A374625 -id:A374625 - cp's OEIS frontend

A115637 In the binary expansion of n+2, transform 0->1 and 1->0 then interpret as base 4.

1, 0, 5, 4, 1, 0, 21, 20, 17, 16, 5, 4, 1, 0, 85, 84, 81, 80, 69, 68, 65, 64, 21, 20, 17, 16, 5, 4, 1, 0, 341, 340, 337, 336, 325, 324, 321, 320, 277, 276, 273, 272, 261, 260, 257, 256, 85, 84, 81, 80, 69, 68, 65, 64, 21, 20, 17, 16, 5, 4, 1, 0, 1365, 1364, 1361, 1360, 1349

Offset: 0

Paul Barry, Jan 27 2006

Row sums of number triangle A115636. Partial sums of A115638.

Old name was "A divide and conquer sequence".

Alois P. Heinz, Table of n, a(n) for n = 0..16384 (first 1025 terms from Antti Karttunen)
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003, see equation 2.4 with a(n) = a_{n+2} for case alpha=4, c=1, d=0.

Cf. A000695, A035327, A115633, A115636, A115638 (first differences), A374625.

b:= n-> 1-(n mod 2)+`if`(n<2, 0, b(iquo(n, 2))*4):
a:= n-> b(n+2):
seq(a(n), n=0..66);  # Alois P. Heinz, Jul 16 2024

A115637[n_] := FromDigits[1 - IntegerDigits[n + 2, 2], 4];
Array[A115637, 100, 0] (* Paolo Xausa, Jul 16 2024 *)

up_to = 1024;
A115633array(n, k) = (((-1)^n)*if(n==k,1, if((k+k+2)==n, -4, if((k+1)==n, -(1+(-1)^k)/2, 0))));
A115637list(up_to) = { my(mA115633=matrix(up_to,up_to,n,k,A115633array(n-1,k-1)), mA115636 = matsolve(mA115633,matid(up_to)), v = vector(up_to)); for(n=1,up_to,v[n] = vecsum(mA115636[n,])); (v); };
v115637 = A115637list(up_to+1);
A115637(n) = v115637[1+n]; \\ Antti Karttunen, Nov 02 2018

a(n) = fromdigits([!b |b<-binary(n+2)], 4); \\ Kevin Ryde, Jul 15 2024

def A115637(n): return int(bin((~(n+2))^(-1<<(n+2).bit_length()))[2:],4) # Chai Wah Wu, Jul 17 2024

G.f.: (1/(1-x))*Sum_{k>=0} 4^k*x^(2^(k+1)-2)/(1+x^(2^k)); the g.f. G(x) satisfies G(x) - 4(1+x)*x^2*G(x^2) = 1/(1-x^2).

a(n) = A000695(A035327(n+2)). - Kevin Ryde, Jul 15 2024

New name from Kevin Ryde, Jul 15 2024

A375156 In the binary expansion of n: expand bits 1 -> 01 and 0 -> x0 from most to least significant, where x is the complement of the previous bit from n.

Original entry on oeis.org

2, 1, 4, 5, 18, 17, 20, 21, 74, 73, 68, 69, 82, 81, 84, 85, 298, 297, 292, 293, 274, 273, 276, 277, 330, 329, 324, 325, 338, 337, 340, 341, 1194, 1193, 1188, 1189, 1170, 1169, 1172, 1173, 1098, 1097, 1092, 1093, 1106, 1105, 1108, 1109, 1322, 1321, 1316, 1317, 1298, 1297, 1300

Offset: 0

Darío Clavijo, Aug 01 2024

At the first bit of n, the previous bit is considered to be 0.
n=0 is a single 0 bit.
Equivalently for n >= 1, between adjacent bits u,v of n, insert bit u NOR v.
This is essentially the MFM(1,3) RLL binary encoding, without leading zeros.
Equivalently, apply (from left to right, in the given order) the following transformations to the binary expansion of n: 10 -> 0100, 1 -> 01, 0 -> 10. - Paolo Xausa, Aug 19 2024

			For n=137,
  n    = binary  1  0  0  0  1  0  0  1
  a(n) = binary 01 00 10 10 01 00 10 01 = 19017

Paolo Xausa, Table of n, a(n) for n = 0..10000
Wikipedia, Run-length limited.

Cf. A008836, A057077, A374625.

A375156[n_] := FromDigits[StringReplace[IntegerString[n, 2], {"10" -> "0100", "1" -> "01", "0" -> "10"}], 2]; Array[A375156,100,0] (* Paolo Xausa, Aug 19 2024 *)

mfm_encode(data)=prev_enc_bit=0;enc=[];for(i=1,#data,enc=concat(enc,(1-data[i])*(1-prev_enc_bit));enc=concat(enc,data[i]);prev_enc_bit=data[i];);enc;
a(n)=if(n==0,return(2));fromdigits(mfm_encode(binary(n)),2);
vector(55,n,a(n-1))

def mfm_encode(data):
    prev_enc_bit, enc = "0", ""
    for i in range(len(data)):
        enc += ("1" if data[i] == "0" and prev_enc_bit == "0" else "0")
        enc += data[i]
        prev_enc_bit = enc[-1]
    return enc
def a(n):
    enc = mfm_encode(bin(n)[2:])
    return int("".join(map(str, enc)), 2)
print([a(n) for n in range(0, 55)])

a(n) = a(n-1) - (-1)^floor(n/2) for n odd.
a(n) mod 2 = 0 for n even.
a(n) = A374625(n) AND a(n) where AND is the bitwise-and.
a(n) OR A374625(n) = A374625(n) where OR is the bitwise-or.

A377029 a(1) = 0; thereafter in the binary expansion of a(n-1), expand bits: 1->01 and 0->10.

Original entry on oeis.org

0, 2, 6, 22, 406, 92566, 6818458006, 26055178074437806486, 540213899028732737068658940860686756246, 163551003506862550406254063077517364557434408527734307437037618419534882498966

Offset: 1

Darío Clavijo, Oct 13 2024

All terms are even and leading zeros omitted in the final encoding.
Conversely the opposite mapping of bits: 0->01 and 1->10 is A133468.
The bit length of a(n) is 2^(n-1)+1.
The count of bits set for a(n) is A094373(n).
a(n) = 2 (mod 4) for n > 1.
Also all the terms align bitwise to the right.
The hamming distance of a(n) and a(n+1) is in A000079.

			For n = 5 a(5) = 406 because:
This encoding results in the following tree:
n | a(n)
--+---------------
1 | 0
  | |\
2 | 1 0
  | | |
3 | 1 10
  | | | \
4 | 1 01 10--
  | | |\  \  \
  | | | \  \  \
5 | 1 10 01 01 10
Which also aligns bitwise to the right:
n | a(n)
--+-----------
1 |         0
2 |        10
3 |       110
4 |     10110
5 | 110010110
And 110010110 in base 10 is 406.

Darío Clavijo, Table of n, a(n) for n = 1..13

Cf. A000051, A000079, A000120, A094373, A133468, A320916, A374625.

NestList[FromDigits[2 - IntegerDigits[#, 2], 4] &, 0, 10] (* Paolo Xausa, Nov 04 2024 *)

from functools import cache
A374625 = lambda n: int(bin(n)[2:].replace('0', '2'), 4)
@cache
def a(n):
  if n == 1: return 0
  return A374625(a(n-1))
print([a(n) for n in range(1, 12)])

a(n) = A320916(2^(n-2)+1) for n > 1.
A000120(a(n+1) XOR a(n)) = A000079(n-2).
a(n) = A374625(a(n-1)) for n > 1. - Paolo Xausa, Nov 04 2024

A383976 In the binary expansion of n, expand bits 1 -> 11 and 0 -> 10.

Original entry on oeis.org

2, 3, 14, 15, 58, 59, 62, 63, 234, 235, 238, 239, 250, 251, 254, 255, 938, 939, 942, 943, 954, 955, 958, 959, 1002, 1003, 1006, 1007, 1018, 1019, 1022, 1023, 3754, 3755, 3758, 3759, 3770, 3771, 3774, 3775, 3818, 3819, 3822, 3823, 3834, 3835, 3838, 3839, 4010, 4011, 4014

Offset: 0

Darío Clavijo, May 16 2025

This is essentially the differential Manchester encoding or FM:(0,1) RLL.
Terms come in groups of two and all terms have even bitsize.
In this case, by convention, 0 is treated as a single 0 bit and leading zeros in other terms are omitted.

Wikipedia, Run-length limited

Cf. A000523, A374625.

a:= n-> 2+(n mod 2)+`if`(n<2, 0, 4*a(iquo(n, 2))):
seq(a(n), n=0..50);  # Alois P. Heinz, May 16 2025

a[n_] := a[n] = 2 + Mod[n, 2] + If[n < 2, 0, 4*a[Floor[n/2]]];Array[a, 51, 0] (* Shenghui Yang, May 21 2025 *)

a = lambda n: int(bin(n)[2:].replace('1','3').replace('0','2'),4)
print([a(n) for n in range(0,51)])

a(n) = a(n-1) + 1 for n odd.
a(n) = 2 + (n mod 2) + 4*a(floor(n/2)) if n >= 2 else 2 + (n mod 2).
a(n) = Sum_{k=0..floor(log_2(n))} 4^k*(2+b_k), where b_k is the k-th bit of n.

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A115637 In the binary expansion of n+2, transform 0->1 and 1->0 then interpret as base 4.

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A375156 In the binary expansion of n: expand bits 1 -> 01 and 0 -> x0 from most to least significant, where x is the complement of the previous bit from n.

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A377029 a(1) = 0; thereafter in the binary expansion of a(n-1), expand bits: 1->01 and 0->10.

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A383976 In the binary expansion of n, expand bits 1 -> 11 and 0 -> 10.

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