login
A346953
a(n) is the number of divisors of A346950(n) ending with 3.
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
1,2
COMMENTS
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
LINKS
EXAMPLE
a(17) = 4 since there are 4 divisors of A346950(17) = 429 ending with 3: 3, 13, 33 and 143.
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
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
PROG
(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
CROSSREFS
Cf. A000005, A017377, A346388 (ending with 5), A346389 (ending with 6), A346950, A346951, A346952.
Sequence in context: A087010 A098220 A297032 * A235508 A271778 A326702
KEYWORD
nonn,base
AUTHOR
Stefano Spezia, Aug 08 2021
STATUS
approved