A159560 -id:A159560 - cp's OEIS frontend

A004760 List of numbers whose binary expansion does not begin 10.

0, 1, 3, 6, 7, 12, 13, 14, 15, 24, 25, 26, 27, 28, 29, 30, 31, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122

Offset: 1

N. J. A. Sloane

For n >= 2 sequence {a(n+2)} is the minimal recursive such that A007814(a(n+2))=A007814(n). - Vladimir Shevelev, Apr 27 2009
A053645(a(n)) = n-1 for n > 0. - Reinhard Zumkeller, May 20 2009
a(n+1) is also the number of nodes in a complete binary tree with n nodes in the bottommost level. - Jacob Jona Fahlenkamp, Feb 01 2023

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Vladimir Shevelev, Several results on sequences which are similar to the positive integers, arXiv:0904.2101 [math.NT], 2009. [_Vladimir Shevelev_, Apr 15 2009]

Cf. A004761, A007814, A010060, A159559, A159560, A159615, A159619, A159629, A159698, A004759.

0,1,seq(seq(3*2^d+x,x=0..2^d-1),d=0..6); # Robert Israel, Aug 03 2016

Select[Range@ 125, If[Length@ # < 2, #, Take[#, 2]] &@ IntegerDigits[#, 2] != {1, 0} &] (* Michael De Vlieger, Aug 02 2016 *)

is(n)=n<2 || binary(n)[2] \\ Charles R Greathouse IV, Sep 23 2012

print1("0, 1");for(i=0,5,for(n=3<Charles R Greathouse IV, Sep 23 2012

a(n) = if(n<=2,n-1, (n-=2) + 2<Kevin Ryde, Jul 22 2022

def A004760(n): return m+(1<0 else n-1 # Chai Wah Wu, Jul 26 2023

maxrow <- 8 # by choice
b01 <- 1
for(m in 0:maxrow){
  b01 <- c(b01,rep(1,2^(m+1))); b01[2^(m+1):(2^(m+1)+2^m-1)] <- 0
}
a <- which(b01 == 1)
# Yosu Yurramendi, Mar 30 2017

For n > 0, a(n) = 3n - 2 - A006257(n-1). - Ralf Stephan, Sep 16 2003
a(0) = 0, a(1) = 1, for n > 0: a(2n) = 2*a(n) + 1, a(2n+1) = 2*a(n+1). - Philippe Deléham, Feb 29 2004
For n >= 3, A007814(a(n)) = A007814(n-2). - Vladimir Shevelev, Apr 15 2009
a(n+2) = min{m>a(n+1): A007814(m)=A007814(n)}; A010060(a(n+2)) = 1-A010060(n). - Vladimir Shevelev, Apr 27 2009
a(1)=0, a(2)=1, a(2^m+k+2) = 2^(m+1) + 2^m+k, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Jul 30 2016
G.f.: x/(1-x)^2 + (x/(1-x))*Sum_{k>=0} 2^k*x^(2^k). - Robert Israel, Aug 03 2016
a(2^m+k) = A004761(2^m+k) + 2^m, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Aug 08 2016
For n > 0, a(n+1) = n + 2^ceiling(log_2(n)) - 1. - Jacob Jona Fahlenkamp, Feb 01 2023

Offset changed to 1, b-file corrected. - N. J. A. Sloane, Aug 07 2016

A159615 The slowest increasing sequence beginning with a(1)=2 such that a(n) and n are both odious or both not odious.

Original entry on oeis.org

2, 4, 5, 7, 9, 10, 11, 13, 15, 17, 19, 20, 21, 22, 23, 25, 27, 29, 31, 33, 35, 37, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 97, 99, 101, 103, 105, 107, 109, 111

Offset: 1

Vladimir Shevelev, Apr 17 2009

			If n=3, then k=1, j=0, therefore a(6)=(10*3-4*0)/3=10.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Vladimir Shevelev, Several results on sequences which are similar to the positive integers, arXiv:0904.2101 [math.NT], 2009.

Cf. A000069, A159559, A159560, A004760.

read("transforms") ; isA000069 := proc(n) option remember ; RETURN( type(wt(n),'odd') ) ; end:
A159615 := proc(n) option remember; if n = 1 then 2; else for a from procname(n-1)+1 do if isA000069(a) = isA000069(n) then RETURN(a) ; fi; od: fi; end:
seq(A159615(n),n=1..120) ; # R. J. Mathar, Aug 17 2009

odiousQ[n_] := OddQ[DigitCount[n, 2, 1]];
a[1] = 2; a[n_] := a[n] = For[k = a[n-1]+1, True, k++, If[FreeQ[Array[a, n-1], k] && odiousQ[n] && odiousQ[k] || !odiousQ[n] && !odiousQ[k], Return[k] ] ];
Array[a, 80] (* Jean-François Alcover, Dec 10 2017 *)

For n>=1, a(n)=min{m>a(n-1): A010060(m)=A010060(n)}.
a(2n+1)=2a(n)+1.
a(2n)=3n+1+j,if n=2^k+j; a(2n)=(10n-4j)/3,if n=2^k+2^(k-1)+j, where 0<=j<=2^(k-1)-1.

Edited and extended by R. J. Mathar, Aug 17 2009

A159619 Slowest increasing sequence beginning with 4 such that n and a(n) are either both evil or both odious.

Original entry on oeis.org

4, 7, 9, 11, 12, 15, 16, 19, 20, 23, 25, 27, 28, 31, 33, 35, 36, 39, 41, 43, 44, 47, 48, 51, 52, 55, 57, 59, 60, 63, 64, 67, 68, 71, 73, 75, 76, 79, 80, 83, 84, 87, 89, 91, 92, 95, 97, 99, 100, 103, 105, 107, 108, 111, 112, 115, 116, 119, 121, 123, 124, 127, 129, 131, 132, 135, 137

Offset: 1

Vladimir Shevelev, Apr 17 2009, Apr 27 2009, May 04 2009

(i) Theorem: For every initial value a(1) > 4, a minimum index n exists such that the a(n) obtained from that initial value coincides with this sequence here. Thus there exist essentially two slowest increasing sequences with this type of evil/odious congruence: A159615 and this one here.
(ii) In connection with this theorem, one can generalize to slowest increasing sequences a_m(n), a_m(1)=m, which let n and a(n) be at the same time in or not in some increasing sequence c(n). (This sequence here is c = A000069, m=4.)
We define a rank r of c as the minimum value a_r(1) such that for sufficiently large n (n depending on m) all sequences a_m(n), m>r, coincide with a_r(n).
In particular, c(n)=A004760(n+1) has rank r=2, and A000069 has rank r=3.
The problems are: 1) to find a sequence of rank r >= 4; 2) to find the rank of primes or to prove that it does not exist (in case of which it could be defined as infinity).
There is a conjecture arising in Sequence Machine that a(n) = A026491(2+n)-1. This appears to be true: Here we start from on odious or evil number and apply a minimum number of van-Eck-Transforms (of A171898) to reach a value larger than a(n-1). The Dekking formula in A026491 says that A026491 is essentially a partial sum of the backward van-Eck-Transforms, and in a (vague) manner this seems to match.
- R. J. Mathar, Jun 24 2021

Amiram Eldar, Table of n, a(n) for n = 1..10000
Hsien-Kuei Hwang, S. Janson and T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint 2016.
Hsien-Kuei Hwang, S. Janson and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Jon Maiga, Computer-generated formulas for A159619, Sequence Machine.
Vladimir Shevelev, Several results on sequences which are similar to the positive integers, arXiv:0904.2101 [math.NT], 2009.

Cf. A000069, A001969, A159615, A007814, A004760, A159559, A159560.

read("transforms") ; isA000069 := proc(n) option remember ; RETURN( type(wt(n), 'odd') ) ; end:
A159619 := proc(n) option remember; if n = 1 then 4; else for a from procname(n-1)+1 do if isA000069(a) = isA000069(n) then RETURN(a) ; fi; od: fi; end:
seq(A159619(n), n=1..120) ; # R. J. Mathar, Mar 25 2010

a[n_] := 2 * n + If[EvenQ[n] || EvenQ[IntegerExponent[n+1, 2]], 3, 2]; Array[a, 100] (* Amiram Eldar, Aug 30 2024 *)

a(n) = 2 * n + if(!(n % 2) || !(valuation(n+1, 2) % 2), 3, 2); \\ Amiram Eldar, Aug 30 2024

a(n) = 2n+3 if n*A007814(n+1) is even, and a(n) = 2n+2 otherwise.

Edited and extended by R. J. Mathar, Mar 25 2010

A004759 Binary expansion starts 111.

Original entry on oeis.org

7, 14, 15, 28, 29, 30, 31, 56, 57, 58, 59, 60, 61, 62, 63, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244

Offset: 1

N. J. A. Sloane

This is the minimal recursive sequence such that a(1)=7, A007814(a(n))= A007814(n) and A010060(a(n))=A010060(n). - Vladimir Shevelev, Apr 23 2009

			30 in binary is 11110, so 30 is in sequence.

Reinhard Zumkeller, Table of n, a(n) for n = 1..4095

Cf. A004754 (10), A004755 (11), A004756 (100), A004757 (101), A004758 (110).
Cf. A004760, A053644, A062050, A076877.
Cf. A007814, A010060, A103204, A159559, A159560, A159615, A159619, A159629, A159698. -Vladimir Shevelev, Apr 23 2009

import Data.List (transpose)
a004759 n = a004759_list !! (n-1)
a004759_list = 7 : concat (transpose [zs, map (+ 1) zs])
                   where zs = map (* 2) a004759_list
-- Reinhard Zumkeller, Dec 03 2015

w = {1, 1, 1}; Select[Range[5, 244], If[# < 2^(Length@ w - 1), True, Take[IntegerDigits[#, 2], Length@ w] == w] &] (* Michael De Vlieger, Aug 10 2016 *)
Sort[FromDigits[#,2]&/@(Flatten[Table[Join[{1,1,1},#]&/@Tuples[{1,0},n],{n,0,5}],1])] (* Harvey P. Dale, Sep 01 2016 *)

PARI
```
a(n)=n+6*2^floor(log(n)/log(2))
```

def A004759(n): return n+(3<Chai Wah Wu, Jul 13 2022

a(2n) = 2a(n), a(2n+1) = 2a(n) + 1 + 6[n==0].
a(n) = n + 6 * 2^floor(log_2(n)) = A004758(n) + A053644(n).
a(n+1) = min{m > a(n): A007814(m) = A007814(n+1) and A010060(m) = A010060(n+1)}. a(2^k) - a(2^k-1) = A103204(k+2), k >= 1. - Vladimir Shevelev, Apr 23 2009
a(2^m+k) = 7*2^m + k, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Aug 08 2016

Edited by Ralf Stephan, Oct 12 2003

A159698 Minimal increasing sequence beginning with 4 such that n and a(n) are either both prime or both nonprime.

Original entry on oeis.org

4, 5, 7, 8, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 29, 30, 32, 33, 37, 38, 39, 40, 42, 44, 47, 48, 53, 54, 55, 56, 57, 58, 59, 60, 62, 63, 67, 68, 71, 72, 74, 75, 79, 80, 81, 82, 84, 85, 89, 90, 91, 92, 93, 94, 97, 98, 101, 102, 104, 105, 106, 108, 109, 110, 111, 112

Offset: 1

Vladimir Shevelev, Apr 20 2009, May 04 2009

For n >= 11, a(n) = A159559(n), which means the two sequences merge.
We may define other sequences a(p-1,n), p prime, which start a(p-1,1)=p-1 and with the same property of n and a(p-1,n) being jointly prime or nonprime.
We find that for p=7, 11 and 13, the sequences a(6,n), a(10,n) and a(12,n) also merge with the current sequence for sufficiently large n. Does this also hold for primes >=17?
It was verified for primes p with 7<=p<=223 that this sequence a(4,n) and a(p-1,n) eventually merge.  The corresponding values of n are 47, 683, 1117, 6257, 390703. - Alois P. Heinz, Mar 09 2011

Alois P. Heinz, Table of n, a(n) for n = 1..20000
V. Shevelev, Several results on sequences which are similar to the positive integers, arXiv:0904.2101 [math.NT], 2009.

Cf. A159559, A159560, A159615, A159619, A159629, A229019, A229132.

a:= proc(n) option remember; local m;
      if n=1 then 4
    else for m from a(n-1)+1 while isprime(m) xor isprime(n)
         do od; m
      fi
    end:
seq(a(n), n=1..80);  # Alois P. Heinz, Nov 21 2010

a[n_] := a[n] = If[n==1, 4, For[m = a[n-1]+1, Xor[PrimeQ[m], PrimeQ[n]], m++]; m]; Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Jan 31 2016, after Alois P. Heinz *)

a(1) = 4; for n>1, a(n) = min { m > a(n-1) : m is prime iff n is prime }.

More terms from Alois P. Heinz, Nov 21 2010

A159629 Slowest increasing sequence beginning with a(1)=4 such that A002828(a(n)) = A002828(n).

Original entry on oeis.org

4, 5, 6, 9, 10, 11, 15, 17, 25, 26, 27, 30, 32, 33, 39, 49, 50, 52, 54, 58, 59, 62, 63, 66, 81, 82, 83, 87, 89, 91, 92, 97, 99, 101, 102, 121, 122, 123, 124, 125, 128, 129, 131, 132, 136, 138, 143, 147, 169, 170, 171, 173, 178, 179, 183, 184, 186, 193, 195, 199, 200, 201, 207

Offset: 1

Vladimir Shevelev, Apr 17 2009, May 04 2009

nonn

Conjecture: For every m>2 there exists a minimum index N(m) such that the minimal increasing recursive sequence S_m(n) beginning with m^2 with the condition A002828(S_m(n)) = A002828(n) coincides with a(n) for all n>N.

V. Shevelev, Several results on sequences which are similar to the positive integers arXiv:0904.2101 [math.NT], 2009.

Cf. A002828, A159619, A159615, A007814, A004760, A159559, A159560.

a2828[n_] := Which[SquaresR[1, n]>0, 1, SquaresR[2, n]>0, 2, SquaresR[3, n] > 0, 3, True, 4];
a[1] = 4; a[n_] := a[n] = For[k = a[n-1]+1, True, k++, If[a2828[k] == a2828[n], Return[k]]];
Array[a, 63] (* Jean-François Alcover, Jul 28 2018 *)

a(n+1) = min { l > a(n) : A002828(l) = A002828(n+1) }.

137 replaced by 136, extended by R. J. Mathar, Sep 17 2009

A172980 a(1)=1, a(2)=3; for n>=3, a(n) is the smallest number larger than a(n-1) such that, for every k

Original entry on oeis.org

1, 3, 4, 9, 11, 12, 13, 15, 16, 33, 37, 42, 43, 117, 154, 159, 163, 168, 173, 231, 338, 555, 557, 558, 649, 1161, 1168, 1209, 1213, 1254, 1259, 1263, 1406, 1467, 1573, 1578, 1579, 2595, 2752, 2805, 2813, 2964, 2969, 2997, 3014, 5013, 5021, 5022, 5057, 5115

Offset: 1

Vladimir Shevelev, Nov 21 2010

nonn

Using the Chinese remainder theorem, it is easy to prove that the sequence is infinite.

Alois P. Heinz, Table of n, a(n) for n = 1..500

Cf. A151976, A159559, A159560, A159615, A159619, A159629, A159698, A160217.

a:= proc(n) option remember;
     local ok, m, k;
     if n<3 then 2*n-1
   else for m from a(n-1)+1 do
          ok:= true;
          for k from 1 to n-1 do
            if igcd(n, k)=1 xor igcd(m, a(k))=1
               then ok:= false; break fi
          od;
          if ok then break fi
        od; m
     fi
    end:
seq (a(n), n=1..50);  # Alois P. Heinz, Nov 21 2010

a[1]=1; a[2]=3; a[n_] := a[n] = For[k = a[n-1]+1, True, k++, If[AllTrue[ Range[n-1], CoprimeQ[k, a[#]] == CoprimeQ[n, #]&], Return[k]]]; Table[ a[n], {n, 1, 100}] (* Jean-François Alcover, Jan 25 2017 *)

More terms from Alois P. Heinz, Nov 21 2010

A172999 a(1)=1, a(2)=4; for n>=3, a(n) is the smallest number larger than a(n-1) such that, for every k

Original entry on oeis.org

1, 4, 5, 6, 7, 10, 11, 12, 25, 28, 29, 30, 31, 44, 175, 178, 179, 180, 181, 182, 275, 348, 349, 360, 371, 372, 395, 396, 397, 420, 421, 422, 725, 1074, 1309, 1310, 1319, 1448, 2945, 2954, 2957, 2970, 2971, 3016, 3325, 4188, 4189, 4190, 4213, 4214, 4475, 4526

Offset: 1

Vladimir Shevelev, Nov 21 2010

nonn

Alois P. Heinz, Table of n, a(n) for n = 1..300

Cf. A172980, A151976, A159559, A159560, A159615, A159619, A159629, A159698, A160217.

a:= proc(n) option remember;
     local ok, m, k;
     if n<3 then 3*n-2
   else for m from a(n-1)+1 do
          ok:= true;
          for k from 1 to n-1 do
            if igcd(n, k)=1 xor igcd(m, a(k))=1
               then ok:= false; break fi
          od;
          if ok then break fi
        od; m
     fi
    end:
seq (a(n), n=1..50);  # Alois P. Heinz, Nov 21 2010

t={1,4}; Do[nxt=t[[-1]]+1; While[CoprimeQ[n,Range[n-1]] != CoprimeQ[nxt,t], nxt++]; AppendTo[t,nxt], {n,3,50}]; t

More terms from Alois P. Heinz, Nov 21 2010

A151976 Minimal recursive sequence beginning with 5 similar to N with respect to property of integer to be or not to be in A079523.

Original entry on oeis.org

5, 6, 8, 10, 13, 14, 17, 18, 21, 22, 24, 26, 29, 30, 32, 34, 37, 38, 40, 42, 45, 46, 49, 50, 53, 54, 56, 58, 61

Offset: 1

Vladimir Shevelev, Jul 12 2009

nonn

			a(2)=6 since 6>5 is the minimal integer such that 2 and 6 simultaneously are not in A079523.

V. Shevelev, Several results on sequences which are similar to the positive integers

Cf. A079523 A035263 A010060 A159559 A159560 A004760.

For n>=1, a(n+1)=min{m>a(n): A035263(m)=A035263(n+1)}

A151994 For k=A079523(n),n>=2, let {S_k} be the minimal recursive sequence beginning with k similar to N with respect to property of integer to be or not to be in A079523. Then a(n) is the point of confluence of {S_k} with {S_5}.

Original entry on oeis.org

5, 13, 13, 29, 29, 61, 61, 61, 61

Offset: 2

Vladimir Shevelev, Jul 12 2009

nonn

			Note that, {S_5} is {5,6,8,10,13,...}(see A162736) and {S_7} is {7,8,10,11,13,...}, then a(3)=13.

Cf. A162736 A079523 A159615 A159619 A159559 A159560 A004760 A035263.

cp's OEIS Frontend

A004760 List of numbers whose binary expansion does not begin 10.

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A159615 The slowest increasing sequence beginning with a(1)=2 such that a(n) and n are both odious or both not odious.

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A159619 Slowest increasing sequence beginning with 4 such that n and a(n) are either both evil or both odious.

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A004759 Binary expansion starts 111.

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A159698 Minimal increasing sequence beginning with 4 such that n and a(n) are either both prime or both nonprime.

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A159629 Slowest increasing sequence beginning with a(1)=4 such that A002828(a(n)) = A002828(n).

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