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a(n) = 8*n + 1.
65

%I #99 May 06 2024 16:53:47

%S 1,9,17,25,33,41,49,57,65,73,81,89,97,105,113,121,129,137,145,153,161,

%T 169,177,185,193,201,209,217,225,233,241,249,257,265,273,281,289,297,

%U 305,313,321,329,337,345,353,361,369,377,385,393,401,409,417,425,433

%N a(n) = 8*n + 1.

%C Cf. A007519 (primes), subsequence of A047522.

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

%C First differences of A054552. - _Wesley Ivan Hurt_, Jul 08 2014

%C 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

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

%C Bisection of A016813. - _L. Edson Jeffery_, Apr 26 2022

%H Vincenzo Librandi, <a href="/A017077/b017077.txt">Table of n, a(n) for n = 0..1000</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H Luis Manuel Rivera, <a href="http://arxiv.org/abs/1406.3081">Integer sequences and k-commuting permutations</a>, arXiv preprint arXiv:1406.3081 [math.CO], 2014.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AntiprismGraph.html">Antiprism Graph</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Clique.html">Clique</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).

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

%F a(n+1) = A004768(n). - _R. J. Mathar_, May 28 2008

%F a(n) = 2*a(n-1)-a(n-2). - _Vincenzo Librandi_, Mar 14 2014

%F E.g.f.: exp(x)*(1 + 8*x). - _Stefano Spezia_, May 13 2021

%e Illustration of initial terms:

%e . o o o

%e . o o o o o o

%e . o o o o o o o o o

%e . o o o o o o o o o o o o

%e . 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

%e . o o o o o o o o o o o o

%e . o o o o o o o o o

%e . o o o o o o

%e . o o o

%e --------------------------------------------------------------

%e . 1 9 17 25 33

%e - _Bruno Berselli_, Feb 28 2014

%p A017077:=n->8*n+1: seq(A017077(n), n=0..50); # _Wesley Ivan Hurt_, Jul 08 2014

%t Table[8 n + 1, {n, 0, 6!}] (* _Vladimir Joseph Stephan Orlovsky_, Mar 10 2010 *)

%t CoefficientList[Series[(1 + 7 x)/(1 - x)^2, {x, 0, 60}], x] (* _Vincenzo Librandi_, Mar 14 2014 *)

%t 8 Range[0, 50] + 1 (* _Wesley Ivan Hurt_, Jul 08 2014 *)

%t LinearRecurrence[{2, -1}, {9, 17}, {0, 20}] (* _Eric W. Weisstein_, Nov 29 2017 *)

%o (Haskell)

%o a017077 = (+ 1) . (* 8)

%o a017077_list = [1, 9 ..] -- _Reinhard Zumkeller_, Dec 28 2012

%o (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

%o (Magma) [8*n+1 : n in [0..50]]; // _Wesley Ivan Hurt_, Jul 08 2014

%o (PARI) a(n)=8*n+1 \\ _Charles R Greathouse IV_, Jul 10 2016

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

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

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

%Y Row 2 of A257852.

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

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_