A195605 a(n) = (4*n*(n+2)+(-1)^n+1)/2 + 1.

2, 7, 18, 31, 50, 71, 98, 127, 162, 199, 242, 287, 338, 391, 450, 511, 578, 647, 722, 799, 882, 967, 1058, 1151, 1250, 1351, 1458, 1567, 1682, 1799, 1922, 2047, 2178, 2311, 2450, 2591, 2738, 2887, 3042, 3199, 3362, 3527, 3698, 3871, 4050, 4231, 4418, 4607, 4802

Offset: 0

Author

Bruno Berselli, Sep 21 2011 - based on remarks and sequences by Omar E. Pol

Keywords

• nonn
• easy

Comments

Sequence found by reading the numbers in increasing order on the vertical line containing 2 of the square spiral whose vertices are the triangular numbers (A000217) - see Pol's comments in other sequences visible in this numerical spiral.

Also A077591 (without first term) and A157914 interleaved.

Links

• Bruno Berselli, Table of n, a(n) for n = 0..1000
• Bruno Berselli, Illustration of initial terms: An origin of A195605.
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Crossrefs

Cf. A000217, A077591, A157914.

Cf. A047621 (contains first differences), A016754 (contains the sum of any two consecutive terms).

Cf. A033585, A069129, A077221, A102083, A139098, A139271-A139277, A139592, A139593, A188135, A194268, A194431, A195241 [incomplete list].

Programs

Magma

[(4*n*(n+2)+(-1)^n+3)/2: n in [0..48]];

Mathematica

CoefficientList[Series[(2 + 3 x + 4 x^2 - x^3) / ((1 + x) (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 19 2013 *)
LinearRecurrence[{2,0,-2,1},{2,7,18,31},50] (* Harvey P. Dale, Jan 21 2017 *)

PARI

for(n=0, 48, print1((4*n*(n+2)+(-1)^n+3)/2", "));

Formula

G.f.: (2+3*x+4*x^2-x^3)/((1+x)*(1-x)^3).

a(n) = a(-n-2) = 2*a(n-1)-2*a(n-3)+a(n-4).

a(n) = A047524(A000982(n+1)).

Sum_{n>=0} 1/a(n) = 1/2 + Pi^2/16 - cot(Pi/(2*sqrt(2)))*Pi/(4*sqrt(2)). - Amiram Eldar, Mar 06 2023