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
Starting from the second most significant bit, we continue complementing every second bit (in this case, not starting before at the thirdmost significant bit), as long as the first zero is encountered, which is also complemented if its distance to the most significant bit is even, after which the remaining bits are left intact. E.g. 121 = 1111001 in binary. Complementing its thirdmost significant bit and the first zero-bit two positions right of it (i.e. bit-2, 4 steps to the most significant bit, bit-6), we obtain "11011.." after which the rest of the bits stay same, so we get 1101101, which is 109's binary representation, thus a(121)=109. On the other hand, 125 = 1111101 in binary and the transducer complements the bits at positions 4 and 2, yielding 11010.. and then switches to the fixing state at the zero encounted at bit-position 1, without complementing it (as it is 5 steps from the msb) and the rest are fixed, so we get 1101001, which is 105's binary representation, thus a(125)=105.
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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