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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. 3
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, 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, 19, 29, 59, 30, 61, 31, 7, 1, 13, 11, 67, 17, 23, 7, 71, 1, 73, 37 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,7
COMMENTS
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.
LINKS
FORMULA
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.
Sum_{k=1..n} a(k) ~ (31/144) * n^2. - Amiram Eldar, Dec 25 2023
EXAMPLE
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.
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.
MAPLE
a:= proc(n) local i, m, r; m:=n;
for i from 0 while irem(m, 30, 'r')=0 do m:=r od;
while irem(m, 2, 'r')=0 do m:=r od;
while irem(m, 3, 'r')=0 do m:=r od;
while irem(m, 5, 'r')=0 do m:=r od;
m*30^i
end:
seq(a(n), n=1..100); # Alois P. Heinz, Jul 04 2013
MATHEMATICA
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 *)
PROG
(Sage)
n=100 #change n for more terms
C=[]
b=30
P = factor(b)
for i in [1..n]:
prod = 1
for j in range(len(P)):
prod = prod * ((P[j][0])^(Integer(i).valuation(P[j][0])))
C.append((b^(Integer(i).valuation(b)) * i) /prod)
CROSSREFS
Sequence in context: A318674 A284118 A165725 * A327670 A178637 A364092
KEYWORD
easy,nonn
AUTHOR
Daniel Juda, Jul 25 2012
STATUS
approved

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Last modified April 25 07:53 EDT 2024. Contains 371964 sequences. (Running on oeis4.)