A355716 a(n) is the smallest number that has exactly n binary palindrome divisors (A006995).
1, 3, 9, 15, 99, 45, 135, 189, 315, 495, 945, 765, 2079, 6237, 3465, 5355, 4095, 8415, 31185, 20475, 25245, 12285, 85995, 58905, 61425, 45045, 69615, 176715, 446985, 225225, 328185, 208845, 135135, 405405, 528255, 1396395, 675675, 2027025, 765765, 5360355, 2993445, 3968055, 3828825
Offset: 1
Examples
a(4) = 15 since 15 has 4 divisors {1, 3, 5, 15} that are all palindromes when written in binary: 1, 11, 101 and 1111; no positive integer smaller than 15 has four divisors that are binary palindromes, hence a(4) = 15.
a(5) = 99 since 99 has 6 divisors {1, 3, 9, 11, 33, 99} of which only 11 is not a palindrome when written in binary: 11_10 = 1011_2; no positive integer smaller than 99 has five divisors that are binary palindromes, hence a(5) = 99.
Links
- Michael S. Branicky, Table of n, a(n) for n = 1..67
Programs
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Mathematica
f[n_] := DivisorSum[n, 1 &, PalindromeQ[IntegerDigits[#, 2]] &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n]; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[25, 10^5] (* Amiram Eldar, Jul 15 2022 *) -
PARI
is(n) = my(d=binary(n)); d==Vecrev(d); \\ A006995 a(n) = my(k=1); while (sumdiv(k, d, is(d)) != n, k++); k; \\ Michel Marcus, Jul 15 2022
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Python
from sympy import divisors from itertools import count, islice def c(n): b = bin(n)[2:]; return b == b[::-1] def f(n): return sum(1 for d in divisors(n, generator=True) if c(d)) def agen(): n, adict = 1, dict() for k in count(1): fk = f(k) if fk not in adict: adict[fk] = k while n in adict: yield adict[n]; n += 1 print(list(islice(agen(), 20))) # Michael S. Branicky, Jul 23 2022
Extensions
More terms from Michael S. Branicky, Jul 15 2022
Comments