-2) + (n-1)*(3n-5), n > 1.

a(n) = n*A016777(n-1) + (n-1)*A016777(n-2).

a(n) = a(n-1) + 12*n-16 (with a(1)=1). - _Vincenzo Librandi_, Nov 24 2010

G.f.: x*(1+x)*(1+5*x)/(1-x)^3. - _Colin Barker_, Jan 09 2012

a(n) = 1 + A033580(n-1). - _Omar E. Pol_, Jul 18 2012

a(n) = A059722(n) - A059722(n-1). - _J. M. Bergot_, Nov 02 2012

a(n) = A000567(n-1) + A000567(n). - _Charlie Marion_, May 29 2024

Mathematica

Table[6n^2-10n+5, {n, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {1, 9, 29}, 50] (* _Harvey P. Dale_, Mar 05 2023 *)

Prog

(PARI) a(n)=6*n^2-10*n+5 \\ _Charles R Greathouse IV_, Nov 29 2011
(Haskell)
a136392 n = 2 * n * (3*n - 5) + 5 -- _Reinhard Zumkeller_, Feb 06 2012

Crossrefs

Cf. A000567, A016777, A033580, A059722, A201279, A204675.

Sequence in context: A031296 A146869 A129397 * A272732 A272740 A162263

Adjacent sequences: A136389 A136390 A136391 * A136393 A136394 A136395

Keyword

nonn,easy

Author

_Gary W. Adamson_, Dec 28 2007

Status

approved