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.
18 = 10010 in binary and after complementing the second, third and fourth most significant bits at positions 3, 2 and 1, we get 1110, at which point we stop (because bit-1 was originally 1) and fix the rest, so we get 11100 (28 in binary), thus a(18)=28. This is the inverse of "binary adding machine". See pages 8, 9 and 103 in the Bondarenko, Grigorchuk, et al. paper. 19 = 10011 in binary. By complementing bits in (zero-based) positions 3, 2 and 1 we get 11101 in binary, which is 29 in decimal, thus a(19)=29.
def ok(n): return n&(n - 1)==0 def a153151(n): return n if n<2 else 2*n - 1 if ok(n) else n - 1 def A(n): return (int(bin(n)[2:][::-1], 2) - 1)/2 def msb(n): return n if n<3 else msb(n/2)*2 def a059893(n): return A(n) + msb(n) def a(n): return 0 if n==0 else a059893(a153151(a059893(n))) # Indranil Ghosh, Jun 09 2017
maxlevel <- 5 # by choice
a <- 1
for(m in 1:maxlevel){
a[2^m ] <- 2^(m+1) - 1
a[2^m + 1] <- 2^(m+1) - 2
for (k in 1:(2^m-1)){
a[2^(m+1) + 2*k ] <- 2*a[2^m + k]
a[2^(m+1) + 2*k + 1] <- 2*a[2^m + k] + 1}
}
a <- c(0,a)
# Yosu Yurramendi, Aug 01 2020
475 = 111011011 in binary. Starting from the second most significant bit and, as we begin with the swapping state a, we complement the bits up to and including the first zero encountered and so the beginning of the binary expansion is complemented as 1001....., then, as we switch to the inactive state b, the following bits are kept same, again up to and including the first zero encountered, after which the binary expansion is 1001110.., after which we switch again to the active state (state a), which complements the two rightmost 1's and we obtain the final answer 100111000, which is 312's binary representation, thus a(475)=312.
from sympy import floor
def a006068(n):
s=1
while True:
ns=n>>s
if ns==0: break
n=n^ns
s<<=1
return n
def a054429(n): return 1 if n==1 else 2*a054429(floor(n/2)) + 1 - n%2
def a(n): return 0 if n==0 else a054429(a006068(a054429(n))) # Indranil Ghosh, Jun 11 2017
maxn <- 63 # by choice a <- c(1,3,2) # If it were a <- 1:3, it would be A180200 for(n in 2:maxn){ a[2*n ] <- 2*a[n] + (a[n]%%2 == 0) a[2*n+1] <- 2*a[n] + (a[n]%%2 != 0) } a # Yosu Yurramendi, Jun 21 2020
312 = 100111000 in binary. Starting from the second most significant bit and, as we begin with the swapping state a, we complement the bits up to and including the first one encountered and so the beginning of the binary expansion is complemented as 1110....., then, as we switch to the inactive state b, the following bits are kept same, up to and including the first zero encountered, after which the binary expansion is 1110110.., after which we switch again to the complementing mode (state a) and we obtain 111011011, which is 475's binary representation, thus a(312)=475.
Function[s, Map[s[[#]] &, BitXor[#, Floor[#/2]] & /@ s]]@ Flatten@ Table[Range[2^(n + 1) - 1, 2^n, -1], {n, 0, 6}] (* Michael De Vlieger, Jun 11 2017 *)
a003188(n) = bitxor(n, n>>1); a054429(n) = 3<<#binary(n\2) - n - 1; a(n) = if(n==0, 0, a054429(a003188(a054429(n)))); \\ Indranil Ghosh, Jun 11 2017
from sympy import floor def a003188(n): return n^(n>>1) def a054429(n): return 1 if n==1 else 2*a054429(floor(n/2)) + 1 - n%2 def a(n): return 0 if n==0 else a054429(a003188(a054429(n))) # Indranil Ghosh, Jun 11 2017
maxn <- 63 # by choice
a <- c(1, 3, 2)
for(n in 2:maxn){
if(n%%2 == 0) {a[2*n] <- 2*a[n]+1 ; a[2*n+1] <- 2*a[n]}
else {a[2*n] <- 2*a[n] ; a[2*n+1] <- 2*a[n]+1}
}
(a <- c(0,a))
# Yosu Yurramendi, Apr 10 2020
25 = 11001 in binary, the first zero-bit at odd distance from the msb is immediately at where we start (at the second most significant bit), so we complement it and fix the rest, yielding 10001 (17 in binary), thus a(25)=17.
maxlevel <- 5 # by choice
a <- 1
for(m in 0:maxlevel) {
for(k in 0:(2^m-1)){
a[2^(m+1) + 2*k ] <- 2*a[2^m + k]
a[2^(m+1) + 2*k + 1] <- 2*a[2^m + k] + 1
}
x <- floor(2^(m+2)/3)
a[2*x ] <- 2*a[x] + 1
a[2*x + 1] <- 2*a[x]
}
(a <- c(0, a))
# Yosu Yurramendi, Oct 12 2020
Starting from the second most significant bit, we continue complementing every second bit (in this case, starting from the second most significant bit), as long as the first zero is encountered, which is also complemented if its distance to the most significant bit is odd, after which the remaining bits are left intact. E.g. 121 = 1111001 in binary. Complementing its second and fourth most significant bits (positions 5 & 3) and stopping at the first zero-bit at position 2 (which is not complemented, as its distance to the msb is 6), we obtain "10100.." after which the rest of the bits stay same, so we get 1010001, which is 81's binary representation, thus a(121)=81. On the other hand, 125 = 1111101 in binary and the transducer complements the bits at positions 5, 3 and also the first zero at the position 1 (because at odd distance from the msb), yielding 101011., after which the remaining bit stays same, thus we get 1010111, which is 87's binary representation, thus a(125)=87.
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