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.

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A355176 a(n) is the smallest index k such that prime(n) divides both A090252(k) and A090252(2*k+1).

Original entry on oeis.org

2, 3, 14, 32, 60, 96, 120, 128, 132, 244, 264, 388, 480, 484, 488, 2064, 1056, 571, 776, 960, 968, 976, 980, 2112, 2128, 1143, 1536, 1552, 1556, 1920, 3872, 1937, 3904, 3920, 1961, 4128, 4256, 3104, 6224, 3113, 3844, 3848, 7808, 7824, 7840, 8256, 8448, 8452
Offset: 1

Views

Author

Thomas Scheuerle, Jun 22 2022

Keywords

Comments

For n > 2, a(n) is not the smallest k such that prime(n) divides A090252(k), but it is the smallest k such that prime(n) divides both A090252(k) and A090252(2*k+1). If k_(0) = a(n) we may find either an infinite or finite range of indices where prime(n) divides A090252 using the recurrence k_(n) = 2*k_(n-1)+1, but there is a caveat: in very rare cases, some k values of this recurrence may be wrong by +-1, and the next iteration will then fit again. This uncertainty is caused by the fact that two terms of A090252 will be governed by the same floor(n/2) history. For yet unknown reasons, there may be an upper limit where such a recurrence may break.
This works because in A090252 the number of primes which do not divide the last floor(n/2) terms is growing faster than they are used up by this sequence. For each prime p then there exists an index k into A090252 where the supply of unused factors is so large that, when p becomes coprime to the last floor(n/2) terms, we can always immediately find a matching second prime to build a yet-unused semiprime or use p as a yet-unused power of itself.

Examples

			prime(1) = 2 divides A090252(2) = 2, A090252(5) = 4, A090252(11) = 8, A090252(23) = 16, A090252(47) = 26, ... .
2*2+1 = 5; 2*5+1 = 11; 2*11+1 = 23; 2*23+1 = 47.
		

Crossrefs

Formula

A090252(a(n)) mod A000040(n) = 0 and a(n) is either even or A090252((a(n)-1)/2) mod A000040(n) > 0 is valid too.
A090252(2*a(n)+1) mod A000040(n) = 0.
A090252(f^m(a(n))) mod A000040(n) = 0, with f(x) = 2*x+1. The range of m is yet unknown.

Extensions

a(41) and beyond (using Russ Cox's gzipped b-file at A090252) from Michael S. Branicky, Jun 23 2022

A355714 Numbers k > 0 such that A090252(A355176(k)) does not equal prime(k)^2.

Original entry on oeis.org

1, 2, 16, 26, 32, 35, 40, 59, 60, 69, 92, 105, 110, 112, 113, 137, 167, 169, 178, 185, 186, 188, 207, 210, 260, 261, 274, 287, 289, 342, 344, 346, 356, 357, 359, 361, 362, 363, 391, 412, 417, 434, 457, 477, 478, 479, 480, 481, 492, 547, 563, 598, 663, 666, 671
Offset: 1

Views

Author

Thomas Scheuerle, Jul 15 2022

Keywords

Comments

In contrast to A354169 (a set theoretic analog of A090252) where we observe many as yet unproved regularities, in A090252 the situation appears to be more complicated. This sequence is intended to help spot these irregularities and perhaps lead to further rules, which probably have no analogy in A354169.
In most cases A090252(A355176(n)) equals prime(a(n))*p where p is prime. Surprisingly p is in many cases greater than prime(a(n)).
Is this sequence infinite?

Crossrefs

Formula

A090252(A355176(a(n))) <> A000040(a(n))^2.
A000040(a(n)) divides A090252(A355176(a(n))).
A000040(a(n)) divides A090252(2*A355176(a(n))+1).

Extensions

More terms from Jinyuan Wang, Jul 15 2022

A084937 Smallest number which is coprime to the last two predecessors and has not yet appeared; a(1)=1, a(2)=2.

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 9, 8, 11, 13, 6, 17, 19, 10, 21, 23, 16, 15, 29, 14, 25, 27, 22, 31, 35, 12, 37, 41, 18, 43, 47, 20, 33, 49, 26, 45, 53, 28, 39, 55, 32, 51, 59, 38, 61, 63, 34, 65, 57, 44, 67, 69, 40, 71, 73, 24, 77, 79, 30, 83, 89, 36, 85, 91, 46, 75, 97, 52, 81
Offset: 1

Views

Author

Reinhard Zumkeller, Jun 13 2003

Keywords

Comments

Equivalently, this is the lexicographically earliest sequence of positive numbers satisfying the condition that each term is relatively prime to the next two terms. - N. J. A. Sloane, Nov 03 2014
Empirically, the points lie roughly on two lines: if n == 2 mod 3 then a(n) ~= 2n/3, otherwise a(n) ~= 4n/3. See A249680-A249683 for the three trisections, and see also the Sigrist scatterplot. - N. J. A. Sloane, Nov 03 2014, Nov 04 2014
All primes and prime powers occur, and the primes occur in their natural order. For any prime p, p occurs before p^2 before p^3, ...
Empirically, this is a permutation of the natural numbers, with inverse A084933: a(A084933(n))=A084933(a(n))=n. It seems that there are no further fixed points after {1,2,3,8,33,39}. Empirically, a(n) mod 2 = A011655(n+1); ABS(a(n)-n) < n; a(3*n+1)>n; a(3*n+2)Reinhard Zumkeller, Dec 16 2007
For a(n) mod 3 see A249603. - N. J. A. Sloane, Nov 03 2014
A249694(n) = GCD(a(n),a(n+3)). - Reinhard Zumkeller, Nov 04 2014

Crossrefs

Cf. A084933 (inverse), A103683, A121216, A247665, A090252, A249603 (read mod 3), A249680, A249681, A249682, A249683 (trisections), A249694, A011655, A249684 (numbers that take a record number of steps to appear), A249685.
Indices of primes: A249602, and of prime powers: A249575.
Running counts of missing numbers: A249686, A250099, A250100; A249777, A249856, A249857.
Where a(3n)>a(3n+1): A249689.
See also A353706, A353709, A353710.

Programs

  • Haskell
    import Data.List (delete)
    a084937 n = a084937_list !! (n-1)
    a084937_list = 1 : 2 : f 2 1 [3..] where
       f x y zs = g zs where
          g (u:us) | gcd y u > 1 || gcd x u > 1 = g us
                   | otherwise = u : f u x (delete u zs)
    -- Reinhard Zumkeller, Jan 28 2012
    
  • Maple
    N:= 1000: # to get a(n) until the first entry > N
    a[1]:= 1: a[2]:= 2:
    R:= {$3..N}:
    for n from 3 while R <> {} do
      success:= false;
      for r in R do
        if igcd(r,a[n-1]) = 1 and igcd(r,a[n-2])=1 then
           a[n]:= r;
           R:= R minus {r};
           success:= true;
           break
        fi
      od:
      if not success then break fi;
    od:
    seq(a[i], i = 1 .. n-1); # Robert Israel, Dec 12 2014
  • Mathematica
    lst={1,2,3}; unused=Range[4,100]; While[n=Select[unused, CoprimeQ[#, lst[[-1]]] && CoprimeQ[#, lst[[-2]]] &, 1]; n != {}, AppendTo[lst, n[[1]]]; unused=DeleteCases[unused, n[[1]]]]; lst
    f[s_] := Block[{k = 1, l = Take[s, -2]}, While[ Union[ GCD[k, l]] != {1} || MemberQ[s, k], k++]; Append[s, k]]; Nest[f, {1, 2}, 67] (* Robert G. Wilson v, Jun 26 2011 *)
  • PARI
    taken(k,t=v[k])=for(i=3,k-1, if(v[i]==t, return(1))); 0
    step(k,g)=while(gcd(k,g)>1, k++); k
    first(n)=local(v=vector(n,i,i)); my(nxt=3,t); for(k=3,n, v[k]=step(nxt, t=v[k-1]*v[k-2]); while(taken(k), v[k]=step(v[k]+1,t)); if(v[k]==t, while(taken(k+1,t++),))); v \\ Charles R Greathouse IV, Aug 26 2016
  • Python
    from math import gcd
    A084937_list, l1, l2, s, b = [1,2], 2, 1, 3, set()
    for _ in range(10**3):
        i = s
        while True:
            if not i in b and gcd(i,l1) == 1 and gcd(i,l2) == 1:
                A084937_list.append(i)
                l2, l1 = l1, i
                b.add(i)
                while s in b:
                    b.remove(s)
                    s += 1
                break
            i += 1 # Chai Wah Wu, Dec 09 2014
    

Extensions

Entry revised by N. J. A. Sloane, Nov 04 2014

A247665 a(1)=2; thereafter a(n) is the smallest number >= 2 not yet used which is compatible with the condition that a(n) is relatively prime to the next n terms.

Original entry on oeis.org

2, 3, 4, 5, 7, 9, 8, 11, 13, 17, 19, 23, 15, 29, 14, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 25, 27, 79, 83, 16, 49, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 149, 151, 157, 163, 167, 169, 173, 179, 181, 191, 85, 193, 57, 197, 199, 211, 223
Offset: 1

Views

Author

N. J. A. Sloane, Oct 06 2014 and Oct 08 2014

Keywords

Comments

It appears that a(k) is even iff k = 2^i-1 (cf. A248379). It also appears that all powers of 2 occur in the sequence. (Amarnath Murthy)
The indices of even terms and their values are [1, 2], [3, 4], [7, 8], [15, 14], [31, 16], [63, 32], [127, 64], [255, 128], [511, 122], ...
Will the numbers 6, 10, 21, 22, ... ever occur? 12, 18, 20, ... are also missing, but if 6 never appears then neither will 12, etc.
A related question: are all terms deficient? - Peter Munn, Jul 20 2017
It appears that the missing numbers are 6, 10, 12, 18, 20, 21, 22, 24, 26, 28, 30, 33, 34, 35, 36, 38, 39, 40, 42, ..., but since there is no proof that any one of these is really missing, this sequence cannot yet be added to the OEIS. - N. J. A. Sloane, May 18 2022

Examples

			a(1) = 2 must be rel. prime to a(2), so a(2)=3.
a(2) = 3 must be rel. prime to a(3) and a(4), so we can take them to be 4 and 5.
a(3) = 4 must be rel. prime to a(5), a(6), so we must take them to be 7,9.
a(4) = 5 must be rel. prime to a(7), a(8), so we must take them to be 8,11.
At each step after the first, we must choose two new numbers, and we must make sure that not only are they rel. prime to a(n), they are also rel. prime to all a(i), i>n, that have been already chosen.
		

References

Crossrefs

Indices of primes and prime powers: A248387, A248918.
Lengths of runs of primes: A249033.
A090252 = similar to A247665 but start with a(1)=1. A249559 starts with a(1)=3.
A249064 is a different generalization.
A064413 is another similar sequence.

Programs

  • Haskell
    a247665 n = a247665_list !! (n-1)
    a247665_list = 2 : 3 : f [3] [4..] where
       f (x:xs) zs = ys ++ f (xs ++ ys) (zs \\ ys) where
         ys = [v, head [w | w <- vs, gcd v w == 1]]
         (v:vs) = filter (\u -> gcd u x == 1 && all ((== 1) . (gcd u)) xs) zs
    -- Reinhard Zumkeller, Oct 09 2014
    
  • PARI
    m=100; v=vector(m); u=vectorsmall(100*m); for(n=1, m, for(i=2, 10^9, if(!u[i], for(j=(n+1)\2, n-1, if(gcd(v[j], i)>1, next(2))); v[n]=i; u[i]=1; break))); v \\ Jens Kruse Andersen, Oct 08 2014
    
  • Python
    from itertools import count, islice
    from math import gcd
    from collections import deque
    def A247665_gen(): # generator of terms
        aset, aqueue, c, f = {2}, deque([2]), 3, True
        yield 2
        while True:
            for m in count(c):
                if m not in aset and all(gcd(m,a) == 1 for a in aqueue):
                    yield m
                    aset.add(m)
                    aqueue.append(m)
                    if f: aqueue.popleft()
                    f = not f
                    while c in aset:
                        c += 1
                    break
    A247665_list = list(islice(A247665_gen(),50)) # Chai Wah Wu, May 19 2022
  • Sage
    # s is the starting point (2 in A247665).
    def gen(s):
        sequence = [s]
        available = list(range(2,2*s))
        available.pop(available.index(s))
        yield s
        while True:
            available.extend(range(available[-1]+1,next_prime(available[-1])+1))
            for i,e in enumerate(available):
                if all(gcd(e, sequence[j])==1 for j in range(-len(sequence)//2,0)):
                    available.pop(i)
                    sequence.append(e)
                    yield(e)
                    break
    g = gen(2)
    [next(g) for i in range(40)]  # (gets first 40 terms of A247665)
    # Nadia Heninger, Oct 28 2014
    

Extensions

More terms from Jens Kruse Andersen, Oct 06 2014
Further terms from Russ Cox, Oct 08 2014
Added condition a(n) >= 2 to definition. - N. J. A. Sloane, May 16 2022

A354169 a(0) = 0, a(1) = 1, a(2) = 2; for k >= 2, given a(k), the sequence is extended by adjoining two terms: a(2*k-1) = smallest m >= 0 not among a(0) .. a(k) such that {m, a(k), a(k+1), ..., a(2*k-2)} are pairwise disjoint in binary, and a(2*k) = smallest m >= 0 not among a(0) .. a(k) such that {m, a(k), ..., a(2*k-1)} are pairwise disjoint in binary.

Original entry on oeis.org

0, 1, 2, 4, 8, 3, 16, 32, 64, 12, 128, 256, 512, 17, 1024, 34, 2048, 4096, 8192, 68, 16384, 136, 32768, 65536, 131072, 768, 262144, 524288, 1048576, 1025, 2097152, 18, 4194304, 2080, 8388608, 16777216, 33554432, 12288, 67108864, 134217728, 268435456, 16388
Offset: 0

Views

Author

N. J. A. Sloane, Jun 05 2022

Keywords

Comments

The paper by De Vlieger et al. (2022) calls this the "binary two-up sequence".
"Pairwise disjoint in binary" means no common 1-bits in their binary representations.
This is a set-theory analog of A090252. It bears the same relation to A090252 as A252867 does to A098550, A353708 to A121216, A353712 to A347113, etc.
A consequence of the definition, and also an equivalent definition, is that this is the lexicographically earliest infinite sequence of distinct nonnegative numbers with the property that the binary representation of a(n) is disjoint from (has no common 1's with) the binary representations of the following n terms.
An equivalent definition is that a(n) is the smallest nonnegative number that is disjoint (in its binary representation) from each of the previous floor(n/2) terms.
For the subsequence 0, 3, 12, 17, 34, ... of the terms that are not powers of 2 see A354680 and A354798.
All terms are the sum of at most two powers of 2 (see De Vlieger et al., 2022). - N. J. A. Sloane, Aug 29 2022

Examples

			After a(2) = 2 = 10_2, a(3) must equal ?0?_2, and the smallest such number we have not seen is a(3) = 100_2 = 4, and a(4) must equal ?00?_2, and the smallest such number we have not seen is a(4) = 1000_2 = 8.
		

Crossrefs

A355889 is a more efficient way to present this sequence.

Programs

Extensions

More terms from Rémy Sigrist, Jun 06 2022

A354790 a(n) is the least positive squarefree number not already used that is coprime to the previous floor(n/2) terms.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 6, 13, 17, 19, 23, 29, 31, 35, 22, 37, 39, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 85, 89, 14, 97, 101, 103, 33, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 65, 233
Offset: 1

Views

Author

Keywords

Comments

A version of the Two-Up sequence A090252 that is restricted to squarefree numbers.

Crossrefs

See A354791 and A354792 for the nonprime terms.
See A355895 for the even terms.

Programs

  • C
    // See Links section.
  • Maple
    # A354790 = Squarefree version of the Two-Up sequence A090252
    # This produces 2*M terms in the array a
    # Assumes b117 is a list of sufficiently many squarefree numbers A005117
    # Test if u is relatively prime to all of a[i], i = i1..i2
    perpq:=proc(u,i1,i2) local i; global a;
    for i from i1 to i2 do if igcd(u,a[i])>1 then return(-1); fi; od: 1; end;
    a:=Array(1..10000,-1);
    hit:=Array(1..10000,-1); # 1 if i has appeared
    a[1]:=1; a[2]:=2; hit[1]:=1; hit[2]:=1;
    M:=100; M1 := 1000;
    for p from 2 to M do
    # step 1 want a[2p-1] relatively prime to a[p] ... a[2p-2]
    sw1:=-1;
    for j from 1 to M1 do
    c:=b117[j];
    if hit[c] = -1 and perpq(c,p,2*p-2) = 1 then a[2*p-1]:=c; hit[c]:=1; sw1:=1; break; fi;
    od: # od j
    if sw1 = -1 then error("no luck, step 1, p =",p ); fi;
    # step 2 want a[2p] relatively prime to a[p] ... a[2p-1]
    sw2:=-1;
    for j from 1 to M1 do
    c:=b117[j];
    if hit[c] = -1 and perpq(c,p,2*p-1) = 1 then a[2*p]:=c; hit[c]:=1; sw2:=1; break; fi;
    od: # od j
    if sw2 = -1 then error("no luck, step 2, p =",p ); fi;
    od: # od p
    [seq(a[i],i=1..2*M)];
  • Mathematica
    nn = 60; pp[] = 1; k = r = 1; c[] = False; a[1] = 1; Do[Set[m, SelectFirst[Union@ Append[Times @@ # & /@ Subsets[#, {2, Infinity}], Prime[r]] &[Prime@ Select[Range[If[k == 1, r, k + 1]], p[Prime[#]] < n &]], ! c[#] &]]; Set[a[n], m]; (c[m] = True; If[PrimeQ[m], r++]; If[n > 1, Map[(Set[p[#], 2 n]; pp[#]++) &, #]]) &@ FactorInteger[m][[All, 1]]; While[pp[Prime[k]] > 2, k++], {n, 2, nn}]; Array[a, nn] (* Michael De Vlieger, Sep 06 2022 *)
  • PARI
    \\ See Links section.
    
  • Python
    from math import lcm, gcd
    from itertools import count, islice
    from collections import deque
    from sympy import factorint
    def A354790_gen(): # generator of terms
        aset, aqueue, c, b, f = {1}, deque([1]), 2, 1, True
        yield 1
        while True:
            for m in count(c):
                if m not in aset and gcd(m,b) == 1 and all(map(lambda n:n<=1,factorint(m).values())):
                    yield m
                    aset.add(m)
                    aqueue.append(m)
                    if f: aqueue.popleft()
                    b = lcm(*aqueue)
                    f = not f
                    while c in aset:
                        c += 1
                    break
    A354790_list = list(islice(A354790_gen(),30)) # Chai Wah Wu, Jul 17 2022
    

Extensions

More terms from Rémy Sigrist, Jul 17 2022

A355150 The Hamming weight of A354169, a(n) = A000120(A354169(n)).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1
Offset: 0

Views

Author

Thomas Scheuerle, Jun 21 2022

Keywords

Comments

All the following conjectures are now known to be true. See De Vlieger et al. (2022). - N. J. A. Sloane, Aug 29 2022
Conjecture: It appears that this sequence may be computed by a fast algorithm:
We begin with an initial sequence 0,1,1,1,1,2. Let n be the index of the last element added. Then extend by the rules:
If a(n) = 2, a((n-3)/2) = 1, and a((n-1)/2) = 2 extend this sequence by 1,2.
If a(n) = 2, a((n-3)/2) = 2, a((n-1)/2) = 1, and a(n-2) = 1, extend this sequence by 1,2.
In all other cases extend this sequence by 1,1,1,2.
This conjecture was verified for n = 0..2^16 against the b-file provided by Michael De Vlieger. - Thomas Scheuerle, Jul 14 2022
[Typos corrected by N. J. A. Sloane, Jul 10 2022 at the suggestion of Michel Dekking.]
From Michel Dekking, Jul 12 2022: (Start)
Conjecture: It appears that this sequence is almost a periodic sequence, with period 6. Let x be the sequence defined below.
If n > 25, n == 2 (mod 6) is not an element of x then (written as words)
a(n)a(n+1)...a(n+5) = 111212.
If n > 25, n == 2 (mod 6) is an element of x then
a(n)a(n+1)...a(n+5) = 121112.
The sequence x = {32, 44, 68, 92, 140, 188, 284, ...} is a sparse sequence defined via the sequence A007283, given by A007283(n)=3*2^n, which has also been encountered in A354169. In fact, x(1) = 32, and
x(2n+2) - x(2n+1) = 3*2^(n+2) for n=0,1,2,....
x(2n+1) - x(2n) = 3*2^(n+2) for n=1,2,.... (End)
From Michel Dekking, Jul 23 2022: (Start)
Extending the sequence x to the right with the four numbers 5,8,14,21 we obtain sequence A354789.
So the sparse positions are given by 9*2^k - 4 for k even, and by 12*2^k - 4 for k odd, for k = 2,3,... (End)

Crossrefs

Programs

  • MATLAB
    function a = A355150( max_n ) % Note: a(0) is omitted here because
                                  % a(1) will be a(1) in the sequence.
        a = [1 1 1 1 2];
        m = length(a);
        while length(a) < max_n
            if (((a((m-3)/2) == 2)&&(a((m-1)/2) == 1)&&(a(m-2) == 1)) ...
                ||((a((m-3)/2) == 1)&&(a((m-1)/2) == 2)))
                a(m+1:m+2) = [1 2];
                m = m+2;
            else
                a(m+1:m+4) = [1 1 1 2];
                m = m+4;
            end
        end
    end

Formula

a(A354767(n)) = 1.
a(A354798(n+1)) != 2.

Extensions

Edited by N. J. A. Sloane, Jul 10 2022

A249064 Lexically first sequence of distinct positive integers such that a(n) is coprime to the next a(n) elements.

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 9, 11, 13, 8, 17, 19, 23, 25, 29, 31, 21, 37, 16, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 22, 109, 113, 27, 35, 127, 131, 137, 139, 149, 151, 157, 163, 167, 169, 173, 179, 181, 191, 193, 197, 199, 211, 32, 121, 223, 227, 229, 233, 239, 241, 51, 251, 257, 263, 269, 271, 277, 281, 283, 49, 95
Offset: 1

Views

Author

Hugo van der Sanden, Oct 20 2014

Keywords

Comments

Described in this form, A090252 would be "lexically first sequence of positive integers such that a(n) is coprime to the next n elements".
(And A247665 would be "lexically first sequence of integers >= 2 such that a(n) is coprime to the next n elements". - N. J. A. Sloane, Nov 01 2014)
All values up to a(1000000) are either prime powers or semiprimes, except when n is in (868, 947, 993, 1069, 1205, 1431, 854300) with values respectively (172, 45, 75, 135, 225, 375, 9475). This suggests the sequence is unlikely to be a permutation of the integers.
If, mimicking A247665, one starts with a(1)=2 and uses the same rule (always using distinct numbers >= 2) one obtains A249064 again, without the leading 1. - N. J. A. Sloane, Nov 01 2014

Crossrefs

Extensions

Added "distinct" to the definition. - Hugo van der Sanden, Oct 28 2014

A354146 Even numbers in A353730 in order of appearance.

Original entry on oeis.org

2, 4, 8, 16, 26, 32, 64, 128, 206, 256, 454, 446, 512, 1024, 2048, 3142
Offset: 1

Views

Author

N. J. A. Sloane, May 21 2022

Keywords

Comments

A090252(1535) = 256 and A090252(3071) = 478 are also even terms in A090252; the latter breaks the correspondence with this sequence. - Michael S. Branicky, May 21 2022

Crossrefs

Extensions

Deleted an incorrect comment. - N. J. A. Sloane, May 25 2022

A354780 a(n) is the bitwise OR of (the binary expansions of) b(n+1) to b(2*n), where b is A354169.

Original entry on oeis.org

2, 12, 27, 115, 252, 1004, 2013, 4031, 16307, 32631, 65279, 261375, 524270, 2096110, 4194253, 8386527, 16773119, 67096575, 134217659, 536854459, 1073741623, 2147450751, 4294901759, 17179672575, 34359737599, 137438690559, 274877382143, 549754765311, 2199022205950, 4398044412927, 8796093022189, 35184367894509, 70368744175567
Offset: 1

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Author

N. J. A. Sloane, Jul 05 2022

Keywords

Comments

If the binary expansion of a(n) has a 1 in the 2^i's bit (for any i >= 0) then A354169(2*n+1) must have a 0 in that bit.
A354169(2*n+1) is the smallest number not yet in A354169 which satisfies that condition (this follows at once from the definition of A354169).
This sequence bears the same relation to A354169 as A355057 does to A090252.

Examples

			Consider n=6. Then b(7) to b(12) are 32, 64, 12, 128, 256, 512. The bitwise OR of those 6 numbers is 1111101100_2 = 1004_10 = a(6). The bitwise complement of 1004_10 is 10011_2 = 19_10 = A354781(6), and A354169(6) = 17_10 = 10001_2.
On the other hand, for n=5, b(6) to b(10) are 16, 32, 64, 12, 128, whose bitwise OR is 11111100_2 = 252_10 = a(5). The bitwise complement of 252_10 is 3_10 = 11_2 = A354781(5). However, 3 has already appeared in A354169, and the smallest available number whose binary expansion is disjoint from 252_10 = 11111100_2 is 2^8 = 100000000_2 = 256_10 = 2^8 = A354169(5).
		

Crossrefs

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