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A277324 Odd bisection of A260443 (the even terms): a(n) = A260443((2*n)+1).

%I A277324 #41 Mar 12 2021 07:53:17
%S A277324 2,6,18,30,90,270,450,210,630,6750,20250,9450,15750,47250,22050,2310,
%T A277324 6930,330750,3543750,1653750,4961250,53156250,24806250,727650,1212750,
%U A277324 57881250,173643750,18191250,8489250,25467750,2668050,30030,90090,40020750,1910081250,891371250,9550406250,455814843750,212713593750
%N A277324 Odd bisection of A260443 (the even terms): a(n) = A260443((2*n)+1).
%C A277324 From _David A. Corneth_, Oct 22 2016: (Start)
%C A277324 The exponents of the prime factorization of a(n) are first nondecreasing, then nonincreasing.
%C A277324 The exponent of 2 in the prime factorization of a(n) is 1. (End)
%H A277324 Antti Karttunen, <a href="/A277324/b277324.txt">Table of n, a(n) for n = 0..1024</a>
%F A277324 a(n) = A260443((2*n)+1).
%F A277324 a(0) = 2; for n >= 1, a(n) = A260443(n) * A260443(n+1).
%F A277324 Other identities. For all n >= 0:
%F A277324 A007949(a(n)) = A005811(n). [See comments in A125184.]
%F A277324 A156552(a(n)) = A277189(n), a(n) = A005940(1+A277189(n)).
%F A277324 A048675(a(n)) = 2n + 1. - _David A. Corneth_, Oct 22 2016
%F A277324 A001222(a(n)) = A007306(1+n).
%F A277324 A056169(a(n)) = A284267(n).
%F A277324 A275812(a(n)) = A284268(n).
%F A277324 A248663(a(n)) = A283975(n).
%F A277324 A000188(a(n)) = A283484(n).
%F A277324 A247503(a(n)) = A284563(n).
%F A277324 A248101(a(n)) = A284564(n).
%F A277324 A046523(a(n)) = A284573(n).
%F A277324 a(n) = A277198(n) * A284008(n).
%F A277324 a(n) = A284576(n) * A284578(n) = A284577(n) * A000290(A284578(n)).
%e A277324 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
%t A277324 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 *)
%o A277324 (Scheme, two versions)
%o A277324 (define (A277324 n) (A260443 (+ 1 n n)))
%o A277324 (define (A277324 n) (if (zero? n) 2 (* (A260443 n) (A260443 (+ 1 n)))))
%o A277324 (Python)
%o A277324 from sympy import factorint, prime, primepi
%o A277324 from operator import mul
%o A277324 def a003961(n):
%o A277324     F=factorint(n)
%o A277324     return 1 if n==1 else reduce(mul, [prime(primepi(i) + 1)**F[i] for i in F])
%o A277324 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)
%o A277324 def a(n): return a260443(2*n + 1)
%o A277324 print([a(n) for n in range(101)]) # _Indranil Ghosh_, Jun 21 2017
%Y A277324 Cf. A005811, A005940, A007306, A007949, A156552, A260443, A277189, A277323, A283484, A283975, A284267, A284268, A284563, A284564, A284573.
%Y A277324 Cf. A277200 (same sequence sorted into ascending order).
%K A277324 nonn
%O A277324 0,1
%A A277324 _Antti Karttunen_, Oct 10 2016
%E A277324 More linking formulas added by _Antti Karttunen_, Apr 16 2017