A212875 Primonacci numbers: composite numbers that appear in the Fibonacci-like sequence generated by their own prime factors.

4, 9, 12, 25, 27, 169, 1102, 7921, 22287, 54289, 103823, 777627, 876897, 2550409, 20854593, 34652571, 144237401, 144342653, 167901581, 267911895, 792504416, 821223649, 1103528482, 2040412557, 2852002829, 3493254541, 6033671841, 15658859018, 116085000401

Offset: 1

Author

Herman Beeksma, May 29 2012

Keywords

• nonn

Comments

Given n, form a sequence that starts with the k prime factors of n in ascending order. After that, each term is the sum of the preceding k terms. If n eventually appears in the sequence, it is a primonacci number. Primes possess this property trivially and are therefore excluded.

Similar to A007629 (repfigit or Keith numbers), but base-independent. If n is in A005478 (Fibonacci primes), then n^2 is a primonacci number.

The only entries that are semiprimes (A001358) are the squares of A005478. - Robert Israel, Mar 08 2016

Examples

			Fibonacci-like sequences for selected values of n:
n=12: 2, 2, 3, 7, 12, ...
n=25: 5, 5, 10, 15, 25, ...
n=1102: 2, 19, 29, 50, 98, 177, 325, 600, 1102, ...
		

Links

• Herman Beeksma, Table of n, a(n) for n = 1..42

Crossrefs

Cf. A000045, A001358, A005478, A007629.

Programs

Maple

with(numtheory): P:=proc(q,h) local a,b,j,k,n,t,v; v:=array(1..h);
for n from 2 to q do if not isprime(n) then b:=ifactors(n)[2]; a:=[];
for k from 1 to nops(b) do for j from 1 to b[k][2] do a:=[op(a),b[k][1]]; od; od; a:=sort([op(a)]);
b:=nops(a);  for k from 1 to b do v[k]:=a[k]; od; t:=b+1; v[t]:=add(v[k], k=1..b);
while v[t]Paolo P. Lava, Mar 08 2016

Mathematica

PrimonacciQ[n_]:=Module[{k,seq},
  seq=FactorInteger[n];
  seq=Map[Table[#[[1]],{#[[2]]}]&, seq];
  seq=Flatten[seq];
  k=Length[seq];
  If[k==1,Return[False]];
  seq=Append[seq,Apply[Plus,seq]];
  While[seq[[-1]]Michael De Vlieger, Mar 08 2016 *)

Python

from sympy import isprime, factorint
from itertools import chain
A212875_list = []
for n in range(2,10**6):
    if not isprime(n):
        x = sorted(chain.from_iterable([p]*e for p,e in factorint(n).items()))
        y = sum(x)
        while y < n:
            x, y = x[1:]+[y], 2*y-x[0]
        if y == n:
            A212875_list.append(n) # Chai Wah Wu, Sep 12 2014