cp's OEIS Frontend

A214685 a(n) is obtained from n by removing 2s, 3s, and 5s from the prime factorization of n that do not contribute to a factor of 30.

%I A214685 #40 Dec 25 2023 01:51:50
%S A214685 1,1,1,1,1,1,7,1,1,1,11,1,13,7,1,1,17,1,19,1,7,11,23,1,1,13,1,7,29,30,
%T A214685 31,1,11,17,7,1,37,19,13,1,41,7,43,11,1,23,47,1,49,1,17,13,53,1,11,7,
%U A214685 19,29,59,30,61,31,7,1,13,11,67,17,23,7,71,1,73,37
%N A214685 a(n) is obtained from n by removing 2s, 3s, and 5s from the prime factorization of n that do not contribute to a factor of 30.
%C A214685 In this sequence, the number 30 exhibits characteristics of a prime number.  It exhibits characteristics of a prime number since all extraneous 2s, 3s, and 5s have been removed from the prime factorizations of all of the numbers.
%H A214685 Alois P. Heinz, <a href="/A214685/b214685.txt">Table of n, a(n) for n = 1..10000</a>
%F A214685 a(n) = (n*30^(v_30(n)))/(2^(v_2(n))*3^(v_3(n))*5^(v_5(n))), where v_k(n) is the k-adic valuation of n. That is, v_k(n) is the largest power of k, a, such that k^a divides n.
%F A214685 Sum_{k=1..n} a(k) ~ (31/144) * n^2. - _Amiram Eldar_, Dec 25 2023
%e A214685 n=15, v_2(15)=0, v_3(15)=1, v_5(15)=1, v_30(15)=0, so a(15) = 30^0*15/(2^0*3^1*5^1) = 1.
%e A214685 n=60, v_2(60)=2, v_3(60)=1, v_5(60)=1, v_30(60)=1, so a(60) = 30^1*60/(2^2*3^1*5^1) = 30.
%p A214685 a:= proc(n) local i, m, r; m:=n;
%p A214685       for i from 0 while irem(m, 30, 'r')=0 do m:=r od;
%p A214685       while irem(m, 2, 'r')=0 do m:=r od;
%p A214685       while irem(m, 3, 'r')=0 do m:=r od;
%p A214685       while irem(m, 5, 'r')=0 do m:=r od;
%p A214685       m*30^i
%p A214685     end:
%p A214685 seq(a(n), n=1..100);  # _Alois P. Heinz_, Jul 04 2013
%t A214685 With[{v = IntegerExponent}, a[n_] := n*30^v[n, 30]/2^v[n, 2]/3^v[n, 3]/5^v[n, 5]; Array[a, 100]] (* _Amiram Eldar_, Dec 09 2020 *)
%o A214685 (Sage)
%o A214685 n=100 #change n for more terms
%o A214685 C=[]
%o A214685 b=30
%o A214685 P = factor(b)
%o A214685 for i in [1..n]:
%o A214685     prod = 1
%o A214685     for j in range(len(P)):
%o A214685         prod = prod * ((P[j][0])^(Integer(i).valuation(P[j][0])))
%o A214685     C.append((b^(Integer(i).valuation(b)) * i) /prod)
%Y A214685 Cf. A214681, A214682.
%K A214685 easy,nonn
%O A214685 1,7
%A A214685 _Daniel Juda_, Jul 25 2012