OFFSET
0,1
COMMENTS
From David A. Corneth, Oct 22 2016: (Start)
The exponents of the prime factorization of a(n) are first nondecreasing, then nonincreasing.
The exponent of 2 in the prime factorization of a(n) is 1. (End)
LINKS
Antti Karttunen, Table of n, a(n) for n = 0..1024
FORMULA
a(n) = A260443((2*n)+1).
Other identities. For all n >= 0:
A048675(a(n)) = 2n + 1. - David A. Corneth, Oct 22 2016
EXAMPLE
A method to find terms of this sequence, explained by an example to find a(7). To find k = a(7), we find k such that A048675(k) = 2*7+1 = 15. 7 has the binary partitions: {[7, 0, 0], [5, 1, 0], [3, 2, 0], [1, 3, 0], [3, 0, 1], [1, 1, 1]}. To each of those, we prepend a 1. This gives the binary partitions of 15 starting with a 1. For example, for the first we get [1, 7, 0, 0]. We see that only [1, 5, 1, 0], [1, 3, 2, 0] and [1, 1, 1, 1] start nondecreasing, then nonincreasing, so we only check those. These numbers will be the exponents in a prime factorization. [1, 5, 1, 0] corresponds to prime(1)^1 * prime(2)^5 * prime(3)^1 * prime(4)^0 = 2430. We find that [1, 1, 1, 1] gives k = 210 for which A048675(k) = 15 so a(7) = 210. - David A. Corneth, Oct 22 2016
MATHEMATICA
a[n_] := a[n] = Which[n < 2, n + 1, EvenQ@ n, Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &@ a[n/2], True, a[#] a[# + 1] &[(n - 1)/2]]; Table[a[2 n + 1], {n, 0, 38}] (* Michael De Vlieger, Apr 05 2017 *)
PROG
(Scheme, two versions)
(Python)
from sympy import factorint, prime, primepi
from operator import mul
def a003961(n):
F=factorint(n)
return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**F[i] for i in F])
def a260443(n): return n + 1 if n<2 else a003961(a260443(n//2)) if n%2==0 else a260443((n - 1)//2)*a260443((n + 1)//2)
def a(n): return a260443(2*n + 1)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 21 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Oct 10 2016
EXTENSIONS
More linking formulas added by Antti Karttunen, Apr 16 2017
STATUS
approved