A274406 -id:A274406 - cp's OEIS frontend

A007494 Numbers that are congruent to 0 or 2 mod 3.

0, 2, 3, 5, 6, 8, 9, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24, 26, 27, 29, 30, 32, 33, 35, 36, 38, 39, 41, 42, 44, 45, 47, 48, 50, 51, 53, 54, 56, 57, 59, 60, 62, 63, 65, 66, 68, 69, 71, 72, 74, 75, 77, 78, 80, 81, 83, 84, 86, 87, 89, 90, 92, 93, 95, 96, 98, 99, 101, 102, 104, 105, 107

Offset: 0

Christopher Lam Cham Kee (Topher(AT)CyberDude.Com)

The map n -> a(n) (where a(n) = 3n/2 if n even or (3n+1)/2 if n odd) was studied by Mahler, in connection with "Z-numbers" and later by Flatto. One question was whether, iterating from an initial integer, one eventually encountered an iterate = 1 (mod 4). - Jeff Lagarias, Sep 23 2002
Partial sums of 0,2,1,2,1,2,1,2,1,... . - Paul Barry, Aug 18 2007
a(n) = numbers k such that antiharmonic mean of the first k positive integers is not integer. A169609(a(n-1)) = 3. See A146535 and A169609. Complement of A016777. - Jaroslav Krizek, May 28 2010
Range of A173732. - Reinhard Zumkeller, Apr 29 2012
Number of partitions of 6n into two odd parts. - Wesley Ivan Hurt, Nov 15 2014
Numbers m such that 3 divides A000217(m). - Bruno Berselli, Aug 04 2017
Maximal length of a snake like polyomino that fits in a 2 X n rectangle. - Alain Goupil, Feb 12 2020

L. Flatto, Z-numbers and beta-transformations, in Symbolic dynamics and its applications (New Haven, CT, 1991), 181-201, Contemp. Math., 135, Amer. Math. Soc., Providence, RI, 1992.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1002.
Attila Máder, The Use of Experimental Mathematics in the Classroom, in Interesting Mathematical Problems in Sciences and Everyday Life - 2011.
Kurt Mahler, An unsolved problem on the powers of 3/2, J. Austral. Math. Soc., Vol. 8 (1968), pp. 313-321.
P. Sabinin and M. G. Stone, Transforming n-gons by Folding the Plane, Amer. Math. Monthly, Vol. 102, No. 7 (1995), pp. 620-627.
Eric Weisstein's World of Mathematics, Folding.
Robert G. Wilson v, Notes with attachment.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000217, A001651, A032766, A035361, A063574, A132462.
Complement of A016777.
Range of A002517.
Cf. A274406. [Bruno Berselli, Jun 26 2016]

a007494 =  flip div 2 . (+ 1) . (* 3) -- Reinhard Zumkeller, Dec 12 2014

[(6*n+1)/4-(-1)^n/4: n in [0..80]]; // Vincenzo Librandi, Aug 20 2011

a[0]:=0:a[1]:=2:for n from 2 to 100 do a[n]:=a[n-2]+3 od: seq(a[n], n=0..71); # Zerinvary Lajos, Mar 16 2008
A007494:=n->floor((3*n+1)/2); seq(A007494(k), k=0..100); # Wesley Ivan Hurt, Sep 27 2013

Flatten[{#,#+2}&/@(3Range[0,40])] (* Harvey P. Dale, May 15 2011 *)
Table[2n - Floor[n/2], {n,0,100}] (* Wesley Ivan Hurt, Sep 27 2013 *)

a(n)=n+(n+1)>>1 \\ Charles R Greathouse IV, Jul 25 2011

a(n) = 3*n/2 if n even, otherwise (3*n+1)/2.
If u(1)=0, u(n) = n + floor(u(n-1)/3), then a(n-1) = u(n). - Benoit Cloitre, Nov 26 2002
G.f.: x*(x+2)/((1-x)^2*(1+x)). - Ralf Stephan, Apr 13 2002
a(n) = 3*floor(n/2) + 2*(n mod 2) = A032766(n) + A000035(n). - Reinhard Zumkeller, Apr 04 2005
a(n) = (6*n+1)/4 - (-1)^n/4; a(n) = Sum_{k=0..n-1} (1 + (-1)^(k/2)*cos(k*Pi/2)). - Paul Barry, Aug 18 2007
A145389(a(n)) <> 1. - Reinhard Zumkeller, Oct 10 2008
a(n) = A002943(n) - A173511(n). - Reinhard Zumkeller, Feb 20 2010
a(n) = 3*n - a(n-1) - 1 (with a(0)=0). - Vincenzo Librandi, Nov 18 2010
a(n) = Sum_{k>=0} A030308(n,k)*A042950(k). - Philippe Deléham, Oct 17 2011
a(n) = n + ceiling(n/2). - Arkadiusz Wesolowski, Sep 18 2012
a(n) = 2n - floor(n/2) = floor((3n+1)/2) = n + (n + (n mod 2))/2. - Wesley Ivan Hurt, Oct 19 2013
a(n) = A000217(n+1) - A099392(n+1). - Bui Quang Tuan, Mar 27 2015
a(n) = n + floor(n/2) + (n mod 2). - Bruno Berselli, Apr 04 2016
a(n) = Sum_{i=1..n} numerator(2/i). - Wesley Ivan Hurt, Feb 26 2017
a(n) = Sum_{k=0..n-1} Sum_{i=0..k} C(i,k)+(-1)^(k-i). - Wesley Ivan Hurt, Sep 20 2017
E.g.f.: (3*exp(x)*x + sinh(x))/2. - Stefano Spezia, Feb 11 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = log(3)/2 - Pi/(6*sqrt(3)). - Amiram Eldar, Dec 04 2021

A014601 Numbers congruent to 0 or 3 mod 4.

Original entry on oeis.org

0, 3, 4, 7, 8, 11, 12, 15, 16, 19, 20, 23, 24, 27, 28, 31, 32, 35, 36, 39, 40, 43, 44, 47, 48, 51, 52, 55, 56, 59, 60, 63, 64, 67, 68, 71, 72, 75, 76, 79, 80, 83, 84, 87, 88, 91, 92, 95, 96, 99, 100, 103, 104, 107, 108, 111, 112, 115, 116, 119, 120, 123, 124

Offset: 0

Eric Rains (rains(AT)caltech.edu)

Discriminants of orders in imaginary quadratic fields (negated). [Comment corrected by Christopher E. Thompson, Dec 11 2016]
Numbers such that Langford-Skolem problem has a solution - see A014552.
Complement of A042963. - Reinhard Zumkeller, Oct 04 2004
Also called skew amenable numbers; a number k is skew amenable if there exist a set {a(i)} of integers satisfying the relations k = Sum_{i=1..k} a(i) = -Product_{i=1..k} a(i). Thus we have 8 = 1 + 1 + 1 + 1 + 1 + 1 - 2 + 4 = -(1*1*1*1*1*1*(-2)*4). - Lekraj Beedassy, Jan 07 2005
Possible nonpositive discriminants of quadratic equation a*x^2 + b*x + c or discriminants of binary quadratic forms a*x^2 + b*x*y + c*y^2. - Artur Jasinski, Apr 28 2008
Also, disregarding the 0 term, positive integers m such that, equivalently,
  (i) +-1 +-2 +-... +-m is even for all choices of signs,
  (ii) +-1 +-2 +-... +-m = 0 for some choices of signs,
  (iii) for all -m <= k <= m, k = +-1 +-2 +-... +-(k-1) +-(k+1) +-(k+2) +-... +-m for at least one choice of signs. - Rick L. Shepherd, Oct 29 2008
A145768(a(n)) is even. - Reinhard Zumkeller, Jun 05 2012
Multiples of 4 interleaved with 1 less than multiples of 4. - Wesley Ivan Hurt, Nov 08 2013
((2*k+0) + (2*k+1) + ... + (2*k+m-1) + (2*k+m)) is even if and only if m = a(n) for some n where k is any nonnegative integer. - Gionata Neri, Jul 24 2015
Numbers whose binary reflected Gray code (A014550) ends with 0. - Amiram Eldar, May 17 2021

			G.f. = 3*x + 4*x^2 + 7*x^3 + 8*x^4 + 11*x^5 + 12*x^6 + 15*x^7 + 16*x^8 + ...

H. Cohen, Course in Computational Alg. No. Theory, Springer, 1993, pp. 514-5.
A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, 5. Aufl., de Gruyter, Berlin, New York, 1973, p. 108.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
S. F. Barger, Solution to problem 10454: Amenable Numbers, Amer. Math. Monthly, Vol. 105, No. 4 (April 1998), p. 368.
Steven R. Finch, Class number theory [Cached copy, with permission of the author]
Heiko Harborth, Solution of Steinhaus's problem with plus and minus signs, Journal of Combinatorial Theory, Series A, Volume 12, Issue 2 (March 1972), Pages 253-259.
Mickaël Launay, Les routes de Numland, L’énigme maths du "Monde" n°2. In French.
Michael Penn, The most beautiful arrangement of numbers, YouTube video, 2024.
Rick L. Shepherd, Binary quadratic forms and genus theory, Master of Arts Thesis, University of North Carolina at Greensboro, 2013.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A004676, A014550, A042948, A079896.

Cf. A274406. - Bruno Berselli, Jun 26 2016

a014601 n = a014601_list !! n
a014601_list = [x | x <- [0..], mod x 4 `elem` [0, 3]]
-- Reinhard Zumkeller, Jun 05 2012

[n: n in [0..200]|n mod 4 in {0,3}]; // Vincenzo Librandi, Dec 24 2010

A014601:=n->3*n-2*floor(n/2); seq(A014601(k), k=0..100); # Wesley Ivan Hurt, Nov 08 2013

aa = {}; Do[Do[Do[d = b^2 - 4 a c; If[d <= 0, AppendTo[aa, -d]], {a, 0, 50}], {b, 0, 50}], {c, 0, 50}]; Union[aa] (* Artur Jasinski, Apr 28 2008 *)
Select[Range[0, 124], Or[Mod[#, 4] == 0, Mod[#, 4] == 3] &] (* Ant King, Nov 18 2010 *)
CoefficientList[Series[2 x/(1 - x)^2 + (1/(1 - x) + 1/(1 + x)) x/2, {x, 0, 100}], x] (* Vincenzo Librandi, May 18 2014 *)
a[ n_] := 2 n + Mod[n, 2]; (* Michael Somos, Jul 24 2015 *)

{a(n) = 2*n + n%2}; /* Michael Somos, Dec 27 2010 */

a(n) = (n + 1)*2 + 1 - n mod 2. - Reinhard Zumkeller, Apr 21 2003
A014494(n) = A000217(a(n)). - Reinhard Zumkeller, Oct 04 2004
a(n) = Sum_{k=1..n} (2 - (-1)^k). - William A. Tedeschi, Mar 20 2008
A139131(a(n)) = A078636(a(n)). - Reinhard Zumkeller, Apr 10 2008
From R. J. Mathar, Sep 25 2009: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 2.
G.f.: x*(3+x)/((1+x)*(x-1)^2). (End)
a(n) = 2*n + (n mod 2). - Paolo Valzasina (p.valzasina(AT)gmail.com), Nov 24 2009
a(n) = (4*n - (-1)^n + 1)/2. - Bruno Berselli, Oct 06 2010
a(n) = 4*n - a(n-1) - 1 (with a(0) = 0). - Vincenzo Librandi, Dec 24 2010
a(n) = -A042948(-n) for all n in Z. - Michael Somos, Dec 27 2010
G.f.: 2*x / (1 - x)^2 + (1 / (1 - x) + 1 / (1 + x)) * x/2. - Michael Somos, Dec 27 2010
a(n) = Sum_{k>=0} A030308(n,k)*b(k) with b(0) = 3 and b(k) = 2^(k+1) for k > 0. - Philippe Deléham, Oct 17 2011
a(n) = ceiling((4/3)*ceiling(3*n/2)). - Clark Kimberling, Jul 04 2012
a(n) = 3n - 2*floor(n/2). - Wesley Ivan Hurt, Nov 08 2013
a(n) = A042948(n+1) - 1 for all n in Z. - Michael Somos, Jul 24 2015
a(n) + a(n+1) = A004767(n) for all n in Z. - Michael Somos, Jul 24 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2)/4 - Pi/8. - Amiram Eldar, Dec 05 2021
E.g.f.: ((4*x + 1)*exp(x) - exp(-x))/2. - David Lovler, Aug 04 2022

A047208 Numbers that are congruent to {0, 4} mod 5.

Original entry on oeis.org

0, 4, 5, 9, 10, 14, 15, 19, 20, 24, 25, 29, 30, 34, 35, 39, 40, 44, 45, 49, 50, 54, 55, 59, 60, 64, 65, 69, 70, 74, 75, 79, 80, 84, 85, 89, 90, 94, 95, 99, 100, 104, 105, 109, 110, 114, 115, 119, 120, 124, 125, 129, 130, 134, 135, 139, 140, 144, 145, 149

Offset: 1

N. J. A. Sloane

Also solutions to 3^x + 5^x == 2 (mod 11). - Cino Hilliard, May 18 2003

G. C. Greubel, Table of n, a(n) for n = 1..5000
Cino Hilliard, solutions to 3^x + 5^x == 2 mod 11/, June 2003.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001622, A010674, A010685 (first differences), A274406.

[(5*(n-1) + 3*((n-1) mod 2))/2: n in [1..100]]; // G. C. Greubel, Nov 23 2021

{#,#+4}&/@(5*Range[0,30])//Flatten (* Harvey P. Dale, Apr 05 2019 *)

forstep(n=0,200,[4,1],print1(n", ")) \\ Charles R Greathouse IV, Oct 17 2011

[(5*(n-1) +3*((n-1)%2))/2 for n in (1..100)] # G. C. Greubel, Nov 23 2021

From R. J. Mathar, Jan 24 2009: (Start)
G.f.: x^2*(4+x)/((1-x)^2*(1+x)).
a(n) = a(n-2) + 5. (End)
a(n) = 5*n - 6 - a(n-1) (with a(1)=0). - Vincenzo Librandi, Nov 18 2010
a(n+1) = Sum_{k>=0} A030308(n,k)*b(k), with b(0)=4 and b(k) = A020714(k-1) = 5*2^(k-1) for k>0. - Philippe Deléham, Oct 17 2011
a(n) = ceiling((5/3)*ceiling(3*n/2)). - Clark Kimberling, Jul 04 2012
a(n) = (5*(n-1) + 3*(n-1 mod 2))/2 = (5*(n-1) + A010674(n-1))/2. - G. C. Greubel, Nov 23 2021
Sum_{n>=2} (-1)^n/a(n) = log(5)/4 + log(phi)/(2*sqrt(5)) - sqrt(1+2/sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 07 2021
E.g.f.: 1 + ((5*x - 7/2)*exp(x) + (3/2)*exp(-x))/2. - David Lovler, Aug 23 2022

A047521 Numbers that are congruent to {0, 7} mod 8.

Original entry on oeis.org

0, 7, 8, 15, 16, 23, 24, 31, 32, 39, 40, 47, 48, 55, 56, 63, 64, 71, 72, 79, 80, 87, 88, 95, 96, 103, 104, 111, 112, 119, 120, 127, 128, 135, 136, 143, 144, 151, 152, 159, 160, 167, 168, 175, 176, 183, 184, 191, 192, 199, 200, 207, 208, 215, 216, 223, 224, 231, 232

Offset: 1

N. J. A. Sloane

Numbers such that the n-th triangular number is divisible by 4. - Charles R Greathouse IV, Apr 07 2011

Except for 0, numbers whose binary reflected Gray code (A014550) ends with 00. - Amiram Eldar, May 17 2021

David Lovler, Table of n, a(n) for n = 1..10000
Lars Pos, Met kleine stapjes grote sprongen make, Pythagoras 61-4. Solutions of returning to the origin after steps of increasing width 1,2,3,.. in the 4 directions on a square grid (in Dutch).
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Union of A008590 and A004771.

Cf. A014550, A030308, A274406.

{#,#+7}&/@(8*Range[0,30])//Flatten (* or *) LinearRecurrence[{1,1,-1},{0,7,8},60] (* Harvey P. Dale, Oct 30 2016 *)

a(n) = 4*n - 5/2 + 3*(-1)^n/2; \\ David Lovler, Jul 25 2022

kmax <- 10 # by choice
a <- c(0,7)
for(k in 3:kmax) a <- c(a, a + 2^k)
a
# Yosu Yurramendi, Jan 18 2022

a(n) = 8*n - a(n-1) - 9 (with a(1)=0). - Vincenzo Librandi, Aug 06 2010
From R. J. Mathar, Oct 08 2011: (Start)
a(n) = 3*(-1)^n/2 - 5/2 + 4*n.
G.f.: x^2*(7+x) / ( (1+x)*(x-1)^2 ). (End)
a(n+1) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=7 and b(k)=2^(k+2) for k > 0. - Philippe Deléham, Oct 17 2011
Sum_{n>=2} (-1)^n/a(n) = log(2)/2 + sqrt(2)*log(sqrt(2)+1)/8 - (sqrt(2)+1)*Pi/16. - Amiram Eldar, Dec 18 2021
E.g.f.: 1 + ((8*x -5)*exp(x) + 3*exp(-x))/2. David Lovler, Aug 22 2022

More terms from Vincenzo Librandi, Aug 06 2010

A193910 Leap centuries in the revised Julian calendar.

Original entry on oeis.org

2, 6, 11, 15, 20, 24, 29, 33, 38, 42, 47, 51, 56, 60, 65, 69, 74, 78, 83, 87, 92, 96, 101, 105, 110, 114, 119, 123, 128, 132, 137, 141, 146, 150, 155, 159, 164, 168, 173, 177, 182, 186, 191, 195, 200, 204, 209, 213, 218, 222, 227, 231, 236, 240, 245, 249

Offset: 1

Frank Ellermann, Aug 09 2011

Terms divided by 100, e.g., 29 indicates year 2900, which is a revised Julian leap year, but not a Gregorian leap year. Values below 20 are "proleptic" (only based on the formula).

			20 mod 9 is 2; 2000 was a leap year in the revised Julian calendar.
24 mod 9 is 6; 2400 and 2000 also happen to be Gregorian leap years.
28 is the first integer greater than 16 only contained in A008586.
29 is the first integer greater than 16 not contained in A008586.

M. Milankovitch, Das Ende des julianischen Kalenders und der neue Kalender der orientalischen Kirchen, Astronomische Nachrichten, volume 220 (1924), 379-384.
Claus Tøndering, Frequently Asked Questions about Calendars, Don't the Greeks do it differently?
Wikipedia, Revised Julian calendar
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

A008586 enumerates "Gregorian leap centuries" (N mod 4 = 0).
A193879 enumerates all differences from A008586.
Cf. A274406: numbers congruent to {0, 8} mod 9; A301451: numbers congruent to {1, 7} mod 9. This sequence lists the numbers congruent to {2, 6} mod 9.

Table[1/4 (18 m - (-1)^m - 11), {m, 56}] (* Farideh Firoozbakht, Oct 08 2014 *)

a(n)=(9*n-5)\2 \\ Charles R Greathouse IV, Aug 23 2011

do C = 0 to 250; J = C // 9; if J = 2 | J = 6 then say C; end C

a(n) = a(n-2) + 9. - Charles R Greathouse IV, Aug 09 2011
a(n) = 2 or 6 (mod 9).
For all positive integers n, a(n) = (1/4)*(18*n-17*(-1)^n-11), which implies a(2*n-1) = 9*n-3 and a(2*n) = 9*n-7. - Farideh Firoozbakht, Oct 08 2014
G.f.: x*(2 + 4*x + 3*x^2)/((1 + x)*(1 - x)^2). - Philippe Deléham, Nov 30 2016

A047335 Numbers that are congruent to {0, 6} mod 7.

Original entry on oeis.org

0, 6, 7, 13, 14, 20, 21, 27, 28, 34, 35, 41, 42, 48, 49, 55, 56, 62, 63, 69, 70, 76, 77, 83, 84, 90, 91, 97, 98, 104, 105, 111, 112, 118, 119, 125, 126, 132, 133, 139, 140, 146, 147, 153, 154, 160, 161, 167, 168

Offset: 1

N. J. A. Sloane

Hsien-Kuei Hwang, S Janson, TH Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016; http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf. Also Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585

Index entries for linear recurrences with constant coefficients, signature (1, 1, -1).

Cf. A274406.

Select[Range[0,200],MemberQ[{0,6},Mod[#,7]]&]  (* Harvey P. Dale, Mar 16 2011 *)

From Bruno Berselli, Oct 06 2010: (Start)
G.f.: x^2*(6+x)/((1+x)*(1-x)^2).
a(n) - a(n-1) - a(n-2) + a(n-3) = 0 (n > 3).
a(n) = (14*n + 5*(-1)^n - 9)/4.
a(n) - a(n-2) = 7 (n > 2).
a(n) - a(n-1) = A010687(k) with n > 1 and k == n-1 (mod 2). (End)
a(n+1) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=6 and b(k) = 7*2^(k-1) = A005009(k-1) for k > 0. - Philippe Deléham, Oct 18 2011

A301451 Numbers congruent to {1, 7} mod 9.

Original entry on oeis.org

1, 7, 10, 16, 19, 25, 28, 34, 37, 43, 46, 52, 55, 61, 64, 70, 73, 79, 82, 88, 91, 97, 100, 106, 109, 115, 118, 124, 127, 133, 136, 142, 145, 151, 154, 160, 163, 169, 172, 178, 181, 187, 190, 196, 199, 205, 208, 214, 217, 223, 226, 232, 235, 241, 244, 250, 253, 259, 262, 268

Offset: 1

Bruno Berselli, Mar 21 2018

First bisection of A056991, second bisection of A242660.

The squares of the terms of A174396 are the squares of this sequence.

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

Cf. A274406: numbers congruent to {0, 8} mod 9.
Cf. A193910: numbers congruent to {2, 6} mod 9.
Subsequence of A016777, A026225, A029739, A033627, A047236, A047259, A055047, A055054, A056991, A153053, A187318, A242660.

a := [1,7,10];; for n in [4..60] do a[n] := a[n-1] + a[n-2] - a[n-3]; od; a;

Magma
```
&cat [[9*n+1, 9*n+7]: n in [0..40]];
```

Table[2 (2 n - 1) + (2 n - 3 (1 - (-1)^n))/4, {n, 1, 60}]
{#+1,#+7}&/@(9*Range[0,30])//Flatten (* or *) LinearRecurrence[{1,1,-1},{1,7,10},60] (* Harvey P. Dale, Nov 08 2020 *)

Vec(x*(1 + 6*x + 2*x^2) / ((1 - x)^2*(1 + x)) + O(x^60)) \\ Colin Barker, Mar 22 2018

[2*(2*n-1)+(2*n-3*(1-(-1)**n))/4 for n in range(1,70)]

[n for n in (1..300) if n % 9 in (1,7)]

O.g.f.: x*(1 + 6*x + 2*x^2)/((1 + x)*(1 - x)^2).
E.g.f.: (3 + 8*exp(x) - 11*exp(2*x) + 18*x*exp(2*x))*exp(-x)/4.
a(n) = a(n-1) + a(n-2) - a(n-3).
a(n) = 2*(2*n - 1) + (2*n - 3*(1 - (-1)^n))/4. Therefore, for n even a(n) = (9*n - 4)/2, otherwise a(n) = (9*n - 7)/2.
a(2n+1) = A017173(n). a(2n) = A017245(n-1). - R. J. Mathar, Feb 28 2019

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A007494 Numbers that are congruent to 0 or 2 mod 3.

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A014601 Numbers congruent to 0 or 3 mod 4.

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A047208 Numbers that are congruent to {0, 4} mod 5.

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A047521 Numbers that are congruent to {0, 7} mod 8.

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A193910 Leap centuries in the revised Julian calendar.

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A047335 Numbers that are congruent to {0, 6} mod 7.

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