A118286 Numbers n such that n == 0 (mod 4) or n == 2 (mod 12).

2, 4, 8, 12, 14, 16, 20, 24, 26, 28, 32, 36, 38, 40, 44, 48, 50, 52, 56, 60, 62, 64, 68, 72, 74, 76, 80, 84, 86, 88, 92, 96, 98, 100, 104, 108, 110, 112, 116, 120, 122, 124, 128, 132, 134, 136, 140, 144, 146, 148, 152, 156, 158, 160, 164, 168, 170, 172, 176, 180, 182

Offset: 1

Author

T. D. Noe, Apr 23 2006

Keywords

• nonn
• easy

Comments

Except for n=2, conjectured n such that A118278(n)=-1 and A118282(n)=-1; n such that there is an infinite set of numbers not representable as the sum of three n-gonal numbers or three generalized n-gonal numbers.

The difference between two consecutive terms follow the 4-period: (2, 4, 4, 2). - Bernard Schott, Feb 25 2019

Links

• Guenther Schrack, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Colin Barker)
• Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Crossrefs

Cf. A047237, A118278, A118282.

Programs

GAP

Filtered([1..190],n->n mod 4=0 or n mod 12=2); # Muniru A Asiru, Feb 22 2019

Magma

[Round((3*n-1) + (Sqrt(-1))^n*(1+(-1)^n)/2): n in [1..70]]; // G. C. Greubel, Feb 21 2019

Maple

select(n->modp(n,4)=0 or modp(n,12)=2,[$1..190]); # Muniru A Asiru, Feb 22 2019

Mathematica

Union[4*Range[50], 2+12*Range[16]]

PARI

a(n) = (-2+(-I)^n+I^n+6*n)/2 \\ Colin Barker, Oct 19 2015

PARI

Vec(2*x*(1+2*x^2)/((1+x^2)*(1-x)^2) + O(x^70)) \\ Colin Barker, Oct 19 2015

PARI

for(n=1, 1e3, if(n%4 == 0 || n%12 == 2, print1(n", "))) \\ Altug Alkan, Oct 19 2015

Sage

[(3*n-1) + I^n*(1+(-1)^n)/2 for n in (1..70)] # G. C. Greubel, Feb 21 2019

Formula

G.f.: 2*x*(1+2*x^2) / ( (1+x^2)*(1-x)^2 ). - R. J. Mathar, Dec 02 2011

a(n) = 2*A047237(n+1) = 3*n - 1 + cos(n*Pi/2). - R. J. Mathar, Dec 02 2011

a(n) = (-2 + (-i)^n + i^n + 6*n)/2, where i = sqrt(-1). - Colin Barker, Oct 19 2015

a(n) = (6*n - 2 + (1 + (-1)^n)*(-1)^(n*(n-1)/2))/2. - Guenther Schrack, Feb 21 2019

E.g.f.: cos(x) + (3*x-1)*exp(x). - G. C. Greubel, Feb 21 2019