A050533 -id:A050533 - cp's OEIS frontend

A057813 a(n) = (2n+1)(4n^2+4n+3)/3.

1, 11, 45, 119, 249, 451, 741, 1135, 1649, 2299, 3101, 4071, 5225, 6579, 8149, 9951, 12001, 14315, 16909, 19799, 23001, 26531, 30405, 34639, 39249, 44251, 49661, 55495, 61769, 68499, 75701, 83391, 91585, 100299, 109549, 119351, 129721, 140675, 152229, 164399

Offset: 0

N. J. A. Sloane, Nov 07 2000

For n>0, 30*a(n) is the sum of the ten distinct products of 2*n-1, 2*n+1, and 2*n+3. For example, when n = 1, we sum the ten distinct products of 1, 3, and 5: 1*1*1 + 1*1*3 + 1*1*5 + 1*3*3 + 1*3*5 + 1*5*5 + 3*3*3 + 3*3*5 + 3*5*5 + 5*5*5 = 330 = 30*11 = 30*a(1). - J. M. Bergot, Apr 06 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

1/12*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

Cf. A019560, A336266.

[(2*n+1)*(4*n^2+4*n+3)/3 : n in [0..50]] // Wesley Ivan Hurt, Apr 22 2014

A057813:=n->(2*n + 1)*(4*n^2 + 4*n + 3)/3; seq(A057813(n), n=0..50); # Wesley Ivan Hurt, Apr 06 2014

Table[(2*n + 1)*(4*n^2 + 4*n + 3)/3, {n, 0, 50}] (* David Nacin, Mar 01 2012 *)

P(x, y, z) = x^3 + x^2*y + x^2*z + x*y^2 + x*y*z + x*z^2 + y^3 + y^2*z + y*z^2 + z^3;
a(n) = P(2*n-1, 2*n+1, 2*n+3)/30; \\ Michel Marcus, Apr 22 2014

a(n) = 2*A050533(n) + 1. - N. J. A. Sloane, Sep 22 2004
G.f.: (1+7*x+7*x^2+x^3)/(1-x)^4. - Colin Barker, Mar 01 2012
G.f. for sequence with interpolated zeros: 1/(8*x)*sinh(8*arctanh(x)) = 1/(16*x)*( ((1 + x)/(1 - x))^4 - ((1 - x)/(1 + x))^4 ) = 1 + 11*x^2 + 45*x^4 + 119*x^6 + .... Cf. A019560. - Peter Bala, Apr 07 2017
E.g.f.: (3 + 30*x + 36*x^2 + 8*x^3)*exp(x)/3. - G. C. Greubel, Dec 01 2017
From Peter Bala, Mar 26 2024: (Start)
12*a(n) = (2*n + 1)*(a(n + 1) - a(n - 1)).
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 3*Pi/16 - 1/2. Cf. A016754 and A336266. (End)

A050492 Thickened cube numbers: a(n) = n(n^2 + (n-1)^2) + (n-1)2n(n-1).

Original entry on oeis.org

1, 14, 63, 172, 365, 666, 1099, 1688, 2457, 3430, 4631, 6084, 7813, 9842, 12195, 14896, 17969, 21438, 25327, 29660, 34461, 39754, 45563, 51912, 58825, 66326, 74439, 83188, 92597, 102690, 113491, 125024, 137313, 150382, 164255, 178956

Offset: 1

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 27 1999

In other words, positive integers k such that 2*k - 1 is a perfect cube. - Altug Alkan, Apr 15 2016

a(n) represents the first term in a sum of (2*n - 1)^3 consecutive integers which equals (2*n - 1)^6. - Patrick J. McNab, Dec 24 2016

			       * *      *      * *
a(2) =  *   +  * *  +   *  = 14.
       * *      *      * *

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A001844, A046092, A050533.

[n*(4*n^2-6*n+3): n in [1..40]]; // Vincenzo Librandi, Oct 03 2011

Table[n(n^2+(n-1)^2)+(n-1)2n(n-1),{n,40}] (* or *) LinearRecurrence[ {4,-6,4,-1},{1,14,63,172},40] (* Harvey P. Dale, Oct 02 2011 *)

a(n)=n*(4*n^2-6*n+3) \\ Charles R Greathouse IV, Nov 10 2015

a(n) = n*(4*n^2-6*n+3).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(1)=1, a(2)=14, a(3)=63, a(4)=172. - Harvey P. Dale, Oct 02 2011
G.f.: x*(1+10*x+13*x^2)/(1-4*x+6*x^2-4*x^3+x^4). - Colin Barker, Jan 04 2012
a(n) = ((2n-1)^3 + 1)/2. - Dave Durgin, May 07 2014
E.g.f.: x*(4*x^2 + 6*x + 1)*exp(x). - G. C. Greubel, Apr 15 2016

cp's OEIS Frontend

A057813 a(n) = (2n+1)(4n^2+4n+3)/3.

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A050492 Thickened cube numbers: a(n) = n(n^2 + (n-1)^2) + (n-1)2n(n-1).

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cp's OEIS Frontend

A057813 a(n) = (2*n+1)*(4*n^2+4*n+3)/3.

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Author

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Magma

Maple

Mathematica

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A050492 Thickened cube numbers: a(n) = n*(n^2 + (n-1)^2) + (n-1)*2*n*(n-1).

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Magma

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Formula

A057813 a(n) = (2n+1)(4n^2+4n+3)/3.

A050492 Thickened cube numbers: a(n) = n(n^2 + (n-1)^2) + (n-1)2n(n-1).