%I #20 Jul 24 2024 15:18:22
%S 1,2,2,3,2,4,2,4,3,2,4,4,6,4,8,4,8,6,2,4,3,2,4,2,4,6,2,2,4,6,8,8,6,18,
%T 4,10,8,6,4,2,4,6,4,2,6,2,2,4,6,12,8,8,12,4,10,8,9,8,4,4,12,4,4,8,4,6,
%U 12,8,8,16,4,12,8,10,12,9,8,16,4,4,2,4,6,2,4,8,2,2,8,4,18,4,10,16,12,4
%N a(n) is the number of divisors of the n-th positive palindromic number.
%H Harvey P. Dale, <a href="/A076888/b076888.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = A000005(A002113(n+1)).
%e a(11) = 4 because there are 4 divisors of 11th positive palindromic number (i.e., 22).
%t palQ[n_]:=Module[{idn=IntegerDigits[n]},idn==Reverse[idn]]; DivisorSigma[ 0,#]&/@ Select[Range[1000],palQ] (* _Harvey P. Dale_, Nov 29 2014 *)
%o (Python)
%o from sympy import divisor_count
%o def A076888(n):
%o y = 10*(x:=10**(len(str(n+1>>1))-1))
%o return divisor_count(int((c:=n+1-x)*x+int(str(c)[-2::-1] or 0) if n+1<x+y else (c:=n+1-y)*y+int(str(c)[::-1] or 0))) # _Chai Wah Wu_, Jul 24 2024
%Y Cf. A000005, A002113.
%K base,nonn
%O 1,2
%A _Shyam Sunder Gupta_, Nov 25 2002