A017077 a(n) = 8*n + 1.

1, 9, 17, 25, 33, 41, 49, 57, 65, 73, 81, 89, 97, 105, 113, 121, 129, 137, 145, 153, 161, 169, 177, 185, 193, 201, 209, 217, 225, 233, 241, 249, 257, 265, 273, 281, 289, 297, 305, 313, 321, 329, 337, 345, 353, 361, 369, 377, 385, 393, 401, 409, 417, 425, 433

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Cf. A007519 (primes), subsequence of A047522.

a(n-1), n >= 1, gives the first column of the triangle A238475 related to the Collatz problem. - Wolfdieter Lang, Mar 12 2014

First differences of A054552. - Wesley Ivan Hurt, Jul 08 2014

An odd number is congruent to a perfect square modulo every power of 2 iff it is in this sequence. Sketch of proof: Suppose the modulus is 2^k with k at least three and note that the only odd quadratic residue (mod 8) is 1. By application of difference of squares and the fact that gcd(x-y,x+y)=2 we can show that for odd x,y, we have x^2 and y^2 congruent mod 2^k iff x is congruent to one of y, 2^(k-1)-y, 2^(k-1)+y, 2^k-y. Now when we "lift" to (mod 2^(k+1)) we see that the degeneracy between a^2 and (2^(k-1)-a)^2 "breaks" to give a^2 and a^2-2^ka+2^(2k-2). Since a is odd, the latter is congruent to a^2+2^k (mod 2^(k+1)). Hence we can form every binary number that ends with '001' by starting modulo 8 and "lifting" while adding digits as necessary. But this sequence is exactly the set of binary numbers ending in '001', so our claim is proved. - Rafay A. Ashary, Oct 23 2016

For n > 3, also the number of (not necessarily maximal) cliques in the n-antiprism graph. - Eric W. Weisstein, Nov 29 2017

Bisection of A016813. - L. Edson Jeffery, Apr 26 2022

Examples

			Illustration of initial terms:
.                                          o       o       o
.                          o     o     o     o     o     o
.              o   o   o     o   o   o         o   o   o
.      o o o     o o o         o o o             o o o
.  o   o o o   o o o o o   o o o o o o o   o o o o o o o o o
.      o o o     o o o         o o o             o o o
.              o   o   o     o   o   o         o   o   o
.                          o     o     o     o     o     o
.                                          o       o       o
--------------------------------------------------------------
.  1       9          17              25                  33
- _Bruno Berselli_, Feb 28 2014
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Tanya Khovanova, Recursive Sequences.
• Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
• Eric Weisstein's World of Mathematics, Antiprism Graph.
• Eric Weisstein's World of Mathematics, Clique.
• Index entries for linear recurrences with constant coefficients, signature (2,-1).

Crossrefs

Cf. A002189 (subsequence), A004768, A007519, A010731 (first differences), A016813, A047522, A054552.

Column 1 of A093565. Column 5 of triangle A130154. Second leftmost column of triangle A281334.

Row 1 of the arrays A081582, A238475, A371095, and A371096.

Row 2 of A257852.

Apart from the initial term, row sums of triangle A278480.

Programs

Haskell

a017077 = (+ 1) . (* 8)
a017077_list = [1, 9 ..]  -- Reinhard Zumkeller, Dec 28 2012

Magma

I:=[1,9]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..60]]; // Vincenzo Librandi, Mar 14 2014

Magma

[8*n+1 : n in [0..50]]; // Wesley Ivan Hurt, Jul 08 2014

Maple

A017077:=n->8*n+1: seq(A017077(n), n=0..50); # Wesley Ivan Hurt, Jul 08 2014

Mathematica

Table[8 n + 1, {n, 0, 6!}] (* Vladimir Joseph Stephan Orlovsky, Mar 10 2010 *)
CoefficientList[Series[(1 + 7 x)/(1 - x)^2, {x, 0, 60}], x] (* Vincenzo Librandi, Mar 14 2014 *)
8 Range[0, 50] + 1 (* Wesley Ivan Hurt, Jul 08 2014 *)
LinearRecurrence[{2, -1}, {9, 17}, {0, 20}] (* Eric W. Weisstein, Nov 29 2017 *)

PARI

a(n)=8*n+1 \\ Charles R Greathouse IV, Jul 10 2016

Formula

G.f.: (1+7*x)/(1-x)^2.

a(n+1) = A004768(n). - R. J. Mathar, May 28 2008

a(n) = 2*a(n-1) - a(n-2). - Vincenzo Librandi, Mar 14 2014

E.g.f.: exp(x)*(1 + 8*x). - Stefano Spezia, May 13 2021

From Elmo R. Oliveira, Apr 10 2025: (Start)

a(n) = a(n-1) + 8 with a(0)=1.

a(n) = A016813(2*n). (End)