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.
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
Offset: 1
Examples
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.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..10000
Programs
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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 *) -
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)
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
Comments