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Numbers n such that A080719(n) divides n.
+20
0
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 20, 30, 40, 50, 60, 70, 80, 90, 100, 105, 112, 120, 200, 210, 220, 240, 300, 330, 342, 360, 400, 408, 453, 462, 500, 504, 573, 600, 700, 720, 800, 900, 924, 1000, 1008, 1010, 1012, 1020, 1040, 1050, 1116, 1120, 1145, 1200, 1320, 1750, 1900, 2000
EXAMPLE
Let us consider n = 342: 3 converted to base 2 is 11, 4 is 100, 2 is 10. Their concatenation is 1110010 that converted to base 10 is 114. So A080719(342) = 114. Finally 114 divides 342, so 342 is a member.
MAPLE
P:=proc(q) local a, b, k, n, x; for n from 1 to q do x:=[]; a:=n;
for k from 1 to length(n) do b:=convert((a mod 10), base, 2);
if b=[] then x:=[op(x), 0]; else x:=[op(x), op(b)]; fi; a:=trunc(a/10); od; a:=0;
for k from 1 to nops(x) do a:=2*a+x[-k]; od; if frac(n/a)=0 then print(n);
fi; od; end: P(2000);
MATHEMATICA
Select[Range[2000], Divisible[#, FromDigits[Flatten[IntegerDigits[#, 2]&/@ IntegerDigits[ #]], 2]]&] (* Harvey P. Dale, Mar 13 2019 *)
In decimal representation of n: replace each digit with its binary representation.
+10
3
0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 10, 11, 110, 111, 1100, 1101, 1110, 1111, 11000, 11001, 100, 101, 1010, 1011, 10100, 10101, 10110, 10111, 101000, 101001, 110, 111, 1110, 1111, 11100, 11101, 11110, 11111, 111000, 111001, 1000, 1001, 10010, 10011
EXAMPLE
. n | dec --> bin | a(n) | A080719(n)
. -----+--------------------------+----------+------------
. 100 | [1,0,0] | [1,0,0] | 100 | 4
. 101 | [1,0,1] | [1,0,1] | 101 | 5
. 102 | [1,0,2] | [1,0,10] | 1010 | 10
. 103 | [1,0,3] | [1,0,11] | 1011 | 11
. 104 | [1,0,4] | [1,0,100] | 10100 | 20
. 105 | [1,0,5] | [1,0,101] | 10101 | 21
. 106 | [1,0,6] | [1,0,110] | 10110 | 22
. 107 | [1,0,7] | [1,0,111] | 10111 | 23
. 108 | [1,0,8] | [1,0,1000] | 101000 | 40
. 109 | [1,0,9] | [1,0,1001] | 101001 | 41
. 110 | [1,1,0] | [1,1,0] | 110 | 6
. 111 | [1,1,1] | [1,1,1] | 111 | 7
. 112 | [1,1,2] | [1,1,10] | 1110 | 14
. 113 | [1,1,3] | [1,1,11] | 1111 | 15
. 114 | [1,1,4] | [1,1,100] | 11100 | 28
. 115 | [1,1,5] | [1,1,101] | 11101 | 29
. 116 | [1,1,6] | [1,1,110] | 11110 | 30
. 117 | [1,1,7] | [1,1,111] | 11111 | 31
. 118 | [1,1,8] | [1,1,1000] | 111000 | 56
. 119 | [1,1,9] | [1,1,1001] | 111001 | 57
. 120 | [1,2,0] | [1,10,0] | 1100 | 12
. 121 | [1,2,1] | [1,10,1] | 1101 | 13
. 122 | [1,2,2] | [1,10,10] | 11010 | 26
. 123 | [1,2,3] | [1,10,11] | 11011 | 27
. 124 | [1,2,4] | [1,10,100] | 110100 | 52
. 125 | [1,2,5] | [1,10,101] | 110101 | 53 .
MATHEMATICA
Table[FromDigits[Flatten[IntegerDigits[#, 2]&/@IntegerDigits[n]]], {n, 0, 50}] (* Harvey P. Dale, Jun 06 2020 *)
PROG
(Haskell)
import Data.Maybe (mapMaybe)
a257831 = foldr (\b v -> 10 * v + b) 0 .
concat . mapMaybe (flip lookup bin) . a031298_row
where bin = zip [0..9] a030308_tabf
(Python)
....return int(''.join((format(int(d), 'b') for d in str(n))))
Square array read by antidiagonals: T(n, k) = 1 if the digits of p = n*k in base 2 are exactly the same as the digits of p when considering the base-2 representations of n, k and p as base-10 numbers, otherwise T(n, k) = 0.
+10
1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
COMMENTS
As n * k = k * n, the array is symmetric.
EXAMPLE
In base 2, 1001 * 10100 = 10110100. In base 10, 1001 * 10100 = 10110100. These digits match and therefore the pairs T(9, 20) and T(20, 9) are a 1 in the sequence (at a(444) and a(455)).
In base 2, the product of 11 * 11 = 1001, whereas 11 * 11 in base 10 yields 121. T(3, 3) is the 24th pair in the sequence and the first to fail. a(24) is thus a 0.
The array begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 1, 1, 0, 1, 1, 0, 0, 1, ...
1, 1, 1, 1, 1, 1, 1, 1, 1, ...
1, 1, 1, 1, 1, 0, 1, 0, 1, ...
1, 1, 1, 0, 1, 1, 0, 0, 1, ...
1, 1, 1, 0, 1, 0, 0, 0, 1, ...
1, 1, 1, 1, 1, 1, 1, 1, 1, ...
PROG
(Python)
def a322674(k):
seq = []
i = 0
while len(seq) <= k:
j = 0
while len(seq) <= k and j < i + 1:
n = i - j
m = j
decn = int(bin(n).replace('0b', ''))
decm = int(bin(m).replace('0b', ''))
binProd = bin(n * m).replace('0b', '')
decProd = str(decn * decm)
seq.append(int(binProd == decProd))
j += 1
i += 1
print(seq)
a322674(100)
(PARI) T(n, k) = fromdigits(binary(n))*fromdigits(binary(k)) == fromdigits(binary(n*k)); \\ Michel Marcus, Apr 03 2019
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