greedy computation of pseudoperfect numbers?

Richard Mathar mathar at strw.leidenuniv.nl
Wed Apr 9 19:17:15 CEST 2008


setsum := proc(S)
	add(i,i=S) ;
	for anxt from op(-1,a)+1 do
		if isA001358(anxt) then
			aset := combinat[powerset](convert(a,set) union {anxt} );
			sset := {} ;
			for s in aset do
				sset := sset union { setsum(s) } ;
			od:
			if nops(sset) = nops(aset) then
				a := [op(a),anxt] ;
				print(a) ;
				break ;
			fi ;
		fi ;
	od:
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Message-ID: <5542af940804091133l45d9f087v82dc470a483e9564 at mail.gmail.com>
Date: Wed, 9 Apr 2008 11:33:41 -0700
From: "Jonathan Post" <jvospost3 at gmail.com>
To: Lallouet <philip.lallouet at orange.fr>, "T. D. Noe" <noe at sspectra.com>
Subject: Re: prime Hyperfibonacci numbers
Cc: seqfan at ext.jussieu.fr
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Thank you!

Here, as I promised (warned) is the first cut at prime hyperlucas
numbers (which are not prime Lucas numbers), i.e. A000040 INTERSECTION
A137176 Hyperlucas number array read by antidiagonals, with k>0 and
n>2:

k=1: {n=7: 73, 1361, 24473, 7881193, ...} = primes in A027961

k=2: {n=3: 13, n=9: 487, 1149769, ...} = primes in A023537

k=3: {n=3: 19, n=4: 47, n=5:101, n=6: 199, n=8: 661, 23993, 12750611,
33383659, ...} = primes in A027963

k=4: {n=4: 73, n=6: 373, n=7: 743, 22193, 33369709, ...} = primes in A027964

k=5: {n=4: 107, n=5: 281, n=8: 2801, n=10: 9859, 30869, 150833, ...} =
primes in A053298

k=6: {n=3: 43, n=5:431, 5283, ...}

k=7.: {n=3: 53, n=7:4201, ...}

and so forth.

I guess that I should firest find if any of these hyperlucas numbers
with k>0 and n>2 are in fact Lucas numbers, to avoid an ambiguity in
definition.

Again, one might speculate that there are a finite number of elements
in each row and column of the hyperlucas array (so long as k>0 and
n>2.  As I recall, we don't know if there are an infinite number of
prime lucas numbers (k=0).  Again, referencing  Ayhan Dil and Istvan
Mezo, Lucas numbers = A000204 (not the only version in OEIS).

And one might wonder which numbers are nontrivially both
hyperfibonacci and hyperlucas (without being Fibonacci nor Lucas).

Best,

Jonathan Vos Post





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