A285100 Numbers k for which A065642(k) = A285109(k).

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 41, 42, 43, 44, 46, 47, 49, 51, 52, 53, 55, 57, 58, 59, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 76, 77, 78, 79, 81, 82, 83, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 97, 99, 101

Offset: 1

Author

Antti Karttunen, Apr 19 2017

Keywords

• nonn

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Complement: A284342.

Positions of ones in A285337.

Subsequences: A000961, A005117.

Cf. A065642, A285109.

Programs

PARI

A020639(n) = if(1==n,n,vecmin(factor(n)[, 1]));
A007947(n) = factorback(factorint(n)[, 1]); \\ From Andrew Lelechenko, May 09 2014
A065642(n) = { my(r=A007947(n)); if(1==n,n,n = n+r; while(A007947(n) <> r, n = n+r); n); };
isA285100(n) = (A065642(n) == n*A020639(n));
n=0; k=1; while(k <= 10000, n=n+1; if(isA285100(n),write("b285100.txt", k, " ", n);k=k+1));

Python

from operator import mul
from sympy import primefactors
from functools import reduce
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a020639(n): return 1 if n==1 else primefactors(n)[0]
def a065642(n):
    if n==1: return 1
    r=a007947(n)
    n += r
    while a007947(n)!=r:
        n+=r
    return n
print([n for n in range(1, 102) if a065642(n) == n*a020639(n)]) # Indranil Ghosh, May 24 2017

Scheme

;; With Antti Karttunen's IntSeq-library.
(define A285100 (MATCHING-POS 1 1 (lambda (n) (= (A065642 n) (A285109 n)))))