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

%I #58 Oct 31 2024 21:25:03

%S 1,9,29,61,105,161,229,309,401,505,621,749,889,1041,1205,1381,1569,

%T 1769,1981,2205,2441,2689,2949,3221,3505,3801,4109,4429,4761,5105,

%U 5461,5829,6209,6601,7005,7421,7849,8289,8741,9205,9681,10169

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

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

%C 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

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

%C 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

%H John Elias, <a href="/A136392/a136392.png">Illustration: Centered nesting cube frames</a>.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

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

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

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

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

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

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

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

%F From _Elmo R. Oliveira_, Oct 31 2024: (Start)

%F E.g.f.: exp(x)*(2*x*(3*x - 2) + 5) - 5.

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

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

%o (PARI) a(n)=6*n^2-10*n+5 \\ _Charles R Greathouse IV_, Nov 29 2011

%o (Haskell)

%o a136392 n = 2 * n * (3*n - 5) + 5 -- _Reinhard Zumkeller_, Feb 06 2012

%Y Cf. A000567, A001318, A016777, A033292, A033580, A059722, A199859, A201279, A204675.

%K nonn,easy,changed

%O 1,2

%A _Gary W. Adamson_, Dec 28 2007