cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A182835 n+floor(sqrt(2n+3)), complement of A179207.

Original entry on oeis.org

3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 39, 40, 41, 42, 43, 44, 45, 46, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92
Offset: 1

Views

Author

Clark Kimberling, Jan 07 2011

Keywords

Crossrefs

Cf. A179207.

Programs

  • Mathematica
    a=2; b=3; Table[n+Floor[(a*n+b)^(1/2)],{n,80}]

Formula

a(n)=n+floor(sqrt(2n+3)).

A109613 Odd numbers repeated.

Original entry on oeis.org

1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, 13, 13, 15, 15, 17, 17, 19, 19, 21, 21, 23, 23, 25, 25, 27, 27, 29, 29, 31, 31, 33, 33, 35, 35, 37, 37, 39, 39, 41, 41, 43, 43, 45, 45, 47, 47, 49, 49, 51, 51, 53, 53, 55, 55, 57, 57, 59, 59, 61, 61, 63, 63, 65, 65, 67, 67, 69, 69, 71, 71, 73
Offset: 0

Views

Author

Reinhard Zumkeller, Aug 01 2005

Keywords

Comments

The number of rounds in a round-robin tournament with n competitors. - A. Timothy Royappa, Aug 13 2011
Diagonal sums of number triangle A113126. - Paul Barry, Oct 14 2005
When partitioning a convex n-gon by all the diagonals, the maximum number of sides in resulting polygons is 2*floor(n/2)+1 = a(n-1) (from Moscow Olympiad problem 1950). - Tanya Khovanova, Apr 06 2008
The inverse values of the coefficients in the series expansion of f(x) = (1/2)*(1+x)*log((1+x)/(1-x)) lead to this sequence; cf. A098557. - Johannes W. Meijer, Nov 12 2009
From Reinhard Zumkeller, Dec 05 2009: (Start)
First differences: A010673; partial sums: A000982;
A059329(n) = Sum_{k = 0..n} a(k)*a(n-k);
A167875(n) = Sum_{k = 0..n} a(k)*A005408(n-k);
A171218(n) = Sum_{k = 0..n} a(k)*A005843(n-k);
A008794(n+2) = Sum_{k = 0..n} a(k)*A059841(n-k). (End)
Dimension of the space of weight 2n+4 cusp forms for Gamma_0(5). - Michael Somos, May 29 2013
For n > 4: a(n) = A230584(n) - A230584(n-2). - Reinhard Zumkeller, Feb 10 2015
The arithmetic function v+-(n,2) as defined in A290988. - Robert Price, Aug 22 2017
For n > 0, also the chromatic number of the (n+1)-triangular (Johnson) graph. - Eric W. Weisstein, Nov 17 2017
a(n-1), for n >= 1, is also the upper bound a_{up}(b), where b = 2*n + 1, in the first (top) row of the complete coach system Sigma(b) of Hilton and Pedersen [H-P]. All odd numbers <= a_{up}(b) of the smallest positive restricted residue system of b appear once in the first rows of the c(2*n+1) = A135303(n) coaches. If b is an odd prime a_{up}(b) is the maximum. See a comment in the proof of the quasi-order theorem of H-P, on page 263 ["Furthermore, every possible a_i < b/2 ..."]. For an example see below. - Wolfdieter Lang, Feb 19 2020
Satisfies the nested recurrence a(n) = a(a(n-2)) + 2*a(n-a(n-1)) with a(0) = a(1) = 1. Cf. A004001. - Peter Bala, Aug 30 2022
The binomial transform is 1, 2, 6, 16, 40, 96, 224, 512, 1152, 2560,.. (see A057711). - R. J. Mathar, Feb 25 2023

Examples

			G.f. = 1 + x + 3*x^2 + 3*x^3 + 5*x^4 + 5*x^5 + 7*x^6 + 7*x^7 + 9*x^8 + 9*x^9 + ...
Complete coach system for (a composite) b = 2*n + 1 = 33: Sigma(33) ={[1; 5], [5, 7, 13; 2, 1, 2]} (the first two rows are here 1 and 5, 7, 13), a_{up}(33) = a(15) = 15. But 15 is not in the reduced residue system modulo 33, so the maximal (odd) a number is 13. For the prime b = 31, a_{up}(31) = a(14) = 15 appears as maximum of the first rows. - _Wolfdieter Lang_, Feb 19 2020
		

References

  • Peter Hilton and Jean Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, 2010, 3rd printing 2012, pp. (260-281).

Crossrefs

Complement of A052928 with respect to the universe A004526. - Guenther Schrack, Aug 21 2018
First differences of A000982, A061925, A074148, A105343, A116940, and A179207. - Guenther Schrack, Aug 21 2018

Programs

Formula

a(n) = 2*floor(n/2) + 1.
a(n) = A052928(n) + 1 = 2*A004526(n) + 1.
a(n) = A028242(n) + A110654(n).
a(n) = A052938(n-2) + A084964(n-2) for n > 1. - Reinhard Zumkeller, Aug 27 2005
G.f.: (1 + x + x^2 + x^3)/(1 - x^2)^2. - Paul Barry, Oct 14 2005
a(n) = 2*a(n-2) - a(n-4), a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 3. - Philippe Deléham, Nov 03 2008
a(n) = A001477(n) + A059841(n). - Philippe Deléham, Mar 31 2009
a(n) = 2*n - a(n-1), with a(0) = 1. - Vincenzo Librandi, Nov 13 2010
a(n) = R(n, -2), where R(n, x) is the n-th row polynomial of A211955. a(n) = (-1)^n + 2*Sum_{k = 1..n} (-1)^(n - k - 2)*4^(k-1)*binomial(n+k, 2*k). Cf. A084159. - Peter Bala, May 01 2012
a(n) = A182579(n+1, n). - Reinhard Zumkeller, May 06 2012
G.f.: ( 1 + x^2 ) / ( (1 + x)*(x - 1)^2 ). - R. J. Mathar, Jul 12 2016
E.g.f.: x*exp(x) + cosh(x). - Ilya Gutkovskiy, Jul 12 2016
From Guenther Schrack, Sep 10 2018: (Start)
a(-n) = -a(n-1).
a(n) = A047270(n+1) - (2*n + 2).
a(n) = A005408(A004526(n)). (End)
a(n) = A000217(n) / A004526(n+1), n > 0. - Torlach Rush, Nov 10 2023

A047838 a(n) = floor(n^2/2) - 1.

Original entry on oeis.org

1, 3, 7, 11, 17, 23, 31, 39, 49, 59, 71, 83, 97, 111, 127, 143, 161, 179, 199, 219, 241, 263, 287, 311, 337, 363, 391, 419, 449, 479, 511, 543, 577, 611, 647, 683, 721, 759, 799, 839, 881, 923, 967, 1011, 1057, 1103, 1151, 1199, 1249, 1299, 1351, 1403
Offset: 2

Views

Author

Michael Somos, May 07 1999

Keywords

Comments

Define the organization number of a permutation pi_1, pi_2, ..., pi_n to be the following. Start at 1, count the steps to reach 2, then the steps to reach 3, etc. Add them up. Then the maximal value of the organization number of any permutation of [1..n] for n = 0, 1, 2, 3, ... is given by 0, 1, 3, 7, 11, 17, 23, ... (this sequence). This was established by Graham Cormode (graham(AT)research.att.com), Aug 17 2006, see link below, answering a question raised by Tom Young (mcgreg265(AT)msn.com) and Barry Cipra, Aug 15 2006
From Dmitry Kamenetsky, Nov 29 2006: (Start)
This is the length of the longest non-self-intersecting spiral drawn on an n X n grid. E.g., for n=5 the spiral has length 17:
1 0 1 1 1
1 0 1 0 1
1 0 1 0 1
1 0 0 0 1
1 1 1 1 1 (End)
It appears that a(n+1) is the maximum number of consecutive integers (beginning with 1) that can be placed, one after another, on an n-peg Towers of Hanoi, such that the sum of any two consecutive integers on any peg is a square. See the problem: http://online-judge.uva.es/p/v102/10276.html. - Ashutosh Mehra, Dec 06 2008
a(n) = number of (w,x,y) with all terms in {0,...,n} and w = |x+y-w|. - Clark Kimberling, Jun 11 2012
The same sequence also represents the solution to the "pigeons problem": maximal value of the sum of the lengths of n-1 line segments (connected at their end-points) required to pass through n trail dots, with unit distance between adjacent points, visiting all of them without overlaping two or more segments. In this case, a(0)=0, a(1)=1, a(2)=3, and so on. - Marco Ripà, Jan 28 2014
Also the longest path length in the n X n white bishop graph. - Eric W. Weisstein, Mar 27 2018
a(n) is the number of right triangles with sides n*(h-floor(h)), floor(h) and h, where h is the hypotenuse. - Andrzej Kukla, Apr 14 2021

Examples

			x^2 + 3*x^3 + 7*x^4 + 11*x^5 + 17*x^6 + 23*x^7 + 31*x^8 + 39*x^9 + 49*x^10 + ...
		

Crossrefs

Complement of A047839. First difference is A052928.
Partial sums: A213759(n-1) for n > 1. - Guenther Schrack, May 12 2018

Programs

  • Magma
    [Floor(n^2/2)-1 : n in [2..100]]; // Wesley Ivan Hurt, Aug 06 2015
  • Maple
    seq(floor((n^2+4*n+2)/2), n=0..20) # Gary Detlefs, Feb 10 2010
  • Mathematica
    Table[Floor[n^2/2] - 1, {n, 2, 60}] (* Robert G. Wilson v, Aug 31 2006 *)
    LinearRecurrence[{2, 0, -2, 1}, {1, 3, 7, 11}, 60] (* Harvey P. Dale, Jan 16 2015 *)
    Floor[Range[2, 20]^2/2] - 1 (* Eric W. Weisstein, Mar 27 2018 *)
    Table[((-1)^n + 2 n^2 - 5)/4, {n, 2, 20}] (* Eric W. Weisstein, Mar 27 2018 *)
    CoefficientList[Series[(-1 - x - x^2 + x^3)/((-1 + x)^3 (1 + x)), {x, 0, 20}], x] (* Eric W. Weisstein, Mar 27 2018 *)
  • PARI
    a(n) = n^2\2 - 1
    

Formula

a(2)=1; for n > 2, a(n) = a(n-1) + n - 1 + (n-1 mod 2). - Benoit Cloitre, Jan 12 2003
a(n) = T(n-1) + floor(n/2) - 1 = T(n) - floor((n+3)/2), where T(n) is the n-th triangular number (A000217). - Robert G. Wilson v, Aug 31 2006
Equals (n-1)-th row sums of triangles A134151 and A135152. Also, = binomial transform of [1, 2, 2, -2, 4, -8, 16, -32, ...]. - Gary W. Adamson, Nov 21 2007
G.f.: x^2*(1+x+x^2-x^3)/((1-x)^3*(1+x)). - R. J. Mathar, Sep 09 2008
a(n) = floor((n^2 + 4*n + 2)/2). - Gary Detlefs, Feb 10 2010
a(n) = abs(A188653(n)). - Reinhard Zumkeller, Apr 13 2011
a(n) = (2*n^2 + (-1)^n - 5)/4. - Bruno Berselli, Sep 14 2011
a(n) = a(-n) = A007590(n) - 1.
a(n) = A080827(n) - 2. - Kevin Ryde, Aug 24 2013
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4), n > 4. - Wesley Ivan Hurt, Aug 06 2015
a(n) = A000217(n-1) + A004526(n-2), for n > 1. - J. Stauduhar, Oct 20 2017
From Guenther Schrack, May 12 2018: (Start)
Set a(0) = a(1) = -1, a(n) = a(n-2) + 2*n - 2 for n > 1.
a(n) = A000982(n-1) + n - 2 for n > 1.
a(n) = 2*A033683(n) - 3 for n > 1.
a(n) = A061925(n-1) + n - 3 for n > 1.
a(n) = A074148(n) - n - 1 for n > 1.
a(n) = A105343(n-1) + n - 4 for n > 1.
a(n) = A116940(n-1) - n for n > 1.
a(n) = A179207(n) - n + 1 for n > 1.
a(n) = A183575(n-2) + 1 for n > 2.
a(n) = A265284(n-1) - 2*n + 1 for n > 1.
a(n) = 2*A290743(n) - 5 for n > 1. (End)
E.g.f.: 1 + x + ((x^2 + x - 2)*cosh(x) + (x^2 + x - 3)*sinh(x))/2. - Stefano Spezia, May 06 2021
Sum_{n>=2} 1/a(n) = 3/2 + tan(sqrt(3)*Pi/2)*Pi/(2*sqrt(3)) - cot(Pi/sqrt(2))*Pi/(2*sqrt(2)). - Amiram Eldar, Sep 15 2022

Extensions

Edited by Charles R Greathouse IV, Apr 23 2010

A112970 A generalized Stern sequence.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 2, 2, 1, 5, 3, 3, 2, 5, 2, 3, 1, 6, 4, 3, 2, 6, 2, 3, 1, 7, 5, 4, 3, 8, 3, 5, 2, 8, 5, 4, 2, 8, 3, 3, 1, 9, 6, 5, 4, 9, 3, 6, 2, 9, 6, 4, 2, 9, 3, 3, 1, 10, 7, 6, 5, 11, 4, 8, 3, 12, 8, 6, 3, 13, 5, 5, 2, 13, 8, 7, 5, 12, 4, 7, 2, 12, 8, 5, 3, 11, 3, 4, 1, 12, 9, 7, 6
Offset: 0

Views

Author

Paul Barry, Oct 07 2005

Keywords

Comments

Conjectures: a(2^n)=a(2^(n+1)+1)=A033638(n); a(2^n-1)=a(3*2^n-1)=1.
The Gi1 and Gi2 triangle sums, see A180662 for their definitions, of Sierpinski's triangle A047999 equal this sequence. The Gi1 and Gi2 sums can also be interpreted as (i + 4*j = n) and (4*i + j = n) sums, see the Northshield reference. Some A112970(2^n-p) sequences, 0<=p<=32, lead to known sequences, see the crossrefs. - Johannes W. Meijer, Jun 05 2011

Crossrefs

Cf. A120562 (Northshield).
Cf. A033638 (p=0), A000012 (p=1), A004526 (p=2, p=3, p=5, p=9, p=17), A002620 (p=4, p=7, p=13, p=25), A000027 (p=6, p=11, p=21), A004116 (p=8, p=15, p=29), A035106 (p=10, p=19), A024206 (p=14, p=27), A007494 (p=18), A014616 (p=22), A179207 (p=26). - Johannes W. Meijer, Jun 05 2011

Programs

  • Maple
    A112970:=proc(n) option remember; if n <0 then A112970(n):=0 fi: if (n=0 or n=1) then 1 elif n mod 2 = 0 then A112970(n/2) + A112970((n/2)-2) else A112970((n-1)/2); fi; end: seq(A112970(n),n=0..99); # Johannes W. Meijer, Jun 05 2011
  • Mathematica
    a[n_] := a[n] = Which[n<0, 0, n==0 || n==1, 1, Mod[n, 2]==0, a[n/2] + a[n/2-2], True, a[(n-1)/2]];
    Table[a[n], {n, 0, 99}] (* Jean-François Alcover, Aug 02 2022 *)

Formula

a(n) = Sum_{k=0..n} mod(sum{j=0..n, (-1)^(n-k)*C(j, n-j)*C(k, j-k)}, 2).
From Johannes W. Meijer, Jun 05 2011: (Start)
a(2*n+1) = a(n) and a(2*n) = a(n) + a(n-2) with a(0) = 1, a(1) = 1 and a(n)=0 for n<=-1.
G.f.: Product_{n>=0} (1 + x^(2^n) + x^(4*2^n)). (End)
G.f. A(x) satisfies: A(x) = (1 + x + x^4) * A(x^2). - Ilya Gutkovskiy, Jul 09 2019
Showing 1-4 of 4 results.