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 1-3 of 3 results.

A096430 Numerator of (9*(n^4 - 2*n^3 + 2*n^2 - n) + 2)/(2*(2*n-1)).

Original entry on oeis.org

1, 28, 38, 703, 1891, 4186, 8128, 2873, 23653, 36856, 54946, 79003, 22043, 149878, 199396, 260281, 334153, 84548, 527878, 651511, 795691, 962578, 230888, 1373653, 1622701, 1904176, 2220778, 515063, 2970703, 3409966, 3896236
Offset: 1

Views

Author

Eric W. Weisstein, Aug 09 2004

Keywords

Examples

			1, 28/3, 38, 703/7, 1891/9, 4186/11, ... = A096430/A096431.

Links

Crossrefs

Cf. A096431 (denominators), A097362.

Programs

  • Magma
    A096430:= func< n | Numerator((9*n*(n^3-2*n^2+2*n-1)+2)/(2*(2*n-1))) >;
[A096430(n): n in [1..50]]; // G. C. Greubel, Oct 14 2024
  • Maple
    A096430:=n->numer((9*(n^4 - 2*n^3 + 2*n^2 - n) + 2)/(2*(2*n-1))): seq(A096430(n), n=1..50); # Wesley Ivan Hurt, Jan 21 2017
  • Mathematica
    Table[Numerator[(9*n*(n^3-2*n^2+2*n-1)+2)/(2*(2*n-1))], {n,50}] (* G. C. Greubel, Oct 14 2024 *)
  • SageMath
    def A096430(n): return numerator((9*n*(n^3-2*n^2+2*n-1)+2)/(2*(2*n-1)))
[A096430(n) for n in range(1,51)] # G. C. Greubel, Oct 14 2024

A207974 Triangle related to A152198.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 2, 2, 1, 1, 5, 2, 4, 1, 1, 1, 6, 3, 6, 3, 2, 1, 1, 7, 3, 9, 3, 5, 1, 1, 1, 8, 4, 12, 6, 8, 4, 2, 1, 1, 9, 4, 16, 6, 14, 4, 6, 1, 1, 1, 10, 5, 20, 10, 20, 10, 10, 5, 2, 1
Offset: 0

Views

Author

Philippe Deléham, Feb 22 2012

Keywords

Comments

Row sums are A027383(n).
Diagonal sums are alternately A014739(n) and A001911(n+1).
The matrix inverse starts
1;
-1,1;
1,-2,1;
1,-1,-1,1;
-1,2,0,-2,1;
-1,1,2,-2,-1,1;
1,-2,-1,4,-1,-2,1;
1,-1,-3,3,3,-3,-1,1;
-1,2,2,-6,0,6,-2,-2,1;
-1,1,4,-4,-6,6,4,-4,-1,1;
1,-2,-3,8,2,-12,2,8,-3,-2,1;
apparently related to A158854. - R. J. Mathar, Apr 08 2013
From Gheorghe Coserea, Jun 11 2016: (Start)
T(n,k) is the number of terms of the sequence A057890 in the interval [2^n,2^(n+1)-1] having binary weight k+1.
T(n,k) = A007318(n,k) (mod 2) and the number of odd terms in row n of the triangle is 2^A000120(n).
(End)

Examples

			Triangle begins :
n\k  [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
[0]  1;
[1]  1,  1;
[2]  1,  2,  1;
[3]  1,  3,  1,  1;
[4]  1,  4,  2,  2,  1;
[5]  1,  5,  2,  4,  1,  1;
[6]  1,  6,  3,  6,  3,  2,  1;
[7]  1,  7,  3,  9,  3,  5,  1,  1;
[8]  1,  8,  4,  12, 6,  8,  4,  2,  1;
[9]  1,  9,  4,  16, 6,  14, 4,  6,  1,  1;
[10] ...

Links

Crossrefs

Cf. Columns : A000012, A000027, A004526, A002620, A008805, A006918, A058187
Cf. Diagonals : A000012, A000034, A052938, A097362
Cf. A007318, A110813, A135278, A152201
Related to thickness: A000120, A027383, A057890, A274036.

Programs

  • Maple
    A207974 := proc(n,k)
    if k = 0 then
        1;
    elif k < 0 or k > n then
        0 ;
    else
        procname(n-1,k-1)-(-1)^k*procname(n-1,k) ;
    end if;
end proc: # R. J. Mathar, Apr 08 2013
  • PARI
    seq(N) = {
  my(t = vector(N+1, n, vector(n, k, k==1 || k == n)));
  for(n = 2, N+1, for (k = 2, n-1,
      t[n][k] = t[n-1][k-1] + (-1)^(k%2)*t[n-1][k]));
  return(t);
};
concat(seq(10))  \\ Gheorghe Coserea, Jun 09 2016
  • PARI
    P(n) = ((2+x+(n%2)*x^2) * (1+x^2)^(n\2) - 2)/x;
concat(vector(11, n, Vecrev(P(n-1)))) \\ Gheorghe Coserea, Mar 14 2017

Formula

T(n,k) = T(n-1,k-1) - (-1)^k*T(n-1,k), k>0 ; T(n,0) = 1.
T(2n,2k) = T(2n+1,2k) = binomial(n,k) = A007318(n,k).
T(2n+1,2k+1) = A110813(n,k).
T(2n+2,2k+1) = 2*A135278(n,k).
T(n,2k) + T(n,2k+1) = A152201(n,k).
T(n,2k) = A152198(n,k).
T(n+1,2k+1) = A152201(n,k).
T(n,k) = T(n-2,k-2) + T(n-2,k).
T(2n,n) = A128014(n+1).
T(n,k) = card {p, 2^n <= A057890(p) <= 2^(n+1)-1 and A000120(A057890(p)) = k+1}. - Gheorghe Coserea, Jun 09 2016
P_n(x) = Sum_{k=0..n} T(n,k)*x^k = ((2+x+(n mod 2)*x^2)*(1+x^2)^(n\2) - 2)/x. - Gheorghe Coserea, Mar 14 2017

A236203 Interleave A005563(n), A028347(n).

Original entry on oeis.org

0, 0, 3, 5, 8, 12, 15, 21, 24, 32, 35, 45, 48, 60, 63, 77, 80, 96, 99, 117, 120, 140, 143, 165, 168, 192, 195, 221, 224, 252, 255, 285, 288, 320, 323, 357, 360, 396, 399, 437, 440, 480, 483, 525, 528, 572, 575, 621, 624, 672, 675, 725, 728, 780, 783, 837, 840, 896
Offset: 2

Views

Author

Paul Curtz, Jan 20 2014

Keywords

Comments

A175628 gives the numerators of interleaved Lyman and Balmer series, i.e., A005563(n)/A000290(n+1) and A061037(n+2)/A061038(n+2).
Difference table of a(n):
-1, -3, 0, 0, 3, 5, 8, 12, 15, 21, 24, ...
-2, 3, 0, 3, 2, 3, 4, 3, 6, 3, 8, ...
5, -3, 3, -1, 1, 1, -1, 3, -3, 5, -5, ...
-8, 6, -4, 2, 0, -2, 4, -6, 8, -10, 12, ...
14, -10, 6, -2, -2, 6, -10, 14, -18, 22, -26, ...
-24, 16, -8, 0, 8, -16, 24, -32, 40, -48, 56, ... .
a(n+2) gives the numerators of 0/1, 0/16, 3/4, 5/36, 8/9, 12/64, 15/16, 21/100, 24/25, 32/144, ... . The denominators are A097362(n+1)^2. (Compare A097362 to A029578.)
Note the particular distribution of a(-n). Example:
a(n-9) = 12,15, 5,8, 0,3, -3,0, -4,-1, -3,0, 0,3, 5,8, 12,15, ... .
a(2n) + a(2n+1) = a(-2n-1) + a(-2n-2) = -4,0,8,20,36,56,80,... = 4*A000096(n-1).
a(2n) + a(2n-1) = a(-2n) + a(-2n-1) = -5,-3,3,13,... = A001105(n) - A010716(n).

Links

Crossrefs

Cf. A005843, A008590, A016825, A109613, A147658.

Programs

  • GAP
    List([2..60], n-> (2*n^2 +2*n -19 -(2*n-11)*(-1)^n)/8 ); # G. C. Greubel, Dec 04 2019
  • Magma
    [(2*n^2 + 2*n - 19 - (2*n - 11)*(-1)^n)/8: n in [2..60]]; // Vincenzo Librandi, Jul 27 2014
  • Maple
    seq( (2*n^2 +2*n -19 -(2*n-11)*(-1)^n)/8, n=2..60); # G. C. Greubel, Dec 04 2019
  • Mathematica
    CoefficientList[Series[x^2(3x^2-2x-3)/((x-1)^3(x+1)^2), {x, 0, 60}], x] (* Vincenzo Librandi, Jul 27 2014 *)
LinearRecurrence[{1,2,-2,-1,1},{0,0,3,5,8},60] (* Harvey P. Dale, Aug 30 2018 *)
  • PARI
    concat([0,0], Vec(x^4*(3*x^2-2*x-3)/((x-1)^3*(x+1)^2) + O(x^60))) \\ Colin Barker, Jan 26 2014
  • Sage
    [(2*n^2 +2*n -19 -(2*n-11)*(-1)^n)/8 for n in (2..60)] # G. C. Greubel, Dec 04 2019

Formula

a(n+2) = (period 8: repeat 1, 16, 1, 1, 1, 4, 1, 1)*A175628(n+1).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(n+4) - a(n-4) = 0, 8, 8, ... = A168397.
From Colin Barker, Jan 26 2014: (Start)
a(n) = (n^2 -4)/4 for n even, a(n) = (n^2 +2*n -15)/4 for n odd.
G.f.: x^4*(3 + 2*x - 3*x^2)/ ((1-x)^3*(1+x)^2). (End)
a(n) = (2*n^2 + 2*n - 19 - (2*n - 11)*(-1)^n)/8. - Luce ETIENNE, Jul 26 2014
Sum_{n>=4} (-1)^n/a(n) = 11/48. - Amiram Eldar, Aug 21 2022

Extensions

More terms from Colin Barker, Jan 26 2014
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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