A151977 Numbers that are congruent to {0, 1} mod 16.

0, 1, 16, 17, 32, 33, 48, 49, 64, 65, 80, 81, 96, 97, 112, 113, 128, 129, 144, 145, 160, 161, 176, 177, 192, 193, 208, 209, 224, 225, 240, 241, 256, 257, 272, 273, 288, 289, 304, 305, 320, 321, 336, 337, 352, 353, 368, 369, 384, 385, 400, 401, 416, 417, 432, 433, 448, 449

Offset: 1

Author

N. J. A. Sloane, Aug 23 2009

Keywords

• nonn
• easy

Comments

Numbers k such that k^2 - k is divisible by 16.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Cf. A321212.

Programs

Magma

[(32*n-14*(-1)^n-46)/4: n in [1..60]]; // Vincenzo Librandi, Jul 11 2012

Mathematica

CoefficientList[Series[x*(1+15*x)/((1+x)(1-x)^2),{x,0,60}],x] (* Vincenzo Librandi, Jul 11 2012 *)
LinearRecurrence[{1,1,-1},{0,1,16},80] (* Harvey P. Dale, Jul 24 2021 *)

PARI

forstep(n=0,200,[1,15],print1(n", ")) \\ Charles R Greathouse IV, Oct 17 2011

PARI

a(n) = (16*n - 7*(-1)^n - 23)/2 \\ David Lovler, Aug 18 2022

Formula

a(n+1) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=1, b(k)=2^(k+3) for k > 0; {b(n)} = 1,16,32,64,128,256,... - Philippe Deléham, Oct 17 2011

G.f.: x^2*(1+15*x)/((1+x)*(1-x)^2). - Vincenzo Librandi, Jul 11 2012

a(n) = (32*n - 14*(-1)^n - 46)/4. - Vincenzo Librandi, Jul 11 2012

From David Lovler, Aug 18 2022: (Start)

a(n) = A321212(n) - 2.

a(n) = a(n-2) + 16.

E.g.f.: 15 + ((16*x - 23)*exp(x) - 7*exp(-x))/2. (End)