A340559 Numbers that are palindromic in base 2 and base 16.
0, 1, 3, 5, 7, 9, 15, 17, 51, 85, 119, 153, 255, 257, 273, 771, 819, 1285, 1317, 1365, 1397, 1799, 1831, 1879, 1911, 2313, 2409, 2457, 2553, 3855, 3951, 3999, 4095, 4097, 4369, 12291, 13107, 20485, 21029, 21845, 22389, 28679, 29223, 30039, 30583, 36873, 38505
Offset: 1
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
Select[Range[0, 10^5], PalindromeQ @ IntegerDigits[#, 2] && PalindromeQ @ IntegerDigits[#, 16] &] (* Amiram Eldar, Jan 11 2021 *)
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PARI
ispal(m, b) = my(d=digits(m, b)); d == Vecrev(d); isok(m) = ispal(m, 2) && ispal(m, 16); \\ Michel Marcus, Jan 20 2021
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Python
def palindrome(x): res = str(x) == str(x)[::-1] return res def dec_to_bin(x): return int(bin(x)[2:]) def dec_to_hex(x): return (hex(x)[2:]) for x in range (1,10000): if palindrome(dec_to_hex(x)) & palindrome(dec_to_bin(x)) == True: print(x) (BASIC:- MM Basic, a modern QBASIC variant, https://www.mmbasic.com/) Function reverse(in_string$) As string Local r$ Local i For i = Len(in_string$) To 1 Step -1 b$=Mid$(in_string$,i,1) r$=r$+b$ Next i reverse=r$ End Function For i = 1 To 10000 If Bin$(i) = reverse(Bin$(i)) Then If Hex$(i) = reverse(Hex$(i)) Then Print i,Bin$(i), Hex$(i) EndIf EndIf Next i