A206853 -id:A206853 - cp's OEIS frontend

A048490 a(n) = T(7,n), array T given by A048483.

1, 9, 25, 57, 121, 249, 505, 1017, 2041, 4089, 8185, 16377, 32761, 65529, 131065, 262137, 524281, 1048569, 2097145, 4194297, 8388601, 16777209, 33554425, 67108857, 134217721, 268435449, 536870905, 1073741817, 2147483641, 4294967289, 8589934585

Offset: 0

Clark Kimberling

n-th difference of a(n), a(n-1), ..., a(0) is (8, 8, 8, ...).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2).

a=1; lst={a}; k=8; Do[a+=k; AppendTo[lst, a]; k+=k, {n, 0, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Dec 16 2008 *)

Vec((6*x+1)/((x-1)*(2*x-1)) + O(x^100)) \\ Colin Barker, Nov 26 2014

a(n) = 8 * 2^n - 7. - Ralf Stephan, Jan 09 2009
Equals binomial transform of [1, 8, 8, 8, ...]. - Gary W. Adamson, Apr 29 2008
a(n) = 2*a(n-1) + 7 for n > 0, a(0)=1. - Vincenzo Librandi, Aug 06 2010
For n>=1, a(n) = 6<+>(n+3), where the operation <+> is defined in A206853. - Vladimir Shevelev, Feb 17 2012
From Colin Barker, Nov 26 2014: (Start)
a(n) = 3*a(n-1) - 2*a(n-2).
G.f.: (6*x+1) / ((x-1)*(2*x-1)). (End)

More terms from Colin Barker, Nov 26 2014

A207063 a(n) is the smallest number larger than a(n-1) with mutual Hamming distance 2 and a(1)=0.

Original entry on oeis.org

0, 3, 5, 6, 10, 12, 15, 23, 27, 29, 30, 46, 54, 58, 60, 63, 95, 111, 119, 123, 125, 126, 190, 222, 238, 246, 250, 252, 255, 383, 447, 479, 495, 503, 507, 509, 510, 766, 894, 958, 990, 1006, 1014, 1018, 1020, 1023, 1535, 1791, 1919, 1983, 2015, 2031, 2039, 2043

Offset: 1

Alois P. Heinz, Feb 14 2012

The binary expansion of a(n) has an even number of 1's. So this is a subsequence of A001969. The odd analog is A206853.

This sequence has 4*k+1 = A016813(k) numbers with exactly 2*k 1's and no number with more than two 0's in their binary expansion.

			| n | a(n) | A007088(a(n))| A000120(a(n))|
+---+------+--------------+--------------+
| 1 |   0  |         0    |       0      |
| 2 |   3  |        11    |       2      |
| 3 |   5  |       101    |       2      |
| 4 |   6  |       110    |       2      |
| 5 |  10  |      1010    |       2      |
| 6 |  12  |      1100    |       2      |
| 7 |  15  |      1111    |       4      |
| 8 |  23  |     10111    |       4      |

Alois P. Heinz, Table of n, a(n) for n = 1..20000

Cf. A182187 (next with Hamming distance 2), A206853 (iterate from 1).

Cf. A000120, A001969, A007088, A016813.

g:= proc(n) option remember; local l; l:= g(n-1);
      `if`(nops(l)=1, [l[1]+1, l[1]-1], `if`(nops(l)=2,
      `if`(l[2]<>0, [l[1], l[2]-1], [l[1]+1, 0, l[1]-1]),
      `if`(l[3]<>1, [l[1], l[2], l[3]-1], [l[1]])))
    end: g(1):= [2, 0, 1]:
a:= n-> (l-> 2^l[1]-1 -add(2^l[i], i=2..nops(l)))(g(n)):
seq(a(n), n=1..300);

def aupton(terms):
    alst = [0]
    for n in range(2, terms+1):
        an = alst[-1] + 1
        while bin(an^alst[-1]).count('1') != 2:  an += 1
        alst.append(an)
    return alst
print(aupton(54)) # Michael S. Branicky, Jul 07 2021

A182209 a(n) is the least m >= n, such that the Hamming distance D(n,m) = 3.

Original entry on oeis.org

7, 6, 5, 4, 9, 8, 8, 9, 15, 14, 13, 12, 16, 17, 18, 19, 23, 22, 21, 20, 25, 24, 24, 25, 31, 30, 29, 28, 36, 37, 38, 39, 39, 38, 37, 36, 41, 40, 40, 41, 47, 46, 45, 44, 48, 49, 50, 51, 55, 54, 53, 52, 57, 56, 56, 57, 63, 62, 61, 60, 76, 77, 78, 79, 71, 70, 69

Offset: 0

Vladimir Shevelev, Apr 18 2012

a(n) = n<+>3 (see comment in A206853).

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A086799 ((n-1)<+>1), A182187 (n<+>2), A182336 (n<+>4).
Cf. A209554 (primes which are not terms nor n<+>2 terms).
Cf. A205509, A205510, A205511, A205302, A205649, A205533, A122565, A206852, A206960, A007814.

HD:= (i, j)-> add(h, h=Bits[Split](Bits[Xor](i, j))):
a:= proc(n) local c;
      for c from n do if HD(n, c)=3 then return c fi od
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Apr 18 2012

a(n) = bitxor(n, if(bitand(n,14)==4, 13, 7<>2+1,2))); \\ Kevin Ryde, Jul 10 2021

def d(n, m): return bin(n^m).count('1')
def a(n):
    m = n+1
    while d(n, m) != 3: m += 1
    return m
print([a(n) for n in range(67)]) # Michael S. Branicky, Jul 06 2021

If n==i mod 8, then a(n) = n-2*i+7, i=0,1,2,3; if n==4 mod 16, then a(n) = n+5; if n==12 mod 16, then a(n) = n+2^(A007814(n+4)-2); if n==5 mod 16, then a(n) = n+3; if n==13 mod 16, then a(n) = n+2^(A007814(n+3)-2); if n==6 mod 8, then a(n) = n+2^(A007814(n+2)-2); if n==7 mod 8, then a(n) = n+2^(A007814(n+1)-2).

Using this formula, we can prove conjecture formulated in comment in A209554 in case k=3. Moreover, one can prove that N could be represented in form n<+>2 or n<+>3 iff N is not a number of the forms 32*t, 32*t+1. - Vladimir Shevelev, Apr 25 2012

A207016 a(1) = 3 , for n > 1 a(n) is the least number greater than a(n-1) such that the Hamming distance D(a(n-1),a(n)) = 3.

Original entry on oeis.org

3, 4, 9, 14, 18, 21, 24, 31, 39, 41, 46, 50, 53, 56, 63, 79, 83, 84, 89, 94, 102, 104, 111, 115, 116, 121, 126, 158, 166, 168, 175, 179, 180, 185, 190, 206, 210, 213, 216, 223, 231, 233, 238, 242, 245, 248, 255, 319, 335, 339, 340, 345, 350, 358, 360, 367, 371

Offset: 1

Vladimir Shevelev, Feb 14 2012

Odious and evil terms are alternating (cf. A000069, A001969).

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Alois P. Heinz)

Cf. A000069, A000225, A001969, A205509, A205510, A205511, A205302, A205649, A205533, A122565, A206852, A206853.

a[1] = 3; a[n_] := a[n] = Module[{k = a[n - 1], m = a[n-1] + 1}, While[DigitCount[BitXor[k, m], 2, 1] != 3, m++]; m]; Array[a, 100] (* Amiram Eldar, Aug 06 2023 *)

list(n)=my(v=vector(n));v[1]=3;for(i=2,n,v[i]=v[i-1]; while(hammingweight(bitxor(v[i-1],v[i]++))!=3,));v \\ Charles R Greathouse IV, Mar 26 2013

A208982 Numbers n such that the next larger number with mutual Hamming distance 1 is prime.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 16, 17, 18, 19, 21, 22, 23, 27, 28, 29, 30, 36, 39, 40, 41, 42, 43, 45, 46, 52, 57, 58, 60, 63, 65, 66, 67, 69, 70, 71, 72, 75, 77, 78, 81, 82, 88, 95, 96, 99, 100, 101, 102, 105, 106, 108, 111, 112, 119, 123, 125, 126, 129, 130, 136, 137, 138, 147, 148, 149, 150

Offset: 1

Vladimir Shevelev, Mar 04 2012

If p is prime, then p-1 is in the sequence.

Using the prime number theorem in arithmetic progressions k*n+b with gcd(k,b)=1 and its uniformity over k<=exp(c*sqrt(log(x))), one can prove that the counting function of a(n)<=x is equivalent to 2*x/log(x), as x tends to infinity.

Cf. A205509, A205510, A205511, A205302, A205649, A205533, A122565, A206852, A206853, A206960, A209085.

isok(n) = my(nextn = n+1); while (hammingweight(bitxor(n, nextn)) != 1, nextn++); isprime(nextn); \\ Michel Marcus, Jul 01 2014

A182336 a(n) is the least m>=n, such that the Hamming distance D(n,m)=4.

Original entry on oeis.org

15, 14, 13, 12, 11, 10, 9, 8, 19, 18, 17, 16, 17, 16, 16, 17, 31, 30, 29, 28, 27, 26, 25, 24, 33, 32, 32, 33, 32, 33, 34, 35, 47, 46, 45, 44, 43, 42, 41, 40, 51, 50, 49, 48, 49, 48, 48, 49, 63, 62, 61, 60, 59, 58, 57, 56, 64, 65, 66, 67, 68, 69, 70, 71, 79, 78, 77, 76, 75, 74

Offset: 0

Vladimir Shevelev, Apr 25 2012

Or (see comment in A206853) a(n)=n<+>4.

Conjecture: for n > 96, n + 1 <= a(n) <= 9n/8 + 1. - Charles R Greathouse IV, Apr 25 2012

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000

Cf. A205509, A205510, A205511, A205302, A205649, A205533, A122565, A206852, A206853, A206960, A207063, A209544, A209554, A007814, A086799, A182187, A182209.

hamming(n)=my(v=binary(n));sum(i=1,#v,v[i])
a(n)=my(k=n);while(hamming(bitxor(n,k++))!=4,);k \\ Charles R Greathouse IV, Apr 25 2012

Terms corrected by Charles R Greathouse IV, Apr 25 2012

A209085 a(n) is the next larger than A208982(n) number with mutual Hamming distance 1.

Original entry on oeis.org

3, 3, 7, 5, 7, 7, 11, 11, 13, 31, 17, 19, 19, 23, 23, 23, 31, 31, 29, 31, 31, 37, 47, 41, 43, 43, 47, 47, 47, 53, 59, 59, 61, 127, 67, 67, 71, 71, 71, 79, 73, 79, 79, 79, 83, 83, 89, 127, 97, 103, 101, 103, 103, 107, 107, 109, 127, 113, 127, 127, 127, 127

Offset: 1

Vladimir Shevelev, Mar 05 2012

All terms are prime by construction of A208982, and prime p occurs k times iff 2^k||p+1. In particular, every prime of the form 4*k+1 occurs 1 time. Thus all odd primes are in the sequence.

			Since 2^4 divides 48, but 2^5 not divides, i.e., 2^4||48, then 47 occurs four times.

Cf. A206853, A208982.

A172252 a(n) = 4*2^n - 9.

Original entry on oeis.org

-1, 7, 23, 55, 119, 247, 503, 1015, 2039, 4087, 8183, 16375, 32759, 65527, 131063, 262135, 524279, 1048567, 2097143, 4194295, 8388599, 16777207, 33554423, 67108855, 134217719, 268435447, 536870903, 1073741815, 2147483639, 4294967287, 8589934583, 17179869175, 34359738359

Offset: 1

Artur Jasinski, Jan 29 2010

Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A000225, A159741, A185346, A206853.

Table[4 2^n - 9, {n, 1, 100}]
LinearRecurrence[{3,-2},{-1,7},30] (* Harvey P. Dale, May 27 2021 *)

my(x='x+O('x^34)); Vec(x*(10*x-1)/((x-1)*(2*x-1))) \\ Elmo R. Oliveira, Jun 15 2025

a(n) = 2*a(n-1) + 9, a(1)= -1. - Vincenzo Librandi, Mar 20 2011
For n >= 3, a(n) = 8<+>(n+2), where operation <+> is defined in A206853. - Vladimir Shevelev, Feb 17 2012
From Elmo R. Oliveira, Jun 15 2025: (Start)
G.f.: x*(10*x-1)/((x-1)*(2*x-1)).
E.g.f.: 5 + exp(x)*(4*exp(x) - 9).
a(n) = A185346(n+2) = 4*A000225(n) - 5.
a(n) = A159741(n-1) - 1 for n > 1. (End)

More terms from Elmo R. Oliveira, Jun 15 2025

A207460 Let a(1) = 4. For n > 1, a(n) is the least number greater than a(n-1) such that the Hamming distance D(a(n-1),a(n)) = 4.

Original entry on oeis.org

4, 11, 16, 31, 35, 44, 49, 62, 70, 73, 82, 93, 97, 110, 112, 127, 143, 145, 158, 162, 173, 176, 191, 199, 200, 211, 220, 224, 239, 241, 254, 286, 290, 301, 304, 319, 327, 328, 339, 348, 352, 367, 369, 382, 398, 400, 415, 419

Offset: 1

Vladimir Shevelev, Feb 18 2012

All terms are odious (A000069).

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000225, A205509, A205510, A205511, A205302, 205649, A205533, A122565, A206852, A206853, A206960, A207016, A207017.

nxt[n_]:=Module[{k=n+1},While[HammingDistance[PadLeft[IntegerDigits[ n,2],IntegerLength[ k,2]],IntegerDigits[k,2]]!=4,k++];k]; NestList[ nxt,4,50] (* Harvey P. Dale, Nov 07 2020 *)

A207472 Let a(1) = 5. For n > 1, a(n) is the least number greater than a(n-1) such that the Hamming distance D(a(n-1),a(n)) = 5.

Original entry on oeis.org

5, 26, 33, 62, 66, 93, 96, 127, 135, 152, 163, 188, 192, 223, 225, 254, 270, 273, 294, 313, 320, 351, 353, 382, 390, 409, 418, 445, 449, 478, 480, 511, 543, 545, 574, 578, 605, 608, 639, 647, 664, 675, 700, 704, 735, 737, 766, 782, 785, 806, 825, 832, 863, 865

Offset: 1

Vladimir Shevelev, Feb 18 2012

Odious and evil terms are alternating (cf. A000069, A001969).

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000069, A000225, A001969, A205509, A205510, A205511, A205302, A205649, A205533, A122565, A206852, A206853, A206960, A207016, A207017.

a[1] = 5; a[n_] := a[n] = Module[{k = a[n - 1], m = a[n-1] + 1}, While[DigitCount[BitXor[k, m], 2, 1] != 5, m++]; m]; Array[a, 100] (* Amiram Eldar, Aug 06 2023 *)

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A048490 a(n) = T(7,n), array T given by A048483.

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A207063 a(n) is the smallest number larger than a(n-1) with mutual Hamming distance 2 and a(1)=0.

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A182209 a(n) is the least m >= n, such that the Hamming distance D(n,m) = 3.

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A207016 a(1) = 3 , for n > 1 a(n) is the least number greater than a(n-1) such that the Hamming distance D(a(n-1),a(n)) = 3.

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A208982 Numbers n such that the next larger number with mutual Hamming distance 1 is prime.

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A182336 a(n) is the least m>=n, such that the Hamming distance D(n,m)=4.

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A209085 a(n) is the next larger than A208982(n) number with mutual Hamming distance 1.

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A172252 a(n) = 4*2^n - 9.

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