[seqfan] Re: phi-antipalindromic numbers

[Vladimir Shevelev]

Excellent! 
 
I see that you use
 1) the additive property over powers of 2: if 0<=i<j<...<k, then  
A171071(2^k+...+2^j+2^i)= A171071(2^k)+ ...+A171071(2^j)+ A171071(2^i)
and
2) A171071(2^t)=F(2*t+1)+F(2*t+3).
 
Regards,
Vladimir



----- Original Message -----
From: Hans Havermann <pxp at rogers.com>
Date: Wednesday, September 8, 2010 3:02
Subject: [seqfan] Re: phi-antipalindromic numbers
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Douglas McNeil:
> 
> > FWIW, I find 25983 such phi-antipalindromic numbers less than 10^6,
> > and I agree with Hans' half-sequence. They have a cute pattern,
> > probably related to the observations in the comments.
> 
> I see that Neil has added my two bisections of A178482 
> (A171070,  
> A171071). Differences of adjacent terms of either bisection are 
> terms  
> of A107328 (which appears to be A065034 with an initial 3 added) 
> and I  
> have used that fact to translate the "cute" pattern into code 
> that  
> will calculate term #n of A171071 fairly quickly - without 
> first  
> calculating any of the previous terms. Assuming that I've done 
> it  
> correctly,
> 
> term #1 = 3
> term #10 = 54
> term #100 = 1183
> term #1000 = 24324
> term #10000 = 771278
> term #100000 = 17680149
> term #1000000 = 364243723
> term #10000000 = 11583057054
> term #100000000 = 238175057907
> term #1000000000 = 5402072437445
> term #10000000000 = 173192141696518
> term #100000000000 = 3578032635908428
> term #1000000000000 = 81149986784320483
> term #10000000000000 = 2600997386890746294
> term #100000000000000 = 53482527632200745828
> term #1000000000000000 = 1213745387417082498749
> 
> term #10^100 =  
> 16243248183500325534450089584837853567324809329361973207684488292891935207090205003873914101783289817252931340703388640057477985409132564969
> 
> 
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 Shevelev Vladimir‎