A133145 -id:A133145 - cp's OEIS frontend

A160700 a(n) = if n<16 then n else a(floor(n/16)) XOR (n mod 16).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 0, 3, 2, 5, 4, 7, 6, 9, 8, 11, 10, 13, 12, 15, 14, 2, 3, 0, 1, 6, 7, 4, 5, 10, 11, 8, 9, 14, 15, 12, 13, 3, 2, 1, 0, 7, 6, 5, 4, 11, 10, 9, 8, 15, 14, 13, 12, 4, 5, 6, 7, 0, 1, 2, 3, 12, 13, 14, 15, 8, 9, 10, 11, 5, 4, 7, 6, 1, 0, 3, 2, 13

Offset: 0

Reinhard Zumkeller, Jun 01 2009

A very simple hash function for the nonnegative integers.

a(A000079(n))=A133145(n); a(A000302(n))=A010685(n); a(A001025(n))=A161452(n); a(A161440(n))=0; a(A161441(n))=1; a(A161442(n))=2; a(A161443(n))=3; a(A161444(n))=4; a(A161445(n))=5; a(A161446(n))=6; a(A161447(n))=7; a(A161448(n))=8; a(A161449(n))=9; a(A161450(n))=10; a(A161451(n))=11; a(A161452(n))=12; a(A161453(n))=13; a(A161454(n))=14; a(A161455(n))=15. - Reinhard Zumkeller, Jun 10 2009

R. Zumkeller, Table of n, a(n) for n = 0..10000

import Data.Bits (xor)
a160700 n = a160700_list !! n
a160700_list = [0..15] ++ map f [16..] where
   f x = a160700 x' `xor` m :: Int where (x', m) = divMod x 16
-- Reinhard Zumkeller, Nov 07 2012

read("transforms") ;
A160700 := proc(n)
    if n < 16 then
        n;
    else
        XORnos(procname(floor(n/16)),modp(n,16))
    end if;
end proc: # R. J. Mathar, Jul 12 2016

a[n_] := a[n] = If[n < 16, n, a[Floor[n/16]] ~BitXor~ Mod[n, 16]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jan 25 2018 *)

load(functs)$
A160700(n):=if n<16 then n else logxor(floor(n/16),mod(n,16))$
makelist(A160700(n),n,0,60); /* Martin Ettl, Nov 05 2012 */

a(n)=my(t=n%16); while(n>15, n>>=4; t=bitxor(t, n%16)); t \\ Charles R Greathouse IV, Jan 25 2018

A083593 Expansion of 1/((1-2x)(1-x^4)).

Original entry on oeis.org

1, 2, 4, 8, 17, 34, 68, 136, 273, 546, 1092, 2184, 4369, 8738, 17476, 34952, 69905, 139810, 279620, 559240, 1118481, 2236962, 4473924, 8947848, 17895697, 35791394, 71582788, 143165576, 286331153, 572662306, 1145324612, 2290649224

Offset: 0

Paul Barry, May 02 2003

Here we let p = 4 to produce the above sequence, but p can be an arbitrary natural number. By letting p = 2, 3, 6, 7 we produce A000975, A033138, A195904 and A117302. We denote by U[p,n,m] the number of cases in which the first player gets killed in a Russian roulette game when p players use a gun with n chambers and m bullets. They never rotate the cylinder after the game starts. The chambers can be represented by the list {1,2,...,n}.
We are going to calculate the following (0), (1), ..., (t) separately. (0) The first player gets killed when one bullet is in the first chamber and the remaining m-1  bullets are in {2,3,...,n}. We have binomial(n-1,m-1) cases for this. (1) The first gets killed when one bullet is in the (p+1)th chamber and the rest of the bullets are in {p+2,...,n}. We have binomial(n-p-1,m-1) cases for this. We continue to calculate and the last is (t), where t = floor((n-m)/p). (t) The first gets killed when one bullet is in the (pt+1)-st chamber and the remaining bullets are in {pt+2,...,n}. We have binomial(n-pt-1,m-1) cases for this. Therefore U[p,n,m] = Sum_{z=0..floor((n-m)/p)} binomial(n-pz-1,m-1). Let A[p,n] be the number of the cases in which the first player gets killed when p players use a gun with n chambers and the number of the bullets can be from 1 to n. Then A[p,n] = Sum_{m=1..n} U[p,n,m]. - Ryohei Miyadera, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006
A001045(n+5) without last digit. - Paul Curtz, Apr 21 2021
a(n) is the number of partitions of n into parts 1 and 4 where there are two colors of part 1 and the order of the colors of parts 1 matters. If the order of colors doesn't matter we get A001972. - Joerg Arndt, Jan 18 2024

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,0,1,-2).

Cf. A000975, A033138, A195904, A117302.

U[p_,n_,m_,v_]:=Block[{t},t=Floor[(1+p-m+n-v)/p];Sum[Binomial[n-v-p*z,m-1],{z,0,t-1}]]; A[p_,n_,v_]:=Sum[U[p,n,k,v],{k,1,n}]; (* Here we let p = 4 to produce the above sequence, but this code can produce A000975, A033138, A195904, A117302 for p=2,3,6,7.*) Table[A[4,n,1], {n,1,20}] (* Ryohei Miyadera, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006 *)
CoefficientList[Series[1/((1-2x)(1-x^4)),{x,0,40}],x] (* Vincenzo Librandi, Apr 04 2012 *)
a[n_] := FromDigits[Table[(Mod[j, 4]/4) // Round, {j, 1, n + 3}], 2] (* Andres Cicuttin, Mar 25 2016 *)
a[n_] := a[n] = 2 a[n - 1] + 1 - Ceiling[Mod[n, 4]/4]; a[0] = 1;
Table[a[n], {n, 0, 31}] (* Andres Cicuttin, Mar 27 2016 *)
LinearRecurrence[{2,0,0,1,-2},{1,2,4,8,17},40] (* Harvey P. Dale, Apr 03 2018 *)

Vec(1/((1-2*x)*(1-x^4))+O(x^99)) \\ Charles R Greathouse IV, May 15 2013

a(n)=(16<Charles R Greathouse IV, Mar 27 2016

def A083593(n): return ((32<Chai Wah Wu, Apr 25 2025

a(n) = 2*a(n-1) + a(n-4) - 2*a(n-5).
If n is a multiple of 4, then a(n) = 2*a(n-1) + 1, otherwise a(n) = 2*a(n-1). - Gerald McGarvey, Oct 14 2008
a(n) = floor((2^(n+5) + 1)/30). - Tani Akinari, Jul 09 2013
From Andres Cicuttin, Mar 29 2016: (Start)
a(n) = 2*a(n-1) + floor(((n-1) mod 4)/3), with a(0)=1.
a(n) = 2*a(n-1) + 1 - ceiling((n mod 4)/4), with a(0)=1. (End)
15*a(n) = 2^(n+4) - A133145(n). - R. J. Mathar, Feb 27 2019
E.g.f.: (3*cos(x) - 5*cosh(x) + 32*cosh(2*x) + 6*sin(x) - 10*sinh(x) + 32*sinh(2*x))/30. - Stefano Spezia, Apr 25 2025

A191497 a(n+1) = 2*a(n) + A014017(n+5), a(0) = 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 4, 8, 15, 30, 60, 120, 241, 482, 964, 1928, 3855, 7710, 15420, 30840, 61681, 123362, 246724, 493448, 986895, 1973790, 3947580, 7895160, 15790321, 31580642, 63161284, 126322568, 252645135

Offset: 0

Paul Curtz, Jun 03 2011

Index entries for linear recurrences with constant coefficients, signature (2,0,0,-1,2).

Cf. A014017, A133145.

A191497 := proc(n): if n=0 then 0 else A191497(n) := 2*A191497(n-1) + A014017(n+4) fi: end: A014017 := proc(n): (1/8)*(-(n mod 8)-((n+3) mod 8)+((n+4) mod 8)+((n+7) mod 8)) end: seq(A191497(n),n=0..32); # Johannes W. Meijer, Jun 28 2011

LinearRecurrence[{2,0,0,-1,2},{0,0,0,0,1},40] (* Harvey P. Dale, Apr 19 2013 *)

a(n+4) = 2^n - a(n).
a(n) = 2*a(n-1) - a(n-4) + 2*a(n-5).
a(4*n+4) = 16*a(4*n) + (-1)^n.
From R. J. Mathar, Jun 23 2011: (Start)
G.f.: -x^4 / ((2*x-1)*(x^4+1)).
a(n) = (2^n - (-1)^floor(n/4)*A133145(n))/17. (End)

cp's OEIS Frontend

A160700 a(n) = if n<16 then n else a(floor(n/16)) XOR (n mod 16).

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A083593 Expansion of 1/((1-2x)(1-x^4)).

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A191497 a(n+1) = 2*a(n) + A014017(n+5), a(0) = 0.

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cp's OEIS Frontend

A160700 a(n) = if n<16 then n else a(floor(n/16)) XOR (n mod 16).

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Keywords

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Programs

Haskell

Maple

Mathematica

Maxima

PARI

A083593 Expansion of 1/((1-2*x)*(1-x^4)).

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Author

Keywords

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Links

Crossrefs

Programs

Mathematica

PARI

PARI

Python

Formula

A191497 a(n+1) = 2*a(n) + A014017(n+5), a(0) = 0.

Views

Author

Keywords

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Crossrefs

Programs

Maple

Mathematica

Formula

A083593 Expansion of 1/((1-2x)(1-x^4)).