A104249 a(n) = (3*n^2 + n + 2)/2.

1, 3, 8, 16, 27, 41, 58, 78, 101, 127, 156, 188, 223, 261, 302, 346, 393, 443, 496, 552, 611, 673, 738, 806, 877, 951, 1028, 1108, 1191, 1277, 1366, 1458, 1553, 1651, 1752, 1856, 1963, 2073, 2186, 2302, 2421, 2543, 2668, 2796, 2927, 3061, 3198, 3338, 3481

Offset: 0

Author

Thomas Wieder, Feb 26 2005

Keywords

• nonn
• easy

Comments

Second differences are all 3.

Related to the sequence of odd numbers A005408 since for these numbers the first differences are all 2.

Column 2 of A114202. - Paul Barry, Nov 17 2005

Equals third row of A167560 divided by 2. - Johannes W. Meijer, Nov 12 2009

A242357(a(n)) = n + 1. - Reinhard Zumkeller, May 11 2014

Also, this sequence is related to A011379, for n>0, by A011379(n) = n*a(n) - Sum_{i=0..n-1} a(i). - Bruno Berselli, Jul 08 2015

The number of Hamiltonian nonisomorphic unfoldings in an n-gonal Archimedean antiprism. See sequence A284647. - Rick Mabry, Apr 10 2021

Examples

			The sequence of first differences delta_a(n) = a(n+1) - a(n) is 2, 5, 8, 11, 14, 17, 20, 23, 26, ...
The sequence of second differences delta_delta_a(n) = a(n+2) - 2*a(n+1) + a(n) is: 3, 3, 3, 3, 3, 3, 3, ... E.g., 78 - 2*58 + 41 = 3.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..3000
• Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
• Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
• Rick Mabry, Fibonacci Numbers, Integer Compositions, and Nets of Antiprisms, The American Mathematical Monthly, Vol. 126 (2019), no. 9, pp. 786-801.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A001399, A002597, A005408, A011379, A016777, A143689.

Counts special cases of A284647.

Programs

Haskell

a104249 n = n*(3*n+1) `div` 2 + 1 -- Reinhard Zumkeller, May 11 2014

Magma

[(3*n^2+n+2)/2: n in [0..50]]; // Vincenzo Librandi, May 09 2011

Maple

a := proc (n) local i, u; option remember; u[0] := 1; u[1] := 3; u[2] := 8; for i from 3 to n do u[i] := -(4*u[i-3]-8*u[i-2]-2*u[i-1]+(-2*u[i-3]+2*u[i-2]-u[i-1])*i)/i end do; [seq(u[i],i = 0 .. n)] end proc;

Mathematica

A104249[n_] := (3*n^2 + n + 2)/2; Table[A104249[n], {n,0,100}] (* Vladimir Joseph Stephan Orlovsky, Jul 22 2011 *)
LinearRecurrence[{3,-3,1},{1,3,8},70] (* Harvey P. Dale, Jul 21 2023 *)

PARI

a(n)=n*(3*n+1)/2+1 \\ Charles R Greathouse IV, Oct 07 2015

Formula

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

Recurrence: (n+3)*u(n+3) + (-5-n)*u(n+2)*(-2+2*n)*u(n+1) + (-2-2*n)*u(n) = 0 for n >= 0 with u(0) = 1, u(1) = 3, and u(2) = 8.

From Paul Barry, Nov 17 2005: (Start)

a(0) = 1, a(n) = a(n-1) + 3*n - 1 for n > 0;

a(n) = Sum_{k=0..n} C(n, k)*C(2, k)*J(k+1), where J(n) = A001045(n). (End)

Binomial transform of [1, 2, 3, 0, 0, 0, ...]. - Gary W. Adamson, Apr 23 2008

E.g.f.: exp(x)*(2 + 4*x + 3*x^2)/2. - Stefano Spezia, Apr 10 2021