A132424 -id:A132424 - cp's OEIS frontend

A130147 a(0)=1. a(n+1) = a(floor(n/a(n))) + 1.

1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 5, 4, 4, 4, 4, 4, 5, 4, 5, 5, 5, 5, 5, 4, 5, 4, 5, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5

Offset: 0

Leroy Quet, Aug 02 2007

Records occur at 0,1,3,10,41,206,1237,8660,69281,623530,6235301, which appears to be A002627. See A132424 for a related sequence. - John W. Layman, Aug 20 2007

N. J. A. Sloane, Table of n, a(n) for n = 0..20000

Cf. A002627, A132424.

See A130193 for "ceiling" version.

a[0]:=1: for n from 0 to 110 do a[n+1]:=1+a[floor(n/a[n])] end do: seq(a[n],n=0..110); # Emeric Deutsch, Aug 19 2007

a[0] = 1; a[n_] := a[n] = 1 + a[Floor[(n - 1)/a[n - 1]]]; Array[a, 105, 0] (* Michael De Vlieger, Sep 03 2017 *)

More terms from Emeric Deutsch and John W. Layman, Aug 19 2007

A286286 a(0) = 0; thereafter, a(n) = (2n-1)a(n-1) + 1.

Original entry on oeis.org

0, 1, 4, 21, 148, 1333, 14664, 190633, 2859496, 48611433, 923617228, 19395961789, 446107121148, 11152678028701, 301122306774928, 8732546896472913, 270708953790660304, 8933395475091790033, 312668841628212651156, 11568747140243868092773

Offset: 0

N. J. A. Sloane, May 15 2017

Seiichi Manyama, Table of n, a(n) for n = 0..404

Conjectured to give indices of records in A132424.

Cf. A001147, A002627 (similar sequence), A000522, A060196.

NestList[{(2 #2 - 1) #1 + 1, #2 + 1} & @@ # &, {0, 1}, 19][[All, 1]] (* Michael De Vlieger, Dec 10 2021 *)

a(n) = (2*n-1)!! * Sum_{k=1..n} 1/(2*k-1)!!. - Seiichi Manyama, Sep 02 2017
a(n) = floor((2*n-1)!!*A060196), for n > 0. - Peter McNair, Dec 10 2021
From Peter Bala, Feb 09 2024: (Start)
a(n) = 2*n*a(n-1) - (2*n - 3)*a(n-2) with a(0) = 0 and a(1) = 1.
The double factorial numbers (2*n-1)!! = A001147(n) satisfy the same recurrence, leading to the generalized continued fraction expansion Limit_{n -> oo} a(n)/(2*n-1)!! = Sum_{k >= 1} 1/(2*k-1)!! = A060196 = 1/(1 - 1/(4 - 3/(6 - 5/(8 - 7/(10 - 9/(12 - ... )))))). (End)

A335901 a(n) = 2*a(floor((n-1)/a(n-1))) with a(1) = 1.

Original entry on oeis.org

1, 2, 2, 2, 4, 2, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 4, 8, 4, 4, 4, 4, 4, 8, 4, 8, 4, 4, 4, 4, 4, 8, 4, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8, 4, 8, 4, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8, 4, 8, 4, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset: 1

Altug Alkan, following a suggestion from Andrew R. Booker, Jun 29 2020

Least k such that a(k) = 2^n are 1, 2, 5, 21, 169, 2705, ... (Conjecture: This sequence is A117261).

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A130147, A132424, A130535, A283207, A288914, A335898.

f:= proc(n) option remember;
  2*procname(floor((n-1)/procname(n-1))) end proc:
f(1):= 1:
map(f, [$1..105]); # Robert Israel, Jul 08 2020

a[1] = 1; a[n_] := a[n] = 2 * a[Floor[(n-1)/a[n-1]]]; Array[a, 100] (* Amiram Eldar, Jun 29 2020 *)

a=vector(10^3); a[1]=1; for(n=2, #a, a[n]=2*a[(n-1)\a[n-1]]); a

cp's OEIS Frontend

A130147 a(0)=1. a(n+1) = a(floor(n/a(n))) + 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A286286 a(0) = 0; thereafter, a(n) = (2n-1)a(n-1) + 1.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A335901 a(n) = 2*a(floor((n-1)/a(n-1))) with a(1) = 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

cp's OEIS Frontend

A130147 a(0)=1. a(n+1) = a(floor(n/a(n))) + 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A286286 a(0) = 0; thereafter, a(n) = (2*n-1)*a(n-1) + 1.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A335901 a(n) = 2*a(floor((n-1)/a(n-1))) with a(1) = 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

A286286 a(0) = 0; thereafter, a(n) = (2n-1)a(n-1) + 1.