A098180 -id:A098180 - cp's OEIS frontend

A042964 Numbers that are congruent to 2 or 3 mod 4.

2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, 42, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 66, 67, 70, 71, 74, 75, 78, 79, 82, 83, 86, 87, 90, 91, 94, 95, 98, 99, 102, 103, 106, 107, 110, 111, 114, 115, 118, 119, 122, 123, 126, 127

Offset: 1

N. J. A. Sloane

Also numbers m such that binomial(m+2, m) mod 2 = 0. - Hieronymus Fischer, Oct 20 2007
Also numbers m such that floor(1+(m/2)) mod 2 = 0. - Hieronymus Fischer, Oct 20 2007
Partial sums of the sequence 2, 1, 3, 1, 3, 1, 3, 1, 3, 1, ... which has period 2. - Hieronymus Fischer, Oct 20 2007
In groups of four add and divide by two the odd and even numbers. - George E. Antoniou, Dec 12 2001
From Jeremy Gardiner, Jan 22 2006: (Start)
Comments on the "mystery calculator". There are 6 cards.
Card 0: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, ... (A005408 sequence).
Card 1: 2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, ... (this sequence).
Card 2: 4, 5, 6, 7, 12, 13, 14, 15, 20, 21, 22, 23, 28, 29, 30, 31, 36, 37, 38, 39, ... (A047566).
Card 3: 8, 9, 10, 11, 12, 13, 14, 15, 24, 25, 26, 27, 28, 29, 30, 31, 40, 41, 42, ... (A115419).
Card 4: 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 48, 49, 50, ... (A115420).
Card 5: 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, ... (A115421).
The trick: You secretly select a number between 1 and 63 from one of the cards. You indicate to me the cards on which that number appears; I tell you the number you selected!
The solution: I add together the first term from each of the indicated cards. The total equals the selected number. The numbers in each sequence all have a "1" in the same position in their binary expansion. Example: You indicate cards 1, 3 and 5. Your selected number is 2 + 8 + 32 = 42.
Numbers having a 1 in position 1 of their binary expansion. One of the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. (End)
Complement of A042948. - Reinhard Zumkeller, Oct 03 2008
Also the 2nd Witt transform of A040000 [Moree]. - R. J. Mathar, Nov 08 2008
In general, sequences of numbers congruent to {a,a+i} mod k will have a closed form of (k-2*i)*(2*n-1+(-1)^n)/4+i*n+a, from offset 0. - Gary Detlefs, Oct 29 2013
Union of A004767 and A016825; Fixed points of A098180. - Wesley Ivan Hurt, Jan 14 2014, Oct 13 2015

David Lovler, Table of n, a(n) for n = 1..10000
Maths Magic, Mystery Calculator.
Pieter Moree, The formal series Witt transform, Discr. Math. no. 295 vol. 1-3 (2005) 143-160.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000040, A133620, A133621, A133622, A133630, A133635.
Cf. A133872, A133882, A133890, A133900, A133910.
Card trick: A005408, A047566, A115419, A115420, A115421.
Cf. A004767, A016825 A040000, A042948, A047406, A098180.

[2*n+((-1)^(n-1)-1)/2 : n in [1..100]]; // Wesley Ivan Hurt, Oct 13 2015

[n: n in [1..150] | n mod 4 in [2, 3]]; // Vincenzo Librandi, Oct 13 2015

A042964:=n->2*n+((-1)^(n-1)-1)/2; seq(A042964(n), n=1..100); # Wesley Ivan Hurt, Jan 07 2014

Flatten[Table[4n + {2, 3}, {n, 0, 31}]] (* Alonso del Arte, Feb 07 2013 *)
Select[Range[200],MemberQ[{2,3},Mod[#,4]]&] (* or *) LinearRecurrence[ {1,1,-1},{2,3,6},90] (* Harvey P. Dale, Nov 28 2018 *)

PARI
```
a(n)=2*n+2-n%2
```

Vec((2+x+x^2)/((1-x)*(1-x^2)) + O(x^100)) \\ Altug Alkan, Oct 13 2015

a(n) = A047406(n)/2.
From Michael Somos, Jan 12 2000: (Start)
G.f.: x*(2+x+x^2)/((1-x)*(1-x^2)).
a(n) = a(n-1) + 2 + (-1)^n. (End)
a(n) = 2n if n is odd, otherwise n = 2n - 1. - Amarnath Murthy, Oct 16 2003
a(n) = (3 + (-1)^(n-1))/2 + 2*(n-1) = 2n + 2 - (n mod 2). - Hieronymus Fischer, Oct 20 2007
A133872(a(n)) = 0. - Reinhard Zumkeller, Oct 03 2008
a(n) = 4*n - a(n-1) - 3 (with a(1) = 2). - Vincenzo Librandi, Nov 17 2010
a(n) = 2*n + ((-1)^(n-1) - 1)/2. - Gary Detlefs, Oct 29 2013
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 - log(2)/4. - Amiram Eldar, Dec 05 2021
E.g.f.: 1 + ((4*x - 1)*exp(x) - exp(-x))/2. - David Lovler, Aug 08 2022

Edited by N. J. A. Sloane, Jun 30 2008 at the suggestion of R. J. Mathar

Corrected by Jaroslav Krizek, Dec 18 2009

A098181 Two consecutive odd numbers separated by multiples of four, repeated twice, between them, written in increasing order.

Original entry on oeis.org

1, 3, 4, 4, 5, 7, 8, 8, 9, 11, 12, 12, 13, 15, 16, 16, 17, 19, 20, 20, 21, 23, 24, 24, 25, 27, 28, 28, 29, 31, 32, 32, 33, 35, 36, 36, 37, 39, 40, 40, 41, 43, 44, 44, 45, 47, 48, 48, 49, 51, 52, 52, 53, 55, 56, 56, 57, 59, 60, 60, 61, 63, 64, 64, 65, 67, 68, 68, 69, 71, 72, 72

Offset: 0

Paul Barry, Aug 30 2004

Essentially partial sums of A007877.

a(n) is the number of odd coefficients of the q-binomial coefficient [n+2 choose 2]. (Easy to prove.) - Richard Stanley, Oct 12 2016

			G.f. = 1 + 3*x + 4*x^2 + 4*x^3 + 5*x^4 + 7*x^5 + 8*x^6 + 8*x^7 + 9*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..10000
P. Barry, On a Generalization of the Narayana Triangle, J. Int. Seq. 14 (2011) # 11.4.5.
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Cf. A098180.

a:=[1,3,4,4];; for n in [5..80] do a[n]:=2*a[n-1]-2*a[n-2]+2*a[n-3] -a[n-4]; od; a; # G. C. Greubel, May 22 2019

R:=PowerSeriesRing(Integers(), 80); Coefficients(R!( (1+x)/((1-x)^2*(1+x^2)) )); // G. C. Greubel, May 22 2019

A:=seq((2*n+3 - cos(Pi*n/2) + sin(Pi*n/2))/2, n=0..50); \\ Bernard Schott, Jun 07 2019

Table[Floor[Binomial[n+3, 2]/2] -Floor[Binomial[n+1, 2]/2], {n, 0, 80}] (* or *) CoefficientList[Series[(1+x)/((1-x)^2*(1+x^2)), {x, 0, 80}], x] (* Michael De Vlieger, Oct 12 2016 *)

{a(n) = n\4*4 + [1, 3, 4, 4][n%4+1]}; /* Michael Somos, Sep 11 2014 */

((1+x)/((1-x)^2*(1+x^2))).series(x, 80).coefficients(x, sparse=False) # G. C. Greubel, May 22 2019

G.f.: (1+x)/((1-x)^2*(1+x^2)).
a(n) = ( (2*n+3) - cos(Pi*n/2) + sin(Pi*n/2) )/2.
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4).
a(n) = floor(C(n+3, 2)/2)-floor(C(n+1, 2)/2). - Paul Barry, Jan 01 2005
a(4*n) = 4*n+1, a(4*n+1) = 4*n+3, a(4*n+2) = a(4*n+3) = 4*n+4. - Philippe Deléham, Apr 06 2007
Euler transform of length 4 sequence [ 3, -2, 0, 1]. - Michael Somos, Sep 11 2014
a(-3-n) = -a(n) for all n in Z. - Michael Somos, Sep 11 2014
a(n) = log_2(|A174882(n+2)|). [Barry] - R. J. Mathar, Aug 18 2017
a(n) = (2*n+3 - (-1)^ceiling(n/2))/2. - Wesley Ivan Hurt, Sep 29 2017

Name edited by G. C. Greubel, Jun 06 2019

A104563 A floretion-generated sequence relating to centered square numbers.

Original entry on oeis.org

0, 1, 3, 5, 8, 13, 19, 25, 32, 41, 51, 61, 72, 85, 99, 113, 128, 145, 163, 181, 200, 221, 243, 265, 288, 313, 339, 365, 392, 421, 451, 481, 512, 545, 579, 613, 648, 685, 723, 761, 800, 841, 883, 925, 968, 1013, 1059, 1105, 1152, 1201, 1251

Offset: 0

Creighton Dement, Mar 15 2005

Floretion Algebra Multiplication Program, FAMP Code: a(n) = 1vesrokseq[A*B] with A = - .5'i - .5i' + .5'ii' + .5e, B = + .5'ii' - .5'jj' + .5'kk' + .5e. RokType: Y[sqa.Findk()] = Y[sqa.Findk()] + Math.signum(Y[sqa.Findk()])*p (internal program code). Note: many slight variations of the "RokType" already exist, such that it has become difficult to assign them all names.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Cf. A001844, A004525, A104564, A098180, A059100, A010000, A047599, A007877.

LinearRecurrence[{3, -4, 4, -3, 1}, {0, 1, 3, 5, 8}, 60] (* Amiram Eldar, Dec 14 2024 *)

concat(0, Vec(x*(1 + x)*(1 - x + x^2) / ((1 - x)^3*(1 + x^2)) + O(x^40))) \\ Colin Barker, Apr 29 2019

G.f.: x*(1 + x^3)/((1 + x^2)*(1 - x)^3).
FAMP result: 2*a(n) + 2*A004525(n+1) = A104564(n) + a(n+1).
Superseeker results:
a(2*n+1) = A001844(n) = 2*n*(n+1) + 1 (Centered square numbers);
a(n+1) - a(n) = A098180(n) (Odd numbers with two times the odd numbers repeated in order between them);
a(n) + a(n+2) = A059100(n+1) = A010000(n+1);
a(n+2) - a(n) = A047599(n+1) (Numbers that are congruent to {0, 3, 4, 5} mod 8);
a(n+2) - 2*a(n+1) + a(n) = A007877(n+3) (Period 4 sequence with initial period (0, 1, 2, 1));
Coefficients of g.f.*(1-x)/(1+x) = convolution of this with A280560 gives A004525;
Coefficients of g.f./(1+x) = convolution of this with A033999 gives A054925.
a(n) = (1/2)*(n^2 + 1 - cos(n*Pi/2)). - Ralf Stephan, May 20 2007
From Colin Barker, Apr 29 2019: (Start)
a(n) = (2 - (-i)^n - i^n + 2*n^2) / 4 where i=sqrt(-1).
a(n) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) for n>4. (End)
a(n) = A011848(n-1)+A011848(n+2). - R. J. Mathar, Sep 11 2019
Sum_{n>=1} 1/a(n) = Pi^2/48 + (Pi/2) * tanh(Pi/2) + (Pi/(4*sqrt(2)) * tanh(Pi/(2*sqrt(2)))). - Amiram Eldar, Dec 14 2024

Stephan's formula corrected by Bruno Berselli, Apr 29 2019

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A042964 Numbers that are congruent to 2 or 3 mod 4.

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A098181 Two consecutive odd numbers separated by multiples of four, repeated twice, between them, written in increasing order.

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A104563 A floretion-generated sequence relating to centered square numbers.

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