A105801 -id:A105801 - cp's OEIS frontend

A181663 a(n) = A105801(n) mod 2.

1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1

Offset: 1

N. J. A. Sloane, Nov 20 2010

nonn

N. J. A. Sloane, Table of n, a(n) for n = 1..3000

Cf. A105801

A122524 Prime numbers in Fibonacci-Collatz sequence A105801.

Original entry on oeis.org

2, 7, 103, 274309092506641744944619304235289, 71913864933263337464162651213311153, 23583673968885646741228783732071635618505511

Offset: 1

Zak Seidov, Sep 17 2006

nonn

Cf. A105801.

A124139 a(n) = A000265(3*(a(n-1) + a(n-2))/2 + 1) starting at a(1)=1, a(2)=11.

Original entry on oeis.org

1, 11, 19, 23, 1, 37, 29, 25, 41, 25, 25, 19, 67, 65, 199, 397, 895, 1939, 1063, 563, 305, 1303, 2413, 5575, 11983, 13169, 37729, 19087, 85225, 156469, 181271, 506611, 64489, 856651, 1381711, 419693, 2702107, 4682701, 11077213, 369373, 2146235, 3773413

Offset: 1

Yasutoshi Kohmoto, Dec 01 2006

nonn

A variant of A105801: The highest power of two is recursively removed from 3x/2+1, where x is the sum of the preceding two elements of the sequence.

Cf. A124138.

A000265 := proc(n) local a,nshft ; a := 1 ; nshft := n ; while nshft mod 2 = 0 do nshft := nshft/2 ; od: nshft ; end:
A124139 := proc(n) option remember ; if n = 1 then 1; elif n = 2 then 11; else A000265(3*(procname(n-1)+procname(n-2))/2 +1) ; fi; end: seq(A124139(n),n=1..60) ; # R. J. Mathar, Jul 02 2009

Edited and extended by R. J. Mathar, Jul 02 2009

A181717 Fibonacci-Collatz sequence: a(1)=0, a(2)=1; for n>2, let fib=a(n-1)+a(n-2); if fib is odd then a(n)=3*fib+1 else a(n)=fib/2.

Original entry on oeis.org

0, 1, 4, 16, 10, 13, 70, 250, 160, 205, 1096, 3904, 2500, 3202, 2851, 18160, 63034, 40597, 310894, 1054474, 682684, 868579, 4653790, 16567108, 10610449, 81532672, 276429364, 178981018, 227705191, 1220058628, 4343291458, 2781675043, 21374899504, 72469723642

Offset: 1

Ralf Stephan, Nov 17 2010

nonn

It is easy to prove that all the terms a(n) with n>=7 are congruent to 7 mod 9. Conjecture: for every k>0 there is an index m such that all the a(n) with n>m have the same residue mod 3^k. - Giovanni Resta, Nov 17 2010

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Cf. A000045, A006370, A105801.

a181717 n = a181717_list !! (n-1)
a181717_list = 0 : 1 : fc 1 0 where
   fc x x' = y : fc y x where y = a006370 (x + x')
-- Reinhard Zumkeller, Oct 09 2011

a:= proc(n) option remember; local f;
      if n<3 then return n-1 fi;
      f:= a(n-1) +a(n-2);
      `if`(irem(f, 2)=0, f/2, 3*f+1)
    end:
seq(a(n), n=1..50); # Alois P. Heinz, Oct 09 2011

nxt[{a_,b_}]:=Module[{fib=a+b},If[OddQ[fib],{b,3fib+1},{b,fib/2}]]; Transpose[NestList[nxt,{0,1},40]][[1]] (* Harvey P. Dale, Mar 21 2012 *)

v=vector(60,n,0); v[2]=1; for(n=3,60,f=v[n-1]+v[n-2]; v[n]=if(f%2,3*f+1,f/2))

@CachedFunction
def a(n):
    if n<3: return n-1
    elif (a(n-1)+a(n-2))%2==1: return 3*(a(n-1)+a(n-2))+1
    else: return (a(n-1)+a(n-2))/2
[a(n) for n in range(1,51)] # G. C. Greubel, Mar 25 2024

A124138 a(n)= A000265(3*(a(n-1)+a(n-2))/2 +1) starting at a(1)=1, a(2)=3.

Original entry on oeis.org

1, 3, 7, 1, 13, 11, 37, 73, 83, 235, 239, 89, 493, 437, 349, 295, 967, 947, 359, 245, 907, 1729, 3955, 8527, 4681, 19813, 18371, 57277, 113473, 128063, 362305, 735553, 411697, 430219, 1262875, 1269821, 3799045, 1900825, 4274903, 9263593, 20307745, 2772313

Offset: 1

Yasutoshi Kohmoto, Dec 01 2006

nonn

A variant of A105801: The highest power of two is recursively removed from 3x/2+1, where x is the sum of the preceding two elements of the sequence.

			Examples which start with s(1)=1 and s(2)=2*k+1:
1,1,1,1,1,1,1,1,1,1,1,1,..... : A000012
1,3,7,1,13,11,37,73,83,235,.... : this sequence
1,5,5,1,5,5,1,5,5,1,5,5,1,5,5,1,5,5,1,5,5.... : periodic
1,7,13,31,67,37,157,73,173,185,269,341,229,107,505,919,2137,4585,....
1,9,1,1,1,1,1,1,1,1,1.1....
1,11,19,23,1,37,29,25,41,25,25,19,67,65,199,397,895,.... : A124139

Edited and extended by R. J. Mathar, Jul 02 2009

A329068 a(1) = 1; thereafter if the sum of digits of all previous terms up to a(n) is even then a(n+1) = (sum of digits of all previous terms)/2, otherwise a(n+1) = (sum of digits of all previous terms)*3 + 1.

Original entry on oeis.org

1, 4, 16, 6, 9, 82, 112, 124, 24, 27, 190, 220, 232, 42, 45, 298, 59, 66, 72, 460, 490, 88, 96, 622, 652, 115, 712, 742, 130, 132, 135, 838, 149, 156, 162, 1000, 167, 174, 180, 1108, 1138, 196, 204, 207, 1270, 1300, 1312, 222, 225, 1378, 239, 246, 252, 1540, 1570, 268, 276

Offset: 1

Bence Bernáth, Nov 03 2019

Bence Bernáth, Table of n, a(n) for n = 1..100001

Cf. A105801, A004207.

clear all;
length_seq=10000;
sequence(1)=1;
seq_for_digits(1)=sequence(1);
for i1=1:1:length_seq
   if  0==mod(sum(seq_for_digits),2)
        sequence(i1+1)=sum(seq_for_digits)/2;
   else
     sequence(i1+1)=sum(seq_for_digits)*3+1;
   end
     append=num2str(sequence(i1+1))-'0';
     seq_for_digits=[seq_for_digits append];
end
result=transpose(sequence);

lista(nn) = {va = vector(nn); va[1] = 1; sd = sumdigits(va[1]); for (n=2, nn, if (sd % 2, va[n] = 3*sd+1, va[n] = sd/2); sd += sumdigits(va[n]);); va;} \\ Michel Marcus, Nov 04 2019

from itertools import islice
def agen(): # generator of terms
    sd, an = 0, 1
    while True:
        yield an
        sd += sum(map(int, str(an)))
        an = 3*sd+1 if sd&1 else sd//2
print(list(islice(agen(), 60))) # Michael S. Branicky, Nov 12 2022

cp's OEIS Frontend

A181663 a(n) = A105801(n) mod 2.

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A122524 Prime numbers in Fibonacci-Collatz sequence A105801.

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A124139 a(n) = A000265(3*(a(n-1) + a(n-2))/2 + 1) starting at a(1)=1, a(2)=11.

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Maple

Extensions

A181717 Fibonacci-Collatz sequence: a(1)=0, a(2)=1; for n>2, let fib=a(n-1)+a(n-2); if fib is odd then a(n)=3*fib+1 else a(n)=fib/2.

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Haskell

Maple

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SageMath

A124138 a(n)= A000265(3*(a(n-1)+a(n-2))/2 +1) starting at a(1)=1, a(2)=3.

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Examples

Extensions

A329068 a(1) = 1; thereafter if the sum of digits of all previous terms up to a(n) is even then a(n+1) = (sum of digits of all previous terms)/2, otherwise a(n+1) = (sum of digits of all previous terms)*3 + 1.

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MATLAB

PARI

Python