A346953 a(n) is the number of divisors of A346950(n) ending with 3.
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, 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, 2, 2, 2, 4, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 4, 2, 2, 4, 2, 2
Offset: 1
Examples
a(17) = 4 since there are 4 divisors of A346950(17) = 429 ending with 3: 3, 13, 33 and 143.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Maple
N:= 10000: # for a(1) .. a(M) where the last term of A346950 less than N is A346950(M) S:= {}: for n from 3 to floor(sqrt(N)) by 10 do S:= S union map(`*`, {seq(i,i= n .. floor(N/n), 10)},n) od: S:= sort(convert(S,list)): map(t -> nops(select(t -> t mod 10 = 3, numtheory:-divisors(t))), S); # Robert Israel, Nov 03 2024
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Mathematica
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
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Python
from sympy import divisors def f(n): return sum(d%10 == 3 for d in divisors(n)[1:-1]) 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))) print(list(map(f, A346950upto(2129)))) # Michael S. Branicky, Aug 11 2021
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