A339508 Number of subsets of {2..n} such that the product of the elements is a decimal palindrome.
1, 1, 2, 4, 6, 7, 8, 10, 11, 13, 13, 33, 43, 56, 70, 73, 99, 114, 134, 151, 151, 185, 333, 372, 445, 456, 558, 565, 672, 1031, 1031, 1220, 1518, 1967, 2161, 2176, 2238, 5399, 5543, 6720, 6720, 8857, 10019, 11819, 16882, 16896, 18072, 19299, 21267, 23405, 23405, 24686
Offset: 0
Examples
a(9) = 13 subsets: {}, {2}, {3}, {4}, {5}, {6}, {7}, {8}, {9}, {2, 3}, {2, 4}, {4, 7, 9} and {2, 3, 6, 7}.
Links
- David A. Corneth, List of factorizations of the a(n) products for n = 1..51.
- David A. Corneth, Lower bounds on a(n) for n = 0..100.
- Eric Weisstein's World of Mathematics, Palindromic Number
- Index entries for sequences related to palindromes
Programs
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Python
from numpy import prod from itertools import combinations def a(n): ans = 1 # empty set for r in range(1, n): for s in combinations(range(2, n+1), r): strsp = str(prod(s)) ans += strsp==strsp[::-1] return ans print([a(n) for n in range(20)]) # Michael S. Branicky, Dec 08 2020 -
Python
from functools import lru_cache, reduce from operator import mul from itertools import combinations @lru_cache(maxsize=None) def A339508(n): nlist = [i for i in range(2,n) if i % 10 != 0] if n == 0 or n == 1: return 1 c = A339508(n-1) if n % 10 != 0: if str(n) == str(n)[::-1]: c += 1 for i in range(1,len(nlist)+1): for d in combinations(nlist,i): s = str(reduce(mul,d)*n) if s == s[::-1]: c += 1 return c # Chai Wah Wu, Dec 08 2020
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Python
from functools import lru_cache def cond(p): sp = str(p); return sp == sp[::-1] @lru_cache(maxsize=None) def b(n, p): if n == 1: return int(cond(p)) return b(n-1, p) + b(n-1, p*n) if p*n%10 else b(n-1, p) def a(n): return 1 if n < 2 else b(n, 1) print([a(n) for n in range(36)]) # Michael S. Branicky, Oct 05 2022
Extensions
a(23)-a(32) from Michael S. Branicky, Dec 08 2020
a(33)-a(42) from Chai Wah Wu, Dec 08 2020
a(43)-a(48) from Chai Wah Wu, Dec 11 2020
a(49)-a(51) from Michael S. Branicky, Oct 05 2022
Comments