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A263172
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Nonpalindromic positive integers whose digits can be rearranged to form a palindrome, after evaluation.
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1
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10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 112, 113, 114, 115, 116, 117, 118, 119, 122, 133, 144, 155, 166, 177, 188, 199, 200, 211, 220, 221, 223, 224, 225, 226, 227, 228, 229, 233, 244, 255, 266, 277, 288, 299, 300, 311, 322, 330
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OFFSET
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1,1
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COMMENTS
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Called Dennis numbers as a tribute to user Dennis for winning the robbers' part of The Programming Language Quiz of the Programming Puzzles & Code Golf Stack Exchange website.
Number of d-digit terms for d in [1..8]: [0, 9, 162, 846, 10044, 55548, 608472, 3911256]. - Jon E. Schoenfield, Oct 11 2015
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LINKS
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EXAMPLE
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10 is in the sequence since rearranging the 1 and 0 gives the digit string "01", which evaluates to 1, which is a palindrome.
16612 is in the sequence since its digits can be rearranged (as 16261 or 61216) to form a palindrome.
212 is not in the sequence since it is already palindromic.
Since all one-digit numbers are already palindromic, they are not in the sequence.
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MATHEMATICA
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palQ[n_] := Block[{d = IntegerDigits@ n}, d == Reverse@ d]; Select[Range@ 330, And[! palQ@ #, AnyTrue[FromDigits /@ Permutations@ IntegerDigits@ #, palQ]] &] (* Michael De Vlieger, Nov 03 2015, Version 10 *)
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PROG
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(Rust) fn d(mut i:u64)->(u64, i32){for n in 1..{let mut o=0; if n.to_string()==n.to_string().chars().rev().collect::<String>(){continue}let mut s=n.to_string().into_bytes(); for a in 0..s.len(){for b in a+1..s.len(){s.swap(a, b); {let t=s.iter().skip_while(|&x|*x==48).collect::<Vec<&u8>>(); if t.iter().cloned().rev().collect::<Vec<&u8>>()==t{o+=1}}s.swap(a, b); }}if o>0{i-=1; if i<1{return(n, o)}}}(0, 0)} // Program written by Programming Puzzles & Code Golf Stack Exchange user "Doorknob" on September 12, 2015
(Python)
def c(s): return sum(s.count(d)&1 for d in set(s)) == len(s)&1
def ok(n): s = str(n); return s != s[::-1] and c(s.replace("0", ""))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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