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 1-7 of 7 results.

A082630 Limit of the sequence obtained from S(0) = (1,1) and, for n > 0, S(n) = I(S(n-1)), where I consists of inserting, for i = 1, 2, 3..., the term a(i) + a(i+1) between any two terms for which 7*a(i+1) <= 11*a(i).

Original entry on oeis.org

1, 2, 5, 8, 19, 30, 71, 112, 265, 418, 989, 1560, 3691, 5822, 13775, 21728, 51409, 81090, 191861, 302632, 716035, 1129438, 2672279, 4215120, 9973081, 15731042, 37220045, 58709048, 138907099, 219105150, 518408351, 817711552, 1934726305, 3051741058, 7220496869
Offset: 1

Views

Author

John W. Layman, May 23 2003

Keywords

Comments

The bisection {1,5,19,265,...} appears to be A001834 and to satisfy the recurrence a(n) = 4*a(n-1) - a(n-2) and the condition that 3*a(n)^2 + 6 is a square. The other bisection {2,8,30,112,...} appears to be A052530 and one-half of this bisection, {1,4,15,56,...}, appears to be A001353 and to satisfy a(n) = 4*a(n-1) - a(n-2) and the condition that 3*a(n)^2 + 1 is a square.
Conjecturally, a(n) = x + y, where these values solve x^2 - floor(y^2/3) = 1, see related sequences and formula below. - Richard R. Forberg, Sep 08 2013
Let alpha be an algebraic integer and define a sequence of integers a(alpha,n) by the condition a(alpha,n) = max { integer d : alpha^n = = 1 (mod d)}. Silverman shows that a(alpha,n) is a strong-divisibility sequence, that is gcd(a(n), a(m)) = a(gcd(n, m)) for all n and m in N; in particular, if n divides m then a(n) divides a(m). This sequence appears to be the strong divisibility sequence a(2 + sqrt(3),n) (Silverman, Example 4). - Peter Bala, Jan 10 2014
This sequence appears as the coefficients of the defining inequalities of a polyhedral realization of the B(infinity) crystal of the Kac-Moody Lie algebra with Cartan matrix [2,-2;-3,2] (see Nakashima-Zelevinsky reference). - Paul E. Gunnells, May 05 2019
From Zhuorui He, Jul 16 2025: (Start)
This sequence is Ratio-determined insertion sequence I(7/11) (see the Layman link below).
If S(0) in the definition is (1,1,a,b,c...) (all numbers >= 0) instead of (1,1), the resulting sequence is still the same.
For a finite sequence S, let k be the least i such that 7*S(i+1) <= 11*S(i). If k didn't exist then I(S)=S. Else, let k' be the least i such that 7*I(S)(i+1) <= 11*I(S)(i). Then k <= k' <= k+1.
This sequence can be generated by this process:
Step 1: Let X=1 and Y=1.
Step 2: If 7*(X+Y)<=11*X, then Y:=X+Y, repeat this step. Else go to step 3.
Step 3: Append X to the sequence. Let X:=X+Y, go back to step 2. (End)

Examples

			Let S(0) = (1,1). Since 7*1 <= 11*1 we obtain S(1) = (1,2,1). Then since 7*2 > 11*1 and 7*1 <= 11*2, we obtain S(2) = (1,2,3,1). Continuing, we get S(3) = (1,2,5,3,4,1), S(4) = (1,2,5,8,3,7,4,5,1), S(5) = (1,2,5,8,11,3,...), S(6) =(1,2,5,8,19,11,...), etc.
		

Crossrefs

Programs

  • Python
    def A082630_list(n):
      a = []
      x = y = 1
      while len(a) < n:
        a.append(x)
        while 7*(x+y) <= 11*x:
          y += x
        x += y
      return a # Zhuorui He, Jul 16 2025

Formula

The sequence appears to satisfy a(n) = 4*a(n-2) - a(n-4).
Apparently a(n)*a(n+3) = -2 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
Conjecturally, a(n) = A143643(n-1) + A005246(n), for n => 2, as derived from comment above. - Richard R. Forberg, Sep 08 2013
If the above conjectures are true, then a(n) = A001353(n)/A005246(n+1). - Andrey Zabolotskiy, Sep 26 2024

Extensions

Edited by M. F. Hasler, Nov 06 2018

A085260 Ratio-determined insertion sequence I(0.0833344) (see the link below).

Original entry on oeis.org

1, 12, 155, 2003, 25884, 334489, 4322473, 55857660, 721827107, 9327894731, 120540804396, 1557702562417, 20129592507025, 260127000028908, 3361521407868779, 43439651302265219, 561353945521579068
Offset: 1

Views

Author

John W. Layman, Jun 23 2003

Keywords

Comments

This sequence is the ratio-determined insertion sequence (RDIS) "twin" to A078362 (see the link for an explanation of "twin"). See A082630 or A082981 for recent examples of RDIS sequences.
a(n) = L(n,13), where L is defined as in A108299. - Reinhard Zumkeller, Jun 01 2005
For n >= 2, a(n) equals the permanent of the (2n-2) X (2n-2) tridiagonal matrix with sqrt(11)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Seems to be positive values of x (or y) satisfying x^2 - 13xy + y^2 + 11 = 0. - Colin Barker, Feb 10 2014
It appears that the b-file, formulas and programs are based on the conjectured, so far apparently unproved recurrence relation. - M. F. Hasler, Nov 05 2018
Nonnegative y values in solutions to the Diophantine equation 11*x^2 - 15*y^2 = -4. The corresponding x values are in A126866. Note that a(n+1)^2 - a(n)*a(n+2) = -11. - Klaus Purath, Mar 21 2025

Crossrefs

Row 13 of array A094954.
Cf. similar sequences listed in A238379.

Programs

  • Magma
    I:=[1,12]; [n le 2 select I[n] else 13*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 18 2018
  • Mathematica
    CoefficientList[Series[(1 - x)/(1 - 13 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)
    LinearRecurrence[{13,-1}, {1,12}, 30] (* G. C. Greubel, Jan 18 2018 *)
  • PARI
    my(x='x+O('x^30)); Vec(x*(1-x)/(1-13*x+x^2)) \\ G. C. Greubel, Jan 18 2018
    

Formula

It appears that the sequence satisfies a(n+1) = 13*a(n) - a(n-1). [Corrected by M. F. Hasler, Nov 05 2018]
If the recurrence a(n+2) = 13*a(n+1) - a(n) holds then for n > 0, a(n)*a(n+3) = 143 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
G.f.: x*(1-x)/(1 - 13*x + x^2). - Philippe Deléham, Nov 17 2008
For n>1, a(n) is the numerator of the continued fraction [1,11,1,11,...,1,11] with (n-1) repetitions of 1,11. - Greg Dresden, Sep 10 2019

A117535 Number of ways of writing n as a sum of powers of 3, each power being used at most 4 times.

Original entry on oeis.org

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

Views

Author

John W. Layman, Mar 27 2006

Keywords

Comments

It seems that this sequence can be calculated by constructing an insertion tree in which the insertion rules depend on the "age" of a term at a particular stage of the calculation. See the link for a discussion of this concept.

Examples

			a(12) = 4 because 12=9+3=9+1+1+1=3+3+3+3=3+3+3+1+1+1.
		

Crossrefs

Cf. A054390.

Programs

  • Maple
    g:= product((1+x^(3^j)+x^(2*(3^j))+x^(3*(3^j))+x^(4*(3^j))), j=0..10): gser:= series(g,x=0,106): seq(coeff(gser,x,n), n=0..103); # Emeric Deutsch, Apr 02 2006
    # second Maple program:
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
          add(`if`(n-j*3^i<0, 0, b(n-j*3^i, i-1)), j=0..4)))
        end:
    a:= n-> b(n, ilog[3](n)):
    seq(a(n), n=0..100);  # Alois P. Heinz, Jun 21 2012
  • Mathematica
    b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 0, 0, Sum[If[n - j*3^i < 0, 0, b[n - j*3^i, i - 1]], {j, 0, 4}]]]; a[n_] := b[n, Floor[Log[3, n]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Dec 22 2016, after Alois P. Heinz *)

Formula

G.f.: product((1+x^(3^j)+x^(2*(3^j))+x^(3*(3^j))+x^(4*(3^j))), j=0..infinity). - Emeric Deutsch, Apr 02 2006
For n>=1, a(3*n+2) = a(n); a(3*n+1) = a(n) + a(n-1); a(3*n) = a(n) + a(n-1). - Tom Edgar, Jun 21 2017
G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4) * A(x^3). - Ilya Gutkovskiy, Jul 09 2019

A082981 Start with the sequence S(0)={1,1} and for k>0 define S(k) to be I(S(k-1)) where I denotes the operation of inserting, for i=1,2,3..., the term a(i)+a(i+1) between any two terms for which 4a(i+1)<=5a(i). The listed terms are the initial terms of the limit of this process as k goes to infinity.

Original entry on oeis.org

1, 2, 3, 4, 9, 14, 19, 24, 53, 82, 111, 140, 309, 478, 647, 816, 1801, 2786, 3771, 4756, 10497, 16238, 21979, 27720, 61181, 94642, 128103, 161564, 356589, 551614, 746639, 941664, 2078353, 3215042, 4351731, 5488420, 12113529, 18738638, 25363747
Offset: 1

Views

Author

John W. Layman, May 28 2003

Keywords

Comments

Conjectures:
(1) the section {a(2n+1)}={1,3,9,19,53,111,...} is A077442, the terms of which are solutions of ax^2+7 = a square,
(2) the section {a(4n+1)}={1,9,53,309,1801,...} is A038761,
(3) the section {a(4n+2)}={2,14,82,478,2786,...} is A077444, the terms of which are solutions of 2x^2+8 = a square,
(4) the sequence {a(4n+2)/2}={1,7,41,239,1393,...} is A002315, the terms of which are solutions of 2x^2+2 = a square,
(5) the section {a(4n+4)}={4,24,140,816,4756,...} is A005319, the terms of which are solutions of 2x^2+4=a square,
(6) the sequence {a(4n+4)/4}={1,6,35,204,1189,...} is A001109, the terms of which are solutions of 8x^2+1=a square.

Crossrefs

Programs

  • Mathematica
    Most@Nest[If[#[[-2]] >= 4 #[[-1]], Append[Most@#, #[[-1]] + #[[-2]]], Insert[#, #[[-1]] + #[[-2]], -2]] &, {1, 1}, 47] (* Ivan Neretin, Apr 27 2017 *)

Formula

It appears that a(n)=6a(n-4)-a(n-8).
Empirical g.f.: x*(x+1)^2*(x^2+1)^2/((x^4-2*x^2-1)*(x^4+2*x^2-1)). - Colin Barker, Nov 06 2014

A085348 Ratio-determined insertion sequence I(0.264) (see the link below).

Original entry on oeis.org

1, 4, 19, 72, 341, 1292, 6119, 23184, 109801, 416020, 1970299, 7465176, 35355581, 133957148, 634430159, 2403763488, 11384387281, 43133785636, 204284540899, 774004377960, 3665737348901, 13888945017644, 65778987739319
Offset: 0

Views

Author

John W. Layman, Jun 24 2003

Keywords

Comments

This is one of the "twin" ratio-determined insertion sequences (RDIS) that are "children" in the next generation below the "parent" sequences I(0.25024) (A004253) and I(0.26816) (A001353) in the recurrence tree of RDIS sequences. The RDIS twin of this sequence is A085349. See the link for an explanation of RDIS twin. See A082630 or A082981 for other recent examples of RDIS sequences.
Assuming that a(n) = 18a(n-2) - a(n-4) is true: For n >= 2, a(n) = (t(i+2n+2) - t(i))/(t(i+n+2) + t(i+n)*(-1)^(n-1)), where (t) is any recurrence of the form (4,1) without regard to initial values. With an additional initional 0 is this sequence the union of A060645 for even n and A049629 for odd n. - Klaus Purath, Sep 22 2024

Crossrefs

Formula

It appears that a(n)=18a(n-2)-a(n-4).
Apparently a(n)a(n+3) = -4 + a(n+1)a(n+2). - Ralf Stephan, May 29 2004
From Klaus Purath, Sep 22 2024: (Start)
Assuming that a(n) = 18a(n-2) - a(n-4) is true:
a(2n) = 5a(2n-1) - a(2n-2), n >= 1.
a(2n+1) = 4a(2n) - a(2n-1), n >= 1. (End)

A085349 Ratio-determined insertion sequence I(0.26688) (see the link below).

Original entry on oeis.org

1, 4, 15, 71, 269, 1274, 4827, 22861, 86617, 410224, 1554279, 7361171, 27890405, 132090854, 500473011, 2370274201, 8980623793, 42532844764, 161150755263, 763220931551, 2891732970941, 13695443923154, 51890042721675
Offset: 1

Views

Author

John W. Layman, Jun 24 2003

Keywords

Comments

This is one of the "twin" ratio-determined insertion sequences (RDIS) that are "children" in the next generation below the "parent" sequences I(0.25024) (A004253) and I(0.26816) (A001353) in the recurrence tree of RDIS sequences. The RDIS twin of this sequence is A085348. See the link for an explanation of RDIS twin. See A082630 or A082981 for other recent examples of RDIS sequences.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{0,18,0,-1},{1,4,15,71},30] (* Harvey P. Dale, Mar 04 2013 *)

Formula

It appears that a(n)=18a(n-2)-a(n-4).
Apparently a(n)a(n+3) = 11 + a(n+1)a(n+2). - Ralf Stephan, May 29 2004

A343273 a(n) is the number of geometrically distinct edge-unfoldings of the regular n-gonal cupola.

Original entry on oeis.org

308, 3030, 29757, 294327, 2911142, 28814940, 285214743, 2823311133, 27947663768, 276653115090, 2738581182417, 27109156615827, 268352962161482, 2656420444277880, 26295851254778283, 260302091898387033, 2576725065493516028, 25506948561006315150
Offset: 3

Views

Author

Rick Mabry, Apr 10 2021

Keywords

Comments

The term "regular" applies only to the regular n-gon and 2n-gon (the "top and bottom" of the cupola), the other faces (the "sides") being n isosceles triangles and n sufficiently long rectangles. For n=3,4,5, regular triangles and squares can be used for the sides. That applies to n=6 if a two-sided (flat) polyhedron is allowed.
The first 25 terms of the auxiliary sequence c(n) in the Formula and Mathematica program match the 25 terms listed for sequence A085376.

Crossrefs

Cf. A085376; see the sequence c(n) in the Formula and Mathematica program, but note that A085376 has only been conjectured to be the same as c(n).

Programs

  • Mathematica
    a[n_]:=Sum[c[k],{k,1,2n-1}]+(1/2)c[2n]+If[OddQ[n],(1/2)c[n],c[n]];
    c[1] = 1; c[2] = 3; c[3] = 11; c[4] = 30;
    c[m_] := c[m] = 10 c[m - 2] - c[m - 4];

Formula

Recursively define the sequence c(m) as follows: Let c(1) = 1, c(2) = 3, c(3) = 11, c(4) = 30, and for m > 4, let c(m) = 10*c(m-2) - c(m-4). Then for all n >= 3, the sequence a(n) can be given by a(n) = (c(2*n+1) + 5*c(2*n) - c(2*n-1) - c(2*n-2) - 5)/8 + (3 + (-1)^n)*c(n)/4.
a(n) = (c(2*n+1) + 5*c(2*n) - c(2*n-1) - c(2*n-2) - 5)/8 + (3 + (-1)^n)*c(n)/4 for n >= 3 where c(m) = 10*c(m-2) - c(m-4) for m > 4 and c(1) = 1, c(2) = 3, c(3) = 11, c(4) = 30.
G.f.: x^3*(308 - 358*x - 3265*x^2 + 3602*x^3 - 360*x^5 + 33 x^6)/(1 - 11*x + x^2 + 109*x^3 - 109*x^4 - x^5 + 11*x^6 - x^7). - Stefano Spezia, Apr 10 2021
Showing 1-7 of 7 results.