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 1-9 of 9 results.

A082109 Third row of number array A082105.

Original entry on oeis.org

1, 13, 33, 61, 97, 141, 193, 253, 321, 397, 481, 573, 673, 781, 897, 1021, 1153, 1293, 1441, 1597, 1761, 1933, 2113, 2301, 2497, 2701, 2913, 3133, 3361, 3597, 3841, 4093, 4353, 4621, 4897, 5181, 5473, 5773, 6081, 6397, 6721, 7053, 7393, 7741, 8097, 8461
Offset: 0

Views

Author

Paul Barry, Apr 03 2003

Keywords

Crossrefs

Column 2 of array A188646.

Programs

Formula

a(n) = 4*n^2 + 8*n + 1.
a(n) = a(n-1) + 8*n + 4, with a(0)=1. - Vincenzo Librandi, Aug 08 2010
G.f.: (1 + 10*x - 3*x^2)/(1-x)^3. - Bruno Berselli, Apr 18 2011
E.g.f.: (1 + 12*x + 4*x^2)*exp(x). - G. C. Greubel, Dec 22 2022
From Amiram Eldar, Jan 18 2023: (Start)
Sum_{n>=0} 1/a(n) = 1/6 - cot(sqrt(3)*Pi/2)*sqrt(3)*Pi/12.
Sum_{n>=0} (-1)^n/a(n) = cosec(sqrt(3)*Pi/2)*sqrt(3)*Pi/12 - 1/6. (End)

A082046 Square array, A(n, k) = (k*n)^2 + 3*k*n + 1, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 11, 11, 1, 1, 19, 29, 19, 1, 1, 29, 55, 55, 29, 1, 1, 41, 89, 109, 89, 41, 1, 1, 55, 131, 181, 181, 131, 55, 1, 1, 71, 181, 271, 305, 271, 181, 71, 1, 1, 89, 239, 379, 461, 461, 379, 239, 89, 1, 1, 109, 305, 505, 649, 701, 649, 505, 305, 109, 1
Offset: 0

Views

Author

Paul Barry, Apr 03 2003

Keywords

Examples

			Array, A(n, k), begins as:
  1,  1,   1,   1,   1,    1,    1,    1, ... A000012;
  1,  5,  11,  19,  29,   41,   55,   71, ... A028387;
  1, 11,  29,  55,  89,  131,  181,  239, ... A082108;
  1, 19,  55, 109, 181,  271,  379,  505, ... A069131;
  1, 29,  89, 181, 305,  461,  649,  869, ... ;
  1, 41, 131, 271, 461,  701,  991, 1331, ... ;
  1, 55, 181, 379, 649,  991, 1405, 1891, ... ;
  1, 71, 239, 505, 869, 1331, 1891, 2549, ... ;
Antidiagonals, T(n, k), begin as:
  1;
  1,  1;
  1,  5,   1;
  1, 11,  11,   1;
  1, 19,  29,  19,   1;
  1, 29,  55,  55,  29,   1;
  1, 41,  89, 109,  89,  41,   1;
  1, 55, 131, 181, 181, 131,  55,  1;
  1, 71, 181, 271, 305, 271, 181, 71,  1;
		

Crossrefs

Programs

  • Magma
    [(k*(n-k))^2 + 3*(k*(n-k)) + 1: k in [0..n], n in [0..13]]; // G. C. Greubel, Dec 22 2022
    
  • Mathematica
    T[n_, k_]:= (k*(n-k))^2 + 3*(k*(n-k)) + 1;
    Table[T[n,k], {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 22 2022 *)
  • SageMath
    def A082046(n,k): return (k*(n-k))^2 + 3*(k*(n-k)) + 1
    flatten([[A082046(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Dec 22 2022

Formula

A(n, k) = (k*n)^2 + 3*k*n + 1 (square array).
A(k, n) = A(n, k).
A(n, n) = T(2*n, n) = A057721(n).
A(n, n+1) = A072025(n).
T(n, k) = (k*(n-k))^2 + 3*k*(n-k) + 1 (antidiagonals).
Sum_{k=0..n} T(n, k) = A082047(n) (antidiagonal sums).
From G. C. Greubel, Dec 22 2022: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = (1/2)*(1 + (-1)^n)*(1 - 2*n).
T(2*n+1, n-1) = T(2*n-1, n-1) = A072025(n-1). (End)

A350470 Array read by ascending antidiagonals. T(n, k) = J(k, n) where J are the Jacobsthal polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 5, 1, 1, 1, 7, 9, 11, 1, 1, 1, 9, 13, 29, 21, 1, 1, 1, 11, 17, 55, 65, 43, 1, 1, 1, 13, 21, 89, 133, 181, 85, 1, 1, 1, 15, 25, 131, 225, 463, 441, 171, 1, 1, 1, 17, 29, 181, 341, 937, 1261, 1165, 341, 1
Offset: 0

Views

Author

Peter Luschny, Mar 19 2022

Keywords

Examples

			Array starts:
n\k 0, 1,  2,  3,   4,    5,    6,     7,      8,      9, ...
---------------------------------------------------------------------
[0] 1, 1,  1,  1,   1,    1,    1,     1,      1,      1, ... A000012
[1] 1, 1,  3,  5,  11,   21,   43,    85,    171,    341, ... A001045
[2] 1, 1,  5,  9,  29,   65,  181,   441,   1165,   2929, ... A006131
[3] 1, 1,  7, 13,  55,  133,  463,  1261,   4039,  11605, ... A015441
[4] 1, 1,  9, 17,  89,  225,  937,  2737,  10233,  32129, ... A015443
[5] 1, 1, 11, 21, 131,  341, 1651,  5061,  21571,  72181, ... A015446
[6] 1, 1, 13, 25, 181,  481, 2653,  8425,  40261, 141361, ... A053404
[7] 1, 1, 15, 29, 239,  645, 3991, 13021,  68895, 251189, ... A350468
[8] 1, 1, 17, 33, 305,  833, 5713, 19041, 110449, 415105, ... A168579
[9] 1, 1, 19, 37, 379, 1045, 7867, 26677, 168283, 648469, ... A350469
      A005408 | A082108 |
           A016813   A014641
		

Crossrefs

Cf. A350467 (main diagonal), A352361 (Fibonacci polynomials), A352362 (Lucas polynomials).

Programs

  • Maple
    J := (n, x) -> add(2^k*binomial(n - k, k)*x^k, k = 0..n):
    seq(seq(J(k, n-k), k = 0..n), n = 0..10);
  • Mathematica
    T[n_, k_] := Hypergeometric2F1[(1 - k)/2, -k/2, -k, -8 n];
    Table[T[n, k], {n, 0, 9}, {k, 0, 9}] // TableForm
    (* or *)
    T[n_, k_] := With[{s = Sqrt[8*n+1]}, ((1+s)^(k+1) - (1-s)^(k+1)) / (2^(k+1)*s)];
    Table[Simplify[T[n, k]], {n, 0, 9}, {k, 0, 9}] // TableForm
  • PARI
    T(n, k) = ([1, 2; k, 0]^n)[1, 1] ;
    export(T)
    for(k = 0, 9, print(parvector(10, n, T(n - 1, k))))

Formula

T(n, k) = Sum_{j=0..k} binomial(k - j, j)*(2*n)^j.
T(n, k) = ((1+s)^(k+1) - (1-s)^(k+1)) / (2^(k+1)*s) where s = sqrt(8*n + 1).
T(n, k) = [x^k] (1 / (1 - x - 2*n*x^2)).
T(n, k) = hypergeom([1/2 - k/2, -k/2], [-k], -8*n).

A206531 a(n) = (2*(n+1)*(2*n+1)-1)*a(n-1) + 2*n*(2*n-1)*a(n-2), a(0)=0, a(1)=2.

Original entry on oeis.org

0, 2, 58, 3250, 292498, 38609738, 7026972314, 1686473355362, 516060846740770, 196103121761492602, 90599642253809582122, 50011002524102889331346, 32507151640666878065374898, 24575406640344159817423422890
Offset: 0

Views

Author

Seiichi Kirikami, Feb 11 2012

Keywords

Comments

The numerators of the fractions limiting to the value of A206533.

References

  • E. W. Cheney, Introduction to Approximation Theory, McGraw-Hill, Inc., 1966.

Crossrefs

Programs

  • Magma
    [n le 2 select 2*(n-1) else (2*n*(2*n-1)-1)*Self(n-1) + 2*(n-1)*(2*n-3)*Self(n-2): n in [1..31]]; // G. C. Greubel, Dec 21 2022
    
  • Mathematica
    RecurrenceTable[{a[n]==(2(n+1)(2n+1)-1)a[n-1]+2n(2n-1)a[n-2],a[0]==0,a[1]==2},a,{n,15}]
  • SageMath
    @CachedFunction
    def a(n): # a = A206531
        if (n<2): return 2*n
        else: return (2*(n+1)*(2*n+1)-1)*a(n-1) + 2*n*(2*n-1)*a(n-2)
    [a(n) for n in range(31)] # G. C. Greubel, Dec 21 2022

Formula

a(n) = A082108(n)*a(n-1) + A002939(n)*a(n-2), a(0) = 0, a(1) = 2.

A206532 a(n) = (2(n+1)(2n+1)-1) * a(n-1) + 2n(2n-1) * a(n-2), a(0) = 1, a(1) = 11.

Original entry on oeis.org

1, 11, 331, 18535, 1668151, 220195931, 40075659443, 9618158266319, 2943156429493615, 1118399443207573699, 516700542761899048939, 285218699604568275014327, 185392154742969378759312551, 140156468985684850342040288555
Offset: 0

Views

Author

Seiichi Kirikami, Feb 11 2012

Keywords

Comments

The denominators of the fractions limiting to the value of A206533.

References

  • E. W. Cheney, Introduction to Approximation Theory, McGraw-Hill, Inc.,1966.

Crossrefs

Programs

  • Maple
    f:= gfun:-rectoproc({a(n) = (2*(n+1)*(2*n+1)-1) * a(n-1) + 2*n*(2*n-1) * a(n-2), a(0) = 1, a(1) = 11},a(n),remember):
    map(f, [$0..40]); # Robert Israel, Sep 16 2018
  • Mathematica
    RecurrenceTable[{a[n]==(2(n+1)(2n+1)-1)a[n-1]+2n(2n-1)a[n-2],a[0]==1,a[1]==11},a,{n,15}]

Formula

a(n) = A082108*a(n-1) + A002939*a(n-2), a(0) = 1, a(1) = 11.
a(n) = -4*n*(-1)^n*(n+1)*LommelS1(2*n+1/2, 3/2, 1)-2*(-1)^n*(n+1)*LommelS1(2*n+3/2, 1/2, 1)+(1-cos(1))*(2*n+2)!+(-1)^n. - Robert Israel, Sep 16 2018

A206533 Decimal expansion of 1/(1-cos(1)).

Original entry on oeis.org

2, 1, 7, 5, 3, 4, 2, 6, 4, 9, 6, 7, 0, 0, 2, 1, 4, 1, 0, 7, 7, 6, 7, 8, 6, 7, 5, 9, 6, 5, 6, 0, 6, 9, 9, 7, 5, 8, 4, 8, 4, 4, 7, 4, 6, 7, 6, 2, 4, 1, 8, 4, 2, 1, 3, 7, 5, 0, 5, 4, 0, 0, 5, 5, 1, 4, 7, 0, 3, 0, 7, 1, 0, 2, 8, 9, 3, 5, 0, 6, 1, 8, 1, 9, 9, 0, 8, 7, 8, 4, 0, 4, 8, 3, 5, 5, 8, 2, 9, 1, 0, 8
Offset: 1

Views

Author

Seiichi Kirikami, Feb 11 2012

Keywords

Comments

The value of the fractional limit of the numerators(A206531+2*A206532) and the denominators(A206532).
Abs(A206532/(1-cos(1)) - (A206531+2*A206532)) -> 0.

Examples

			2.17534264967002141077678675965606997584844...
		

References

  • E. W. Cheney, Introduction to Approximation Theory, McGraw-Hill, Inc., 1966.

Crossrefs

Programs

  • Mathematica
    RealDigits[N[1/(1-Cos[1]), 150]][[1]]
  • PARI
    1/(1 - cos(1)) \\ Stefano Spezia, Apr 21 2025

Formula

Equals 1/(1-A049470).
A206531/A206532+2 -> 1/(1-A049470).
Equals 1/A371936. - Hugo Pfoertner, Apr 21 2025

Extensions

Incorrect a(86)=9 removed by Georg Fischer, Apr 04 2020

A082112 a(n) = 4*n^2 + 10*n + 1.

Original entry on oeis.org

1, 15, 37, 67, 105, 151, 205, 267, 337, 415, 501, 595, 697, 807, 925, 1051, 1185, 1327, 1477, 1635, 1801, 1975, 2157, 2347, 2545, 2751, 2965, 3187, 3417, 3655, 3901, 4155, 4417, 4687, 4965, 5251, 5545, 5847, 6157, 6475, 6801, 7135, 7477, 7827, 8185, 8551
Offset: 0

Views

Author

Paul Barry, Apr 04 2003

Keywords

Comments

A row of number array A082110.

Crossrefs

Programs

Formula

a(n) = a(n-1) + 8*n + 6 (with a(0)=1). - Vincenzo Librandi, Aug 08 2010
G.f.: (1+12*x-5*x^2) / (1-x)^3. - R. J. Mathar, Dec 03 2014
a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - Wesley Ivan Hurt, Dec 22 2021
E.g.f.: (1 + 14*x + 4*x^2)*exp(x). - G. C. Greubel, Dec 22 2022

A102094 a(n) = (2*n-1)*(2*n+1)^2.

Original entry on oeis.org

9, 75, 245, 567, 1089, 1859, 2925, 4335, 6137, 8379, 11109, 14375, 18225, 22707, 27869, 33759, 40425, 47915, 56277, 65559, 75809, 87075, 99405, 112847, 127449, 143259, 160325, 178695, 198417, 219539, 242109, 266175, 291785, 318987, 347829, 378359, 410625
Offset: 1

Views

Author

Gerald McGarvey, Feb 13 2005

Keywords

Comments

Numbers which are both the sum of 2n+1 consecutive odd integers and, after skipping one odd integer, the sum of the 2n-1 immediately higher consecutive odd integers. See A082108(n-1) for the smallest of the 2n+1 odd integers, and A054569(n+1) for the skipped number. Odd integer counterpart to A059270. - Charlie Marion, Apr 30 2020

References

  • G. Boros and V. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, Cambridge, 2004, p. 123.
  • J. Ewell, An Eulerian Method for Representing Pi^2 by Series, The Rocky Mountain Journal of Mathematics 1992 v.22, pp. 165-168.

Crossrefs

Cf. A002388.

Programs

  • GAP
    List([1..40], n-> (2*n-1)*(2*n+1)^2); # G. C. Greubel, Oct 27 2019
  • Magma
    [(2*n-1)*(2*n+1)^2: n in [1..40]]; // G. C. Greubel, Oct 27 2019
    
  • Maple
    seq((2*n-1)*(2*n+1)^2, n=1..40); # G. C. Greubel, Oct 27 2019
  • Mathematica
    Table[(2n-1)(2n+1)^2,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{9,75,245,567},40] (* Harvey P. Dale, Jul 24 2012 *)
  • PARI
    vector(40, n, (2*n-1)*(2*n+1)^2) \\ G. C. Greubel, Oct 27 2019
    
  • Sage
    [(2*n-1)*(2*n+1)^2 for n in (1..40)] # G. C. Greubel, Oct 27 2019
    

Formula

Sum_{n>=1} 1/a(n) = (12 - Pi^2)/16.
Sum_{n>=1} n/a(n) = (Pi^2 - 4)/32. - Sign flipped by Bernard Schott, May 06 2020
From Harvey P. Dale, Jul 24 2012: (Start)
a(1)=9, a(2)=75, a(3)=245, a(4)=567, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: (9 + 39*x - x^2 + x^3)/(1-x)^4. (End)
E.g.f.: 1 + (-1 + 10*x + 28*x^2 + 8*x^3)*exp(x). - G. C. Greubel, Oct 27 2019

Extensions

More terms from Harvey P. Dale, Jul 24 2012

A324017 Square array A(m,n) (m>=1, n>=1) read by antidiagonals: A(m,n) = (2*n - 1)^^m mod (2*n)^m (see Comments for definition of ^^).

Original entry on oeis.org

1, 3, 1, 5, 11, 1, 7, 29, 59, 1, 9, 55, 29, 59, 1, 11, 89, 119, 1109, 827, 1, 13, 131, 289, 3703, 3701, 2875, 1, 15, 181, 563, 5289, 7799, 34805, 15163, 1, 17, 239, 965, 16115, 45289, 138871, 128117, 31547, 1, 19, 305, 1519, 25661, 57587, 745289, 1711735, 687989, 97083, 1
Offset: 1

Views

Author

Davis Smith, Mar 28 2019

Keywords

Comments

Tetration (x^^n) is defined as x^^0 = 1 and x^^n = x^(x^^(n - 1)). Another way to put this is that x^^n = x^x^x^...x with n x's.
Conjecture: For any three integers (greater than 1), m, n, and k, such that (2*n - 1)^^m == k (mod (2*n)^m), (2*n - 1)^k == k (mod (2*n)^m). For example, 5^^2 == 29 (mod 6^2) and 5^29 == 29 (mod 6^2).
Conjecture: For n > 1 and m >= 2, floor(((2*n - 1)^^m)/(2*n)) == 2*(n - 1) (mod 2*n). For example, floor((13^^3)/14) == 12 (mod 14) and floor((15^^4)/16) == 14 (mod 16).
Conjecture: For m > 1, where (2*n - 1)^^m == j (mod (2*n)^(m + 1)), A(m + 1,n) = j. For example, A(6,3) = 563 and A(6,4) = 16115; 11^^3 == 563 (mod 12^3) and 11^^3 == 16115 (mod 12^4).

Examples

			Square array A(m,n) begins:
  \n  1     2      3       4        5          6         7          8 ...
  m\
   1| 1     3      5       7        9         11        13         15 ...
   2| 1    11     29      55       89        131       181        239 ...
   3| 1    59     29     119      289        563       965       1519 ...
   4| 1    59   1109    3703     5289      16115     25661      13807 ...
   5| 1   827   3701    7799    45289      57587    332989     669167 ...
   6| 1  2875  34805  138871   745289    1799411   4635581     669167 ...
   7| 1 15163 128117 1711735  2745289   25687283  49812797   67778031 ...
   8| 1 31547 687989 8003191 92745289  419837171 155226301 3557438959 ...
.
Examples of columns in this array:
A(m,1) = A000012(m - 1).
A(m,5) = A306686(m) with a note about how this sequence repeats terms rather than skipping.
Examples of rows in this array:
A(1,n) = A005408(n - 1).
A(2,n) = A082108(n - 1).
		

Crossrefs

Programs

  • PARI
    tetrmod(b,n,m)=my(t=b);i=0;while(i<=n, i++&&if(i>1, t=lift(Mod(b,m)^t), t)); t
    tetrmatrix(lim)= matrix(lim,lim,x,y,tetrmod((2*y)-1,x,(2*y)^x))
Showing 1-9 of 9 results.