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A346953 a(n) is the number of divisors of A346950(n) ending with 3.

%I A346953 #15 Nov 04 2024 01:40:05
%S A346953 1,2,2,2,2,2,1,2,2,2,2,2,2,2,2,2,4,2,2,2,1,2,2,2,2,2,2,2,2,2,4,2,4,2,
%T A346953 2,2,2,2,2,2,2,2,2,2,3,2,2,2,4,2,2,2,2,2,2,2,2,4,4,2,2,2,2,2,4,2,2,2,
%U A346953 2,2,2,4,2,2,2,1,2,2,2,2,2,2,4,2,2,4,2,2
%N A346953 a(n) is the number of divisors of A346950(n) ending with 3.
%C A346953 a(n) = 1 if A346950(n) = k^2 where k is either a prime ending with 3 or the product of a prime ending with 7 and a prime ending with 9. - _Robert Israel_, Nov 03 2024
%H A346953 Robert Israel, <a href="/A346953/b346953.txt">Table of n, a(n) for n = 1..10000</a>
%e A346953 a(17) = 4 since there are 4 divisors of A346950(17) = 429 ending with 3: 3, 13, 33 and 143.
%p A346953 N:= 10000: # for a(1) .. a(M) where the last term of A346950 less than N is A346950(M)
%p A346953 S:= {}:
%p A346953 for n from 3 to floor(sqrt(N)) by 10 do
%p A346953   S:= S union map(`*`, {seq(i,i= n .. floor(N/n), 10)},n)
%p A346953 od:
%p A346953 S:= sort(convert(S,list)):
%p A346953 map(t -> nops(select(t -> t mod 10 = 3, numtheory:-divisors(t))), S); # _Robert Israel_, Nov 03 2024
%t A346953 b={}; For[n=0, n<=450, n++, For[k=0, k<=n, k++, If[Mod[10*n+9, 10*k+3]==0 && Mod[(10*n+9)/(10*k+3), 10]==3 && 10*n+9>Max[b], AppendTo[b, 10*n+9]]]]; (* A346950 *) a={}; For[i =1, i<=Length[b], i++, AppendTo[a, Length[Drop[Select[Divisors[Part[b, i]], (Mod[#, 10]==3&)]]]]]; a
%o A346953 (Python)
%o A346953 from sympy import divisors
%o A346953 def f(n): return sum(d%10 == 3 for d in divisors(n)[1:-1])
%o A346953 def A346950upto(lim): return sorted(set(a*b for a in range(3, lim//3+1, 10) for b in range(a, lim//a+1, 10)))
%o A346953 print(list(map(f, A346950upto(2129)))) # _Michael S. Branicky_, Aug 11 2021
%Y A346953 Cf. A000005, A017377, A346388 (ending with 5), A346389 (ending with 6), A346950, A346951, A346952.
%K A346953 nonn,base
%O A346953 1,2
%A A346953 _Stefano Spezia_, Aug 08 2021