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 1-6 of 6 results.

A085376 Ratio-dependent insertion sequence I(0.36704) (see the link below).

Original entry on oeis.org

1, 3, 11, 30, 109, 297, 1079, 2940, 10681, 29103, 105731, 288090, 1046629, 2851797, 10360559, 28229880, 102558961, 279447003, 1015229051, 2766240150, 10049731549, 27382954497, 99482086439, 271063304820, 984771132841
Offset: 1

Views

Author

John W. Layman, Jun 26 2003

Keywords

Comments

This sequence is the ratio-determined insertion sequence (RDIS) "twin" of I(0.37802)=A080874 and "child" of I(0.33344)=A001835 and I(0.38208)=A001906 in the RDIS recurrence tree (see the link for an explanation of terms). See A082630, A082981, A085348 and A085349 for recent examples of RDIS sequences.
Conjecture: partial sums of A129445. - Sean A. Irvine, Jul 14 2022

Crossrefs

Formula

It is conjectured that a(n) = 10*a(n-2) - a(n-4).
Apparently a(n)*a(n+3) = -3 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004

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

A303827 Number of ways of writing n as a sum of powers of 4, each power being used at most 5 times.

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, May 01 2018

Keywords

Examples

			a(17) = 3 because 17=16+1=4+4+4+4+1=4+4+4+1+1+1+1+1.
		

Crossrefs

Number of ways of writing n as a sum of powers of b, each power being used at most b+1 times: A117535 (b=3), this sequence (b=4), A303828 (b=5).
Cf. A277872.

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
          add(b(n-j*4^i, i-1), j=0..min(5, n/4^i))))
        end:
    a:= n-> b(n, ilog[4](n)):
    seq(a(n), n=0..120);  # Alois P. Heinz, May 01 2018
  • Mathematica
    m = 100; A[_] = 1;
    Do[A[x_] = (1+x+x^2+x^3+x^4+x^5) * A[x^4] + O[x]^m // Normal, {m}];
    CoefficientList[A[x], x] (* Jean-François Alcover, Oct 06 2019, after Ilya Gutkovskiy *)

Formula

G.f.: Product_{k>=0} (1-x^(6*4^k))/(1-x^(4^k)).
G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4 + x^5) * A(x^4). - Ilya Gutkovskiy, Jul 09 2019

A303828 Number of ways of writing n as a sum of powers of 5, each power being used at most 6 times.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 3, 3, 2, 2, 2, 4, 4, 2, 2, 2, 3, 3, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 3, 3, 2, 2, 2, 4, 4, 2, 2, 2, 3, 3, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 3, 3, 2, 2, 2, 4, 4, 2, 2, 2, 3, 3, 1, 1, 1, 2, 2
Offset: 0

Views

Author

Seiichi Manyama, May 01 2018

Keywords

Examples

			a(26) = 3 because 26=25+1=5+5+5+5+5+1=5+5+5+5+1+1+1+1+1+1.
		

Crossrefs

Number of ways of writing n as a sum of powers of b, each power being used at most b+1 times: A117535 (b=3), A303827 (b=4), this sequence (b=5).
Cf. A277873.

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
          add(b(n-j*5^i, i-1), j=0..min(6, n/5^i))))
        end:
    a:= n-> b(n, ilog[5](n)):
    seq(a(n), n=0..120);  # Alois P. Heinz, May 01 2018
  • Mathematica
    m = 100; A[_] = 1;
    Do[A[x_] = Total[x^Range[0, 6]] A[x^5] + O[x]^m // Normal, {m}];
    CoefficientList[A[x], x] (* Jean-François Alcover, Oct 19 2019 *)

Formula

G.f.: Product_{k>=0} (1-x^(7*5^k))/(1-x^(5^k)).
G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6) * A(x^5). - Ilya Gutkovskiy, Jul 09 2019
Showing 1-6 of 6 results.