[seqfan] Re: Atomic number of n-th element in the "neptunium series"

Frank Adams-watters franktaw at netscape.net
Thu Apr 19 01:55:13 CEST 2018


If this is added, I think each term should be the atomic number of the most common element at that step.
This "stabilizes " the sequence: otherwise at step n, if you discovered their was another decay product 
occurring only with, say, probability 0.001%, the sequence would have to be changed so that a(n) = 0. 

Second, why Neptunium 237? Is there something special about it?

Franklin T. Adams-Watters


-----Original Message-----
From: Felix Fröhlich <felix.froe at gmail.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Wed, Apr 18, 2018 7:41 am
Subject: [seqfan] Atomic number of n-th element in the "neptunium series"

Dear SeqFans,here is an idea for a sequence related to radioactive decay of a chemical
element:
Atomic number of n-th element in the "neptunium series", the decay chain of
neptunium-237, or 0 if the n-th link of the chain consists of more than one
element. For an overview, please see
https://en.wikipedia.org/wiki/Decay_chain#Neptunium_series
The sequence starts 93, 91, 92, 90, 88, 89, 87, 85, 83, 0, 82, 83, 81a(10) 
(if the offset is 1) is 0, because bismuth-213 decays into
polonium-213 and thallium-209, both of which in turn decay into lead-209.
I don't know if this sequence is interesting. The "0" term may be a bit
unsatisfactory, but I am not sure of a better way to resolve the issue of an
 isotope decaying into two daughter isotopes at the moment.

Best regardsFelix Fröhlich

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