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 1-5 of 5 results.

A322595 a(n) = (n^3 + 9*n + 14*n + 9)/3.

Original entry on oeis.org

3, 11, 21, 35, 55, 83, 121, 171, 235, 315, 413, 531, 671, 835, 1025, 1243, 1491, 1771, 2085, 2435, 2823, 3251, 3721, 4235, 4795, 5403, 6061, 6771, 7535, 8355, 9233, 10171, 11171, 12235, 13365, 14563, 15831, 17171, 18585, 20075, 21643, 23291, 25021, 26835
Offset: 0

Views

Author

Keywords

Comments

For n >= 6, a(n) is the number of evaluating points on the hypersphere in R^n in Stoyanovas's degree 7 cubature rule.

Crossrefs

First differences: A027693.

Programs

  • Magma
    [(n^3 + 9*n + 14*n + 9)/3: n in [0..45]]; // Vincenzo Librandi, Jun 05 2019
  • Mathematica
    Table[(n^3 + 9*n + 14*n + 9)/3, {n, 0, 50}]
    LinearRecurrence[{4,-6,4,-1},{3,11,21,35},50] (* Harvey P. Dale, Aug 19 2020 *)
  • Maxima
    makelist((n^3 + 9*n + 14*n + 9)/3, n, 0, 50);
    

Formula

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n >= 4.
a(n) = 2*binomial(n + 1, 3) + 6*binomial(n + 1, 2) + 2*binomial(n + 1, 1) + 1.
G.f.: (3 - x - 5*x^2 + 5*x^3)/(1 - x)^4. [Corrected by Georg Fischer, May 23 2019]
E.g.f.: (1/3)*(9 + 24*x + 12*x^2 + x^3)*exp(x).

A027724 Numbers k such that k^2+k+8 is a palindrome.

Original entry on oeis.org

0, 29, 202, 285, 2925, 2935, 20377, 29570, 297119, 2834699, 2837875, 2990390, 2997334, 287010920, 2926428849, 202542945597, 295431039629, 21495814697072, 21614586653852
Offset: 1

Views

Author

Keywords

Comments

From Matthew L. LaSelle, Feb 23 2025: (Start)
k^2+k+8 only ends in a 0, 4, or 8; thus, if it is a palindrome, it only begins and ends in a 4 or 8.
a(20) > 2.174 * 10^13. (End)

Crossrefs

Programs

  • Mathematica
    palQ[n_] := Block[{d = IntegerDigits[n]}, d == Reverse[d]]; f[n_] := n^2 + n + 8; Select[Range[0, 3*10^5], palQ@ f@ # &] (* Giovanni Resta, Aug 29 2018 *)
  • PARI
    isok(k) = my(d=digits(k^2+k+8)); d == Vecrev(d); \\ Michel Marcus, Feb 24 2025

Extensions

a(14)-a(19) from Giovanni Resta, Aug 29 2018

A027725 Palindromes of form k^2 + k + 8.

Original entry on oeis.org

8, 878, 41014, 81518, 8558558, 8617168, 415242514, 874414478, 88279997288, 8035521255308, 8053537353508, 8942435342498, 8984014104898, 82375268486257328, 8563985811185893658, 41023644811311844632014, 87279499176567199497278, 462070049490878094940070264, 467190356216898612653091764
Offset: 1

Views

Author

Keywords

Comments

Palindromes h such that 4*h - 31 is a square. - Bruno Berselli, Aug 29 2018

Crossrefs

Programs

  • Mathematica
    palQ[n_] := Block[{d = IntegerDigits[n]}, d == Reverse[d]]; f[n_] := n^2 + n + 8; Select[f@ Range[0, 3*10^5], palQ] (* Giovanni Resta, Aug 29 2018 *)

Extensions

a(14)-a(19) from Giovanni Resta, Aug 29 2018

A281699 Sierpinski stellated octahedron numbers: a(n) = 2*(-3*2^(n+1) + 2^(2n+3) + 5).

Original entry on oeis.org

14, 50, 218, 938, 3914, 16010, 64778, 260618, 1045514, 4188170, 16764938, 67084298, 268386314, 1073643530, 4294770698, 17179475978, 68718690314, 274876334090, 1099508482058, 4398040219658, 17592173461514, 70368719011850, 281474926379018, 1125899806179338, 4503599426043914, 18014398106828810
Offset: 0

Views

Author

Steven Beard, Jan 27 2017

Keywords

Comments

Stella octangula with Sierpinski recursion.

Crossrefs

Programs

  • Mathematica
    Table[8 (2^(2 n + 1) + 2) - 6 (2^(n + 1) + 1), {n, 0, 25}] (* or *)
    LinearRecurrence[{7, -14, 8}, {14, 50, 218}, 26] (* or *)
    CoefficientList[Series[2 (7 - 24 x + 32 x^2)/((1 - x) (1 - 2 x) (1 - 4 x)), {x, 0, 25}], x] (* Michael De Vlieger, Jan 28 2017 *)
  • PARI
    Vec(2*(7 - 24*x + 32*x^2) / ((1 - x)*(1 - 2*x)*(1 - 4*x)) + O(x^30)) \\ Colin Barker, Jan 28 2017
    
  • PARI
    a(n) = 16*4^n - 12*2^n + 10 \\ Charles R Greathouse IV, Jan 29 2017

Formula

a(n) = 8*(2^(2*n+1)+2) - 6*(2^(n+1)+1).
From Colin Barker, Jan 28 2017: (Start)
a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3) for n>2.
G.f.: 2*(7 - 24*x + 32*x^2) / ((1 - x)*(1 - 2*x)*(1 - 4*x)).
(End)

A214870 Natural numbers placed in table T(n,k) layer by layer. The order of placement: at the beginning filled odd places of layer clockwise, next - even places counterclockwise. T(n,k) read by antidiagonals.

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 10, 9, 8, 13, 17, 16, 6, 14, 21, 26, 25, 11, 12, 22, 31, 37, 36, 18, 15, 20, 32, 43, 50, 49, 27, 24, 23, 30, 44, 57, 65, 64, 38, 35, 19, 33, 42, 58, 73, 82, 81, 51, 48, 28, 29, 45, 56, 74, 91, 101, 100, 66, 63, 39, 34, 41, 59, 72, 92, 111
Offset: 1

Views

Author

Boris Putievskiy, Mar 11 2013

Keywords

Comments

Permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.
Layer is pair of sides of square from T(1,n) to T(n,n) and from T(n,n) to T(n,1).
Enumeration table T(n,k) layer by layer. The order of the list:
T(1,1)=1;
T(1,2), T(2,1), T(2,2);
. . .
T(1,n), T(3,n), ... T(n,3), T(n,1); T(n,2), T(n,4), ... T(4,n), T(2,n);
. . .

Examples

			The start of the sequence as table:
   1   2   5  10  17  26 ...
   3   4   9  16  25  36 ...
   7   8   6  11  18  27 ...
  13  14  12  15  24  35 ...
  21  22  20  23  19  28 ...
  31  32  30  33  29  34 ...
  ...
The start of the sequence as triangle array read by rows:
   1;
   2,  3;
   5,  4,  7;
  10,  9,  8, 13;
  17, 16,  6, 14, 21;
  26, 25, 11, 12, 22, 31;
  ...
		

Crossrefs

Programs

  • Python
    t=int((math.sqrt(8*n-7) - 1)/ 2)
    i=n-t*(t+1)/2
    j=(t*t+3*t+4)/2-n
    if i > j:
       result=i*i-i+(j%2)*(2-(j+1)/2)+((j+1)%2)*(j/2+1)
    else:
       result=j*j-2*(i%2)*j + (i%2)*((i+1)/2+1) + ((i+1)%2)*(-i/2+1)

Formula

As table
T(n,k) = k*k-2*(n mod 2)*k+(n mod 2)*((n+1)/2+1)+((n+1) mod 2)*(-n/2+1), if n<=k;
T(n,k) = n*n-n+(k mod 2)*(2-(k+1)/2)+((k+1) mod 2)*(k/2+1), if n>k.
As linear sequence
a(n) = j*j-2*(i mod 2)*j+(i mod 2)*((i+1)/2+1)+((i+1) mod 2)*(-i/2+1), if i<=j;
a(n) = i*i-i+(j mod 2)*(2-(j+1)/2)+((j+1) mod 2)*(j/2+1), if i>j; where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2).
Showing 1-5 of 5 results.