A240454 Smallest prime divisors of the palindromes with an even number of digits.
11, 2, 3, 2, 5, 2, 7, 2, 3, 7, 11, 3, 11, 11, 3, 11, 7, 3, 11, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 11, 11, 3, 11, 11, 3, 7, 11, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 3, 5, 5, 3, 5, 5, 3, 5, 5, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 7, 11, 3, 11, 11, 3, 11, 7, 3, 11, 2, 2, 2
Offset: 1
Examples
a(10) = 7 because the concatenation of 10 and 01 is 1001 = 7*11*13 where 7 is the smallest divisor of 1001.
Links
- Michel Lagneau, Table of n, a(n) for n = 1..10000
Programs
-
Maple
with(numtheory):for n from 1 to 100 do:x:=convert(n,base,10):n1:=nops(x): s:=sum('x[i]*10^(n1-i)', 'i'=1..n1):y:=n*10^n1+s:z:=factorset(y):n2:=nops(z):d:=z[1]:printf(`%d, `,d):od:
-
Mathematica
d[n_]:=IntegerDigits[n];Table[FactorInteger[FromDigits[Join[x=d[n],Reverse[x]]]][[1,1]],{n,1,100}] Table[FactorInteger[#][[1,1]]&/@Select[Range[10^n,10^(n+1)-1],PalindromeQ],{n,1,3,2}]//Flatten (* Harvey P. Dale, Jul 19 2021 *)
-
Python
from sympy import primefactors def a(n): s = str(n); return min(primefactors(int(s + s[::-1]))) print([a(n) for n in range(1, 83)]) # Michael S. Branicky, Nov 02 2021
Comments