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 A253594 Numbers n that have more than one palindromic representation in bases 2 <= b <= n-2. 2
 10, 15, 16, 17, 18, 20, 21, 24, 26, 27, 28, 30, 31, 32, 33, 34, 36, 38, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 60, 62, 63, 64, 65, 66, 67, 68, 70, 72, 73, 74, 75, 76, 78, 80, 81, 82, 84, 85, 86, 88, 90, 91, 92, 93, 96, 98, 99, 100, 102, 104, 105, 107, 108, 109, 110, 111, 112, 114, 116, 117, 118, 119, 120, 121, 122, 123, 124 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS We do not include base n-1, because every n>2 is written '11' in base n-1. LINKS Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1685 from Christian Perfect) EXAMPLE 10 is written '101' in base 3, and '22' in base 4. 12 is written '22' in base 5, but is not a palindrome in any other base up to 10, so does not belong to this sequence. PROG (Python) from itertools import count . def base(n, b): ...while n: ......m = n%b ......yield m ......n = (n-m)//b . def is_palindrome(seq): ...seq = list(seq) ...l = len(seq)//2 ...return seq[:l] == seq[-1:-l-1:-1] . def a(): ...for n in count(2): ......base_representations = [(b, list(base(n, b))) for b in range(2, n-1)] ......pals = [(b, s) for b, s in base_representations if is_palindrome(s)] ......if len(pals)>1: .........yield n (Python) from sympy.ntheory import digits def ok(n): c = 0 for b in range(2, n-1): d = digits(n, b)[1:] c += int(d == d[::-1]) if c == 2: return True return c > 1 print([k for k in range(125) if ok(k)]) # Michael S. Branicky, Feb 05 2024 CROSSREFS Sequence in context: A074391 A324527 A164865 * A330698 A342593 A046424 Adjacent sequences: A253591 A253592 A253593 * A253595 A253596 A253597 KEYWORD nonn,base,easy AUTHOR Christian Perfect, Jan 05 2015 STATUS approved

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Last modified June 13 17:32 EDT 2024. Contains 373391 sequences. (Running on oeis4.)