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.

Previous Showing 11-20 of 29 results. Next

A080039 a(n) = floor((1+sqrt(2))^n).

Original entry on oeis.org

1, 2, 5, 14, 33, 82, 197, 478, 1153, 2786, 6725, 16238, 39201, 94642, 228485, 551614, 1331713, 3215042, 7761797, 18738638, 45239073, 109216786, 263672645, 636562078, 1536796801, 3710155682, 8957108165, 21624372014, 52205852193
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Jan 21 2003

Keywords

Comments

a(n) = P(n) - (1+(-1)^n)/2, where P(n) is the Pell sequence (A000129) with initial conditions 2, 2.
For n>0 a(n) is the maximum element in the continued fraction for P(n)*sqrt(2) where P=A000129 - Benoit Cloitre, Jun 19 2005

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-t^2+2t^3)/(1-2t-2t^2+2t^3+t^4), {t, 0, 30}], t]
    Floor[(1+Sqrt[2])^Range[0,40]] (* or *) LinearRecurrence[{2,2,-2,-1},{1,2,5,14},40] (* Harvey P. Dale, Aug 10 2021 *)
  • PARI
    t='t+O('t^50); Vec((1-t^2+2*t^3)/(1-2*t-2*t^2+2*t^3+t^4)) \\ G. C. Greubel, Jul 05 2017

Formula

G.f.: (1-t^2+2*t^3)/(1-2*t-2*t^2+2*t^3+t^4).
From Hieronymus Fischer, Jan 02 2009: (Start)
The fractional part of (1+sqrt(2))^n equals (1+sqrt(2))^(-n), if n odd. For even n, the fractional part of (1+sqrt(2))^n is equal to 1-(1+sqrt(2))^(-n).
fract((1+sqrt(2))^n) = (1/2)*(1+(-1)^n)-(-1)^n*(1+sqrt(2))^(-n) = (1/2)*(1+(-1)^n)-(1-sqrt(2))^n.
See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).
a(n) = (sqrt(2)+1)^n - (1/2) + (-1)^n*((sqrt(2)-1)^n - (1/2)) for n>0. (End)

A175105 Triangle T(n,k) read by rows. T(n,1)=1; T(n,k) = Sum_{i=1..k-1} ( T(n-i,k-1) + T(n-i,k) ), k>1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 4, 1, 1, 4, 10, 6, 1, 1, 5, 21, 22, 8, 1, 1, 6, 40, 64, 38, 10, 1, 1, 7, 72, 163, 140, 58, 12, 1, 1, 8, 125, 382, 442, 256, 82, 14, 1, 1, 9, 212, 846, 1259, 954, 420, 110, 16, 1, 1, 10, 354, 1800, 3334, 3166, 1794, 640, 142, 18, 1, 1, 11, 585, 3719, 8366, 9657, 6754, 3074, 924, 178, 20, 1
Offset: 1

Views

Author

Mats Granvik, Feb 10 2010

Keywords

Comments

Are there closed forms for diagonals and columns?
With the definition of the array, I note that the sequence (phi(k)) (phi(k)= g.f. of the column number k) is given by the recurrence relation: phi(k+1)=phi(k)*(1-z^k)/(1-2*z+z^(k+1)). The consequence is: the sequence number k+1 column is the convolution of the k-one and a "-acci like" sequence whose g.f. is given by (1-z^k)/(1-2*z+z^(k+1)). E.g., the 2-column is the convolution of the 1-column and the sequence 1, 2, 3, 5, ... classical Fibonacci sequence without the first 1. The 3-column is the convolution of the 2-column and 1, 2, 4, 7, 13, ... tribonacci like-sequence (exactly: A000073 without beginning 0, 0, 1). - Richard Choulet, Feb 19 2010
Relation to metallic means:
T(n,1)=1, k>1: T(n,k) = Sum_{i=1..k-1} T(n-i,k-1) + 0*Sum_{i=1..k-1} T(n-i,k)
has antidiagonal sums for which the limiting ratio tends to the golden ratio, A001622.
T(n,1)=1, k>1: T(n,k) = Sum_{i=1..k-1} T(n-i,k-1) + 1*Sum_{i=1..k-1} T(n-i,k)
has antidiagonal sums for which the limiting ratio tends to the silver ratio, A014176.
T(n,1)=1, k>1: T(n,k) = Sum_{i=1..k-1} T(n-i,k-1) + 2*Sum_{i=1..k-1} T(n-i,k)
has antidiagonal sums for which the limiting ratio tends to the bronze ratio, A098316.
A similar point can be made about variations of the Pascal triangle.

Examples

			Table begins:
  n/k| 1    2    3    4    5    6    7    8    9   10   11
  ---+-----------------------------------------------------
   1 | 1
   2 | 1    1
   3 | 1    2    1
   4 | 1    3    4    1
   5 | 1    4   10    6    1
   6 | 1    5   21   22    8    1
   7 | 1    6   40   64   38   10    1
   8 | 1    7   72  163  140   58   12    1
   9 | 1    8  125  382  442  256   82   14    1
  10 | 1    9  212  846 1259  954  420  110   16    1
  11 | 1   10  354 1800 3334 3166 1794  640  142   18    1
Example: T(8,4) = 163 because it is the sum of the numbers:
  10    6
  21   22
  40   64
For k=1, we obtain phi(k)(z)=1/(1-z) which is clear; for k=2, we obtain phi(k)(z)=1/(1-z)^2. For k=3, we obtain phi(3)(z)=(1+z)/((1-2*z+z^3)*(1-z)); this is A001891 without the beginning zero. - _Richard Choulet_, Feb 19 2010
		

Crossrefs

Cf. A172119, A051731, A001891 (column k=3), A176084 (row sums).
(1-((-1)^T(n, k)))/2 = T(n, k) mod 2 = A051731.
Cf. A179807=antidiagonal sums. A179748 has simpler recurrence.

Programs

  • Excel
    =if(column()=1;1;if(row()>=column();sum(indirect(address(row()-column()+1;column()-1;4)&":"&address(row()-column()+column()-1;column()-1;4);4))+sum(indirect(address(row()-column()+1;column();4)&":"&address(row()-column()+column()-1;column();4);4));0)) ' Mats Granvik, Mar 28 2010
  • Maple
    A175105 := proc(n,k) if k =1 then 1; elif k > n or k< 1 then 0 ; else    add(procname(n-i,k-1)+procname(n-i,k),i=1..k-1) ; end if; end proc; # R. J. Mathar, Feb 16 2011
  • Mathematica
    T[_, 1] = 1;
    T[n_, k_] /; 1, ] = 0;
    Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 19 2019 *)

Formula

The g.f of the number k column is phi(k)(z) = (1/(1-z))*Product_{i=1..k-1}(1-z^i)/(1-2*z+z^(i+1)). - Richard Choulet, Feb 19 2010

Extensions

Corrected and edited by Mats Granvik, Jul 28 2010, Dec 09 2010
Choulet formulas indices shifted (to adapt to the new column index) by R. J. Mathar, Dec 13 2010

A295330 Decimal expansion of sqrt(13)/2.

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, Nov 20 2017

Keywords

Comments

In a regular hexagon inscribed in a circle of radius R the largest distance between any vertex and a midpoint of a side, after division of R, is sqrt(13)/2. The two smaller distance ratios are sqrt(7)/2 = A242703 and 1/2.
The regular period 6 continued fraction of sqrt(13)/2 is [1; 1, 4, 14, 4, 1, 2], and the convergents are given in A295331/A295332.
Essentially the same as A223139, A209927, A098316 and A085550. - R. J. Mathar, Nov 23 2017

Examples

			1.8027756377319946465596106337352479731256482869226231063552265281135834741465...
		

Crossrefs

Programs

A244593 Decimal expansion of z_c = phi^5 (where phi is the golden ratio), a lattice statistics constant which is the exact value of the critical activity of the hard hexagon model.

Original entry on oeis.org

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

Views

Author

Jean-François Alcover, Jul 01 2014

Keywords

Comments

Essentially the same digit sequence as A239798, A019863 and A019827. - R. J. Mathar, Jul 03 2014
The minimal polynomial of this constant is x^2 - 11*x - 1. - Joerg Arndt, Jan 01 2017

Examples

			11.09016994374947424102293417182819058860154589902881431067724311352630...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 5.12.1 Phase transitions in Lattice Gas Models, p. 347.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 138-139.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 83.

Crossrefs

Programs

  • Mathematica
    RealDigits[GoldenRatio^5, 10, 103] // First
  • PARI
    (5*sqrt(5)+11)/2 \\ Charles R Greathouse IV, Aug 10 2016

Formula

Equals ((1 + sqrt(5))/2)^5 = (11 + 5*sqrt(5))/2.
Equals phi^5 = 11 + 1/phi^5 = 3 + 5*phi, an integer in the quadratic number field Q(sqrt(5)). - Wolfdieter Lang, Nov 11 2023
Equals lim_{n->infinity} S(n, 5*(-1 + 2*phi))/ S(n-1, 5*(-1 + 2*phi)), with the S-Chebyshev polynomials (see A049310). - Wolfdieter Lang, Nov 15 2023

A304249 Triangle T(n,k) = 3*T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2), with T(0,0) = 1 and T(n,k) = 0 for n < 0 or k < 0, read by rows.

Original entry on oeis.org

1, 3, 9, 1, 27, 6, 81, 27, 1, 243, 108, 9, 729, 405, 54, 1, 2187, 1458, 270, 12, 6561, 5103, 1215, 90, 1, 19683, 17496, 5103, 540, 15, 59049, 59049, 20412, 2835, 135, 1, 177147, 196830, 78732, 13608, 945, 18, 531441, 649539, 295245, 61236, 5670, 189, 1
Offset: 0

Views

Author

Zagros Lalo, May 08 2018

Keywords

Comments

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n), cf. formula.
The coefficients in the expansion of 1/(1-3x-x^2) are given by the sequence generated by the row sums.
The sequence of the row sums are the "Bronze Fibonacci numbers" A006190, and the limit of their ratio is 3.30277563773... (Bronze ratio), see A098316.

Examples

			Triangle begins:
          1;
          3;
          9,         1;
         27,         6;
         81,        27,         1;
        243,       108,         9;
        729,       405,        54,        1;
       2187,      1458,       270,       12;
       6561,      5103,      1215,       90,        1;
      19683,     17496,      5103,      540,       15;
      59049,     59049,     20412,     2835,      135,       1;
     177147,    196830,     78732,    13608,      945,      18;
     531441,    649539,    295245,    61236,     5670,     189,      1;
    1594323,   2125764,   1082565,   262440,    30618,    1512,     21;
    4782969,   6908733,   3897234,  1082565,   153090,   10206,    252,    1;
   14348907,  22320522,  13817466,  4330260,   721710,   61236,   2268,   24;
   43046721,  71744535,  48361131, 16888014,  3247695,  336798,  17010,  324,  1;
  129140163, 229582512, 167403915, 64481508, 14073345, 1732104, 112266, 3240, 27;
		

References

  • Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 86, 363.

Crossrefs

Row sums give A006190.
Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), this sequence (q=2), A317497 (q=3), A318773 (q=4).

Programs

  • Magma
    [3^(n-2*k)*Binomial(n-k,k): k in [0..Floor(n/2)], n in [0..24]]; // G. C. Greubel, May 12 2021
    
  • Mathematica
    T[0, 0] = 1; T[n_, k_]:= If[n<0 || k<0, 0, 3T[n-1, k] + T[n-2, k-1]];
    Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}]//Flatten
    With[{q=2}, Table[3^(n-q*k)*Binomial[n-(q-1)*k, k], {n,0,24}, {k,0,Floor[n/q]}] ]//Flatten (* G. C. Greubel, May 12 2021 *)
  • PARI
    T(n,k)=if( n>0 && k>0, 3*T(n-1, k) + T(n-2, k-1), !n && !k)
    tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 10 2018
    
  • Sage
    flatten([[3^(n-2*k)*binomial(n-k,k) for k in (0..n//2)] for n in (0..24)]) # G. C. Greubel, May 12 2021

Formula

T(n,k) = A013610(n-k, n-2k). - M. F. Hasler, Jun 01 2018

A223139 Decimal expansion of (sqrt(13) - 1)/2.

Original entry on oeis.org

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

Views

Author

Jaroslav Krizek, Apr 02 2013

Keywords

Comments

Apart from a(1) the same as A209927 and A085550. [Joerg Arndt, Sep 17 2013]
Decimal expansion of sqrt(3 - sqrt(3 - sqrt(3 - sqrt(3 - ... )))).
Sequence with a(1) = 2 is decimal expansion of sqrt(3 + sqrt(3 + sqrt(3 + sqrt(3 + ... )))) - A209927.
This is the positive root of x^2 + x - 3, and the negative one is -A209927. - Wolfdieter Lang, Aug 29 2022

Examples

			1.3027756377319946465...
		

Crossrefs

CF. A122553 (continued fraction).

Programs

  • Mathematica
    RealDigits[(Sqrt[13] - 1)/2, 10, 130]
  • PARI
    (sqrt(13)-1)/2 \\ Altug Alkan, Oct 02 2018

Formula

Closed form: (sqrt(13) - 1)/2 = A209927-1 = A098316-2.
sqrt(3 - sqrt(3 - sqrt(3 - sqrt(3 - ... )))) + 1 = sqrt(3 + sqrt(3 + sqrt(3 + sqrt(3 + ... )))). See A209927.
Equals 2*(cos(2*Pi/13) + cos(6*Pi/13) + cos(8*Pi/13)) (see Sury link). - Michel Marcus, Aug 21 2015

A080081 Beatty sequence for (3+sqrt(13))/2.

Original entry on oeis.org

3, 6, 9, 13, 16, 19, 23, 26, 29, 33, 36, 39, 42, 46, 49, 52, 56, 59, 62, 66, 69, 72, 75, 79, 82, 85, 89, 92, 95, 99, 102, 105, 108, 112, 115, 118, 122, 125, 128, 132, 135, 138, 142, 145, 148, 151, 155, 158, 161, 165, 168, 171, 175, 178, 181, 184, 188, 191, 194, 198, 201, 204, 208, 211, 214, 217, 221, 224, 227, 231, 234, 237, 241, 244, 247, 251, 254, 257, 260
Offset: 1

Views

Author

N. J. A. Sloane, Mar 15 2003

Keywords

Comments

a(1)=3; for n>1, a(n+1)=a(n)+4 if n is already in the sequence, a(n+1)=a(n)+3 otherwise.
a(n)-1 is the index of the n-th 1 in sequence A276397. - M. F. Hasler, Oct 07 2016

Crossrefs

Cf. A098316.
Differs from A059550 at 76th term.

Programs

A165310 a(0)=1, a(1)=3, a(n) = 7*a(n-1) - 9*a(n-2) for n > 1.

Original entry on oeis.org

1, 3, 12, 57, 291, 1524, 8049, 42627, 225948, 1197993, 6352419, 33684996, 178623201, 947197443, 5022773292, 26634636057, 141237492771, 748950724884, 3971517639249, 21060066950787, 111676809902268, 592197066758793
Offset: 0

Views

Author

Philippe Deléham, Sep 14 2009

Keywords

Comments

a(n)/a(n-1) tends to (7+sqrt(13))/2 = 5.30277563... = 2+A098316.
For n >= 2, a(n) equals 3^n times the permanent of the (2n-2) X (2n-2) matrix with 1/sqrt(3)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011

Crossrefs

Programs

  • Magma
    I:=[1,3]; [n le 2 select I[n] else 7*Self(n-1)-9*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 24 2011
  • Mathematica
    LinearRecurrence[{7,-9},{1,3},30] (* Harvey P. Dale, Sep 23 2011 *)

Formula

G.f.: (1-4x)/(1-7x+9x^2).
a(n) = Sum_{k=0..n} A165253(n,k)*3^(n-k).
a(n) = ((13-sqrt(13))*(7+sqrt(13))^n+(13+sqrt(13))*(7-sqrt(13))^n )/(26*2^n). - Klaus Brockhaus, Sep 26 2009

A261391 a(n) = n^5 + 5*n^3 + 5*n.

Original entry on oeis.org

0, 11, 82, 393, 1364, 3775, 8886, 18557, 35368, 62739, 105050, 167761, 257532, 382343, 551614, 776325, 1069136, 1444507, 1918818, 2510489, 3240100, 4130511, 5206982, 6497293, 8031864, 9843875, 11969386, 14447457, 17320268, 20633239, 24435150, 28778261, 33718432, 39315243, 45632114
Offset: 0

Views

Author

Raphael Ranna, Aug 17 2015

Keywords

Comments

Also numbers of the form (n-th metallic mean)^5 - 1/(n-th metallic mean)^5, see link to Wikipedia.

Crossrefs

Programs

  • Mathematica
    Array[#^5 + 5 #^3 + 5 # &, 34] (* Michael De Vlieger, Aug 18 2015 *)
    Table[n^5 + 5*n^3 + 5*n, {n,0, 50}] (* G. C. Greubel, Aug 21 2015 *)
    LinearRecurrence[{6,-15,20,-15,6,-1},{0,11,82,393,1364,3775},40] (* Harvey P. Dale, May 07 2018 *)
  • PARI
    concat(0, Vec(x*(11*x^4+16*x^3+66*x^2+16*x+11)/(x-1)^6 + O(x^100))) \\ Colin Barker, Aug 18 2015

Formula

a(n) = ( (n+sqrt(n^2+4))/2 )^5 - 1/( (n+sqrt(n^2+4))/2 )^5.
a(n) = -a(-n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Colin Barker, Aug 18 2015
G.f.: x*(11*x^4+16*x^3+66*x^2+16*x+11) / (x-1)^6. - Colin Barker, Aug 18 2015
E.g.f.: (x^5 + 15*x^4 + 70*x^3 + 120*x^2 + 71*x + 11)*e^x. - G. C. Greubel, Aug 21 2015

Extensions

Offset changed from 1 to 0, initial 0 added and b-file adapted from Bruno Berselli, Aug 25 2015

A275860 a(n) = floor(c*s*a(n-1)) + floor(d*r*a(n-2)), where r = (3 + sqrt(13))/2, s = r/(r-1), c = 3, d = 1, a(0) = 1, a(1) = 1.

Original entry on oeis.org

1, 1, 7, 33, 164, 813, 4039, 20063, 99665, 495099, 2459470, 12217747, 60693301, 301502133, 1497752387, 7440286381, 36960623072, 183606865105, 912091791531, 4530938620963, 22508046862781, 111811749387479, 555439900107962, 2759222392297991, 13706808258965257
Offset: 0

Views

Author

Clark Kimberling, Aug 12 2016

Keywords

Crossrefs

Programs

  • Magma
    Q:=Rationals(); R:=PowerSeriesRing(Q, 40); Coefficients(R!((1-4*x+2*x^2-2*x^3+3*x^4-3*x^5+x^6)/(1-5*x+4*x^4-x^6+x^7))); // G. C. Greubel, Feb 08 2018
  • Mathematica
    c = 3; d = 1; z = 40;
    r = (c + Sqrt[c^2 + 4 d])/2; s = r/(r - 1); a[0] = 1; a[1] = 1;
    a[n_] := a[n] = Floor[c*s*a[n - 1]] + Floor[d*r*a[n - 2]];
    t = Table[a[n], {n, 0, z}]
    CoefficientList[Series[(1-4*x+2*x^2-2*x^3+3*x^4-3*x^5+x^6)/(1-5*x+4*x^4-x^6+x^7), {x,0, 50}], x] (* G. C. Greubel, Feb 08 2018 *)
    LinearRecurrence[{5,0,0,-4,0,1,-1},{1,1,7,33,164,813,4039},30] (* Harvey P. Dale, Sep 17 2024 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-4*x+2*x^2-2*x^3+3*x^4-3*x^5+x^6)/(1-5*x +4*x^4-x^6+x^7)) \\ G. C. Greubel, Feb 08 2018
    

Formula

a(n) = floor(c*s*a(n-1)) + floor(d*r*a(n-2)), where r = (3 + sqrt(13))/2 (A098316), s = r/(r-1), c = 3, d = 1, a(0) = 1, a(1) = 1.
G.f.: (1 -4*x +2*x^2 -2*x^3 +3*x^4 -3*x^5 +x^6)/(1 -5*x +4*x^4 -x^6 +x^7).
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