A048715 -id:A048715 - cp's OEIS frontend

A115422 Integers n > 0 such that n XOR 20n = 21n.

1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 33, 36, 48, 64, 65, 66, 67, 72, 73, 96, 97, 128, 129, 130, 131, 132, 134, 144, 146, 192, 193, 194, 195, 256, 257, 258, 259, 260, 262, 264, 265, 268, 288, 289, 292, 384, 385, 386, 387, 388, 390, 512, 513, 514, 515, 516, 518

Offset: 1

Paul D. Hanna, Jan 22 2006

nonn

n is in the sequence iff 2*n is in the sequence. - Robert Israel, Nov 11 2016

Ivan Neretin, Table of n, a(n) for n = 1..10203

Cf. A003714 (Fibbinary numbers), A048715, A048718, A115423, A115424.

select(n -> Bits:-Xor(n,20*n)=21*n, [$1..1000]); # Robert Israel, Nov 11 2016

Select[Range[600],BitXor[#,20#]==21#&] (* Harvey P. Dale, Apr 21 2018 *)

isok(n) = bitxor(n, 20*n) == 21*n; \\ Michel Marcus, Nov 11 2016

$cnt=0;
foreach(1..1_000){
    print ++$cnt," $\n" if ((20*$)^$)==21*$;
}
# Ivan Neretin, Nov 10 2016

This sequence also seems to satisfy:
5*a(n) XOR 16*a(n) = 21*a(n);
5*a(n) XOR 17*a(n) = 20*a(n); etc.
a(A224809(n+4)) = 2^n. - Gheorghe Coserea, Nov 11 2016

A115423 Integers n > 0 such that n XOR 30n = 31n.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 33, 64, 65, 66, 128, 129, 130, 132, 256, 257, 258, 260, 264, 512, 513, 514, 516, 520, 528, 1024, 1025, 1026, 1028, 1032, 1040, 1056, 1057, 2048, 2049, 2050, 2052, 2056, 2064, 2080, 2081, 2112, 2113, 2114, 4096, 4097, 4098, 4100, 4104, 4112

Offset: 1

Paul D. Hanna, Jan 22 2006

nonn

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A003714 (Fibbinary numbers), A048715, A048718, A115422, A115424.

Select[Range[4200],BitXor[#,30#]==31#&] (* Harvey P. Dale, May 23 2017 *)

isok(n) = bitxor(n, 30*n) == 31*n; \\ Michel Marcus, Nov 11 2016

$cnt=0;
foreach(1..4_000){
    print ++$cnt," $\n" if ((30*$)^$)==31*$;
}
# Ivan Neretin, Nov 11 2016

This sequence also seems to satisfy:
3*a(n) XOR 21*a(n) = 22*a(n);
5*a(n) XOR 19*a(n) = 22*a(n);
6*a(n) XOR 19*a(n) = 21*a(n); etc.
a(A003520(n+4)) = 2^n. - Gheorghe Coserea, Nov 11 2016

A342697 For any number n with binary expansion Sum_{k >= 0} b(k) * 2^k, the binary expansion of a(n) is Sum_{k >= 0} floor((b(k) + b(k+1) + b(k+2))/2) * 2^k.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 3, 3, 0, 0, 2, 3, 6, 7, 7, 7, 0, 0, 0, 1, 4, 5, 7, 7, 12, 12, 14, 15, 14, 15, 15, 15, 0, 0, 0, 1, 0, 1, 3, 3, 8, 8, 10, 11, 14, 15, 15, 15, 24, 24, 24, 25, 28, 29, 31, 31, 28, 28, 30, 31, 30, 31, 31, 31, 0, 0, 0, 1, 0, 1, 3, 3, 0, 0, 2, 3, 6

Offset: 0

Rémy Sigrist, Mar 18 2021

The value of the k-th bit in a(n) corresponds to the most frequent value in the bit triple starting at the k-th bit in n.

			The first terms, in decimal and in binary, are:
  n   a(n)  bin(n)  bin(a(n))
  --  ----  ------  ---------
   0     0       0          0
   1     0       1          0
   2     0      10          0
   3     1      11          1
   4     0     100          0
   5     1     101          1
   6     3     110         11
   7     3     111         11
   8     0    1000          0
   9     0    1001          0
  10     2    1010         10
  11     3    1011         11
  12     6    1100        110
  13     7    1101        111
  14     7    1110        111
  15     7    1111        111

Rémy Sigrist, Table of n, a(n) for n = 0..8192
Michael De Vlieger, Log log scatterplot of a(n), n = 0..2^20.
Index entries for sequences related to binary expansion of n

Cf. A048715, A048730, A048733, A342698, A342700.

A342697[n_] := Quotient[7*n - BitXor[n, 2*n, 4*n], 8];
Array[A342697, 100, 0] (* Paolo Xausa, Aug 06 2025 *)

a(n) = sum(k=0, #binary(n), ((bittest(n, k)+bittest(n, k+1)+bittest(n, k+2))>=2) * 2^k)

a(n) = 0 iff n belongs to A048715.

a(n) = floor(A048730(n)/8) = floor(A048733(n)/2). - Kevin Ryde, Mar 26 2021

A350311 Replace 2^k in the binary expansion of n with A000930(k+2), Narayana's cows sequence.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 6, 7, 8, 9, 9, 10, 11, 12, 10, 11, 12, 13, 13, 14, 15, 16, 9, 10, 11, 12, 12, 13, 14, 15, 13, 14, 15, 16, 16, 17, 18, 19, 15, 16, 17, 18, 18, 19, 20, 21, 19, 20, 21, 22, 22, 23, 24, 25, 13, 14, 15, 16, 16, 17

Offset: 0

A.H.M. Smeets, Dec 24 2021

A048715(n) = m, if and only if a(n) = m and for all k > n a(k) > m.

Rémy Sigrist, Table of n, a(n) for n = 0..8191

Cf. A000930, A048715, A350215, A350312.

Cf. A022290 (analog for Fibonacci numbers).

b:= (n, i, j, k)->`if`(n=0, 0, k*irem(n, 2, 'q')+b(q, j, k, i+k)):
a:= n-> b(n, 1$3):
seq(a(n), n=0..100);  # Alois P. Heinz, Jan 26 2022

my(p=Mod('x,'x^3-'x^2-1)); a(n) = vecsum(Vec(lift(subst(Pol(binary(n))*'x^2,'x,p)))); \\ Kevin Ryde, Dec 26 2021

def Interpretation(n):
    f0, f1, f2, r = 1, 1, 1, 0
    while n > 0:
        if n%2 == 1:
            r = r+f0
        n, f0, f1, f2 = n//2, f0+f2, f0, f1
    return r
n = 0
while n <= 69:
    print(Interpretation(n), end = ", ")
    n += 1

A350312 Narayana weighted representation of n (the bottom version). Also binary representation of numbers not containing 00 or 010 as a substring.

Original entry on oeis.org

0, 1, 10, 11, 101, 110, 111, 1011, 1101, 1110, 1111, 10110, 10111, 11011, 11101, 11110, 11111, 101101, 101110, 101111, 110110, 110111, 111011, 111101, 111110, 111111, 1011011, 1011101, 1011110, 1011111, 1101101, 1101110, 1101111, 1110110, 1110111, 1111011

Offset: 0

A.H.M. Smeets, Dec 24 2021

a(n) equals binary representation of m, if and only if A350311(m) = n and for all k < m, A350311(k) < n.

A.H.M. Smeets, The design of greedy number representations

Cf. A000930, A048715, A350215 (top version), A350311.

Fibonacci representations: A014417 (Zeckendorf), A104326 (dual Zeckendorf).

q[n_] := SequenceCount[IntegerDigits[n, 2], #] & /@ {{0, 0}, {0, 1, 0}} == {0, 0}; bin[n_] := FromDigits[IntegerDigits[n, 2]]; bin /@ Select[Range[0, 120], q] (* Amiram Eldar, Jan 27 2022 *)

# first method (as from definition)
def A101(n):
    f0, f1, f2, r = 1, 1, 1, 0
    while n > 0:
        if n%2 == 1:
            r = r+f0
        n, f0, f1, f2 = n//2, f0+f2, f0, f1
    return r
n, a = 0, 0
while n < 36:
    if A101(a) == n:
        print(bin(a)[2:], end = ", ")
        n += 1
    a += 1

# second method (as from regular expression)
def test(n):
    s, i, n1 = bin(n)[2:], 0, 2
    while i < len(s):
        if s[i] == "0":
            if n1 < 2:
                return 0
            n1 = 0
        else:
            n1 += 1
        i += 1
    return 1
n, a = 0, 0
while n < 36:
    if test(a):
        print(bin(a)[2:], end = ", ")
        n += 1
    a += 1

Regular expression: 0|11*(0111*)*(0|01|011*)?.

cp's OEIS Frontend

A115422 Integers n > 0 such that n XOR 20*n = 21*n.

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A115423 Integers n > 0 such that n XOR 30*n = 31*n.

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A342697 For any number n with binary expansion Sum_{k >= 0} b(k) * 2^k, the binary expansion of a(n) is Sum_{k >= 0} floor((b(k) + b(k+1) + b(k+2))/2) * 2^k.

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A350311 Replace 2^k in the binary expansion of n with A000930(k+2), Narayana's cows sequence.

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Author

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Programs

Maple

PARI

Python

A350312 Narayana weighted representation of n (the bottom version). Also binary representation of numbers not containing 00 or 010 as a substring.

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Author

Keywords

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Programs

Mathematica

Python

Python

Formula

A115422 Integers n > 0 such that n XOR 20n = 21n.

A115423 Integers n > 0 such that n XOR 30n = 31n.