A367084
Indices k such that A367083(k) and A367083(k+1) are both odd.
Original entry on oeis.org
0, 7, 16, 25, 34, 43, 50, 59, 68, 77, 86, 95, 102, 111, 120, 129, 138, 145, 154, 163, 172, 181, 190, 197, 206, 215, 224, 233, 240, 249, 258, 267, 276, 285, 292, 301, 310, 319, 328, 335, 344, 353, 362, 371, 380, 387, 396, 405, 414, 423, 430, 439, 448, 457, 466, 475, 482, 491, 500
Offset: 0
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A367084_upto(N) = [n|n<-A=A367083_upto(N), A[n+1]==Mod(A[n],2)]
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from itertools import islice
def A367084_gen(): # generator of terms
a, b, c = 1, 4, -1
while True:
while (a:=a*3)A367084_list = list(islice(A367084_gen(),30)) # Chai Wah Wu, Nov 08 2023
A367085
3-valuation r(n) of the terms A367083(A367084(n)+1) = 3^r(n) which are the odd terms preceded by another odd term.
Original entry on oeis.org
1, 5, 10, 15, 20, 25, 29, 34, 39, 44, 49, 54, 58, 63, 68, 73, 78, 82, 87, 92, 97, 102, 107, 111, 116, 121, 126, 131, 135, 140, 145, 150, 155, 160, 164, 169, 174, 179, 184, 188, 193, 198, 203, 208, 213, 217, 222, 227, 232, 237, 241, 246, 251, 256, 261, 266, 270, 275, 280, 285, 290, 294, 299
Offset: 0
The first group (3^r, 4^s, ..., 3^r') in A367083 starts with A367083(1) = 3 = 3^1 (following the odd term A367083(A367084(0)) = 3^0 = 1), therefore a(0) = 1.
The second such group starts with A367083(8) = 3^5 (following the odd term A367083(A367084(1) = 7) = 3^4), therefore a(1) = 5.
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A367085(n)=valuation(A367083(A367084(n)+1),3) \\ or Axxx[.+1] if vectors are used instead of the 0-indexed functions/sequences.
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/* more efficiently: */
A367085_upto(N)={my(r=1, s=1, L3=log(3), L4=log(4), A=List(r)); until(r>=N, listput(A, r += 1-s+s+=((r+4)*L3 > (s+3)*L4)+3)); Vec(A)}
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from itertools import islice
def A367085_gen(): # generator of terms
a, b, c = 1, 4, 0
while True:
while (a:=a*3)A367085_list = list(islice(A367085_gen(),30)) # Chai Wah Wu, Nov 09 2023
A367086
Exponents k > 0 such that the interval [4^(k-1), 4^k] contains two powers of 3.
Original entry on oeis.org
1, 4, 8, 12, 16, 20, 23, 27, 31, 35, 39, 43, 46, 50, 54, 58, 62, 65, 69, 73, 77, 81, 85, 88, 92, 96, 100, 104, 107, 111, 115, 119, 123, 127, 130, 134, 138, 142, 146, 149, 153, 157, 161, 165, 169, 172, 176, 180, 184, 188, 191, 195, 199, 203, 207, 211, 214, 218, 222, 226, 230, 233, 237, 241
Offset: 0
The smallest power 4^s such that the interval [4^(s-1), 4^s] contains two powers of 3 is 4^1, i.e., s = 1, where [4^0, 4^1] contains 3^0 and 3^1. Hence a(0) = 1. (This is also the exponent of the smallest power of 4 in the first group of the form (3^r, 4^s, ..., 3^r') in A367083, namely: (3^1, 4^1, 3^2, 4^2, 3^3, 4^3, 3^4).)
The next larger power of 4 with this property is 4^4, hence a(1) = 4, where [4^3, 4^4] contains 3^4 and 3^5. This is also the least exponent of a power of 4 in the second group (3^5, 4^4, 3^6, 4^5, ..., 3^9), which is marked on the left in the table below.
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Numbers of the forms
3^r 4^s
------ ------
...
| 16
| 27 __________ the interval
| 64 | [4^3, 4^4]
\____ 81 | includes two
/ 243 | powers of 3,
2 | ____ 256 _| so 4 is a term
n | 729 of this sequence
d | 1024
| 2187
g | 4096
r | 6561 __________ the interval
p | 16384 | [4^7, 4^8]
\_ 19683 | includes two
/ 59049 | powers of 3,
| __ 65536 _| so 8 is a term
| 177147 of this sequence
| 262144
| 531441
...
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A367086_upto(N)={my(r=1, s=1, L3=log(3), L4=log(4), A=List(s)); until(r>=N, listput(A, s-=1+r-r+=((r+4)*L3 > (s+3)*L4)+4)); Vec(A)}
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from itertools import islice
def A367086_gen(): # generator of terms
a, b, c, i = 1, 4, -1, 1
while True:
while (a:=a*3)A367086_list = list(islice(A367086_gen(),30)) # Chai Wah Wu, Nov 18 2023
Showing 1-3 of 3 results.
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