Fiddling around with some recursion and the dragon curve and what should pop out but this exact sequence.
https://oeis.org/A072339
While it wasn't at all obvious to me that it would match a sequence like this, it is easy to demonstrate the connection. Take the odd numbers and use them as right turns and the even numbers as left turns.
Shame I don't have access to Knuth's work to see if the connection is described there.
@Unprovable
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Going from the dragon curve to that sequence is slightly more tricky but only involves counting the number of times a turn is flipped left/right or right/left at each iteration.
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Maybe he didn't. :)
https://youtu.be/v678Em6qyzk
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[#]bbcmicrobot 馃殌MO.0:GC.3,1
P."Any number n can be written in the form n = 2^k1 - 2^k2 + 2^k3 - ... + 2^kx"'"where the signs alternate and k1 > k2 > k3 > ... >kx >= 0."'"E.G. a(6)=2 since 6=2^3-2^1. Sequence gives the minimal value of x."
V.28,59,12,79,4
DIMX(3),Y(3):D=0
X(0)=4:X(2)=-4:Y(1)=4:Y(3)=-4
MOVE359,719
REP.L=1+L
PL.1,X(D),Y(D)
N=L:T=N:C=0:R=1:Z=1
REP.T=T/2:C=C+1:U.T<=1
M=2^(C-1):E=2^C
REP.
IFN>M ANDN<E R=R+1:Z=2EORZ:N=E-N
E=E/2:M=M/2:U.M<2
D=(D+Z)MOD4
P.';"a("L") = "R;
U.0
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I ran @geoffl's program and got this.
Source: https://bbcmic.ro/?t=beHPj #bbcbasic
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[#]bbcmicrobot 馃幀MO.0:GC.3,1
P."Any number n can be written in the form n = 2^k1 - 2^k2 + 2^k3 - ... + 2^kx"'"where the signs alternate and k1 > k2 > k3 > ... >kx >= 0."'"E.G. a(6)=2 since 6=2^3-2^1. Sequence gives the minimal value of x."
V.28,59,12,79,4
DIMX(3),Y(3):D=0
X(0)=4:X(2)=-4:Y(1)=4:Y(3)=-4
MOVE359,719
REP.L=1+L
PL.1,X(D),Y(D)
N=L:T=N:C=0:R=1:Z=1
REP.T=T/2:C=C+1:U.T<=1
M=2^(C-1):E=2^C
REP.
IFN>M ANDN<E R=R+1:Z=2EORZ:N=E-N
E=E/2:M=M/2:U.M<2
D=(D+Z)MOD4
P.';"a("L") = "R;
U.0
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I ran @geoffl's program and got this.
Source: https://bbcmic.ro/?t=beHRB #bbcbasic
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@bbcmicrobot This is an easier, faster, and less memory intensive method to generate the same sequence than the code at https://oeis.org/A072339
It also generates the dragon curve as a side effect.
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[#]bbcmicrobot MO.1:@%=8:X=0:Y=1:P.'"Numbers that are the sum of two squares"'X;:REP.L=XX+YY:R=L*2+1:Q%=1+SQRL:P.L;:F.E%=1TOQ%:F.F%=Q%-E% TOQ%-(E%/3):T=E%*E%+F%*F%:IFT>L ANDT<R R=T:U=E%:V=F%
N.,:X=U:Y=V:U.0
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I ran @geoffl's program and got this.
Source: https://bbcmic.ro/?t=bfEzt #bbcbasic
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[#]bbcmicrobot @geoffl CLS:N%=1E4:DIMB%N%:@%=5:P.'"Numbers that are the sum of two squares"
F.X%=0TOSQRN%:T%=0:F.Y%=0TOX%:A%=S%+T%:IFA%<N%A%?B%=1:T%=T%+Y%-NOTY%:N.EL.Y%=X%:N.
S%=S%+X%-NOTX%:N.
F.L%=0TON%-1:IFL%?B%=1P.L%;
N.
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I ran @rheolism's program and got this.
Source: https://bbcmic.ro/?t=bfGyW #bbcbasic
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@bbcmicrobot The runtime here is (just) dominated by printing the answers - this completes in 44.50 seconds, but removing that PRINT it only takes 22.19.
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@rheolism #bbcmicrobot CLS:N%=1E4:DIMB%N%:@%=5:P.'"Numbers that are the sum of two squares"
F.X%=0TOSQRN%:T%=0:F.Y%=0TOX%:A%=S%+T%:IFA%<N%A%?B%=-1:T%=T%+Y%-NOTY%:N.EL.Y%=X%:N.
S%=S%+X%-NOTX%:N.
F.L%=B%TOB%+N%-1:IF?L%P.L%-B%;
N.
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I ran @geoffl's program and got this.
Source: https://bbcmic.ro/?t=bfMaz #bbcbasic
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@rheolism
Nice algorythm.
Setting the memory locations to -1 makes the last IF quicker as it's implicit true rather than =0, and as there are more misses than hits adding B% to the loop then subtracting it from the number printed is faster than adding B% to every lookup. Saving about 2.34 seconds.
Edit: Oops, that introduces an error that prints several false positives. Too tired to work out why right now.
Edit 2: The memory on tbe beeb isn"t initialised fully empty. Thats why.
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@rheolism
[#]bbcmicrobot MO.0:MO.7:N%=1E4:B%=&3000:@%=5:P.'"Numbers that are the sum of two squares"
F.X%=0TOSQRN%:T%=0:F.Y%=0TOX%:A%=S%+T%:IFA%<N%A%?B%=-1:T%=T%+Y%-NOTY%:N.EL.Y%=X%:N.
S%=S%+X%-NOTX%:N.
F.L%=B%TOB%+N%-1:IF?L%P.L%-B%;
N.
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I ran @geoffl's program and got this.
Source: https://bbcmic.ro/?t=bfMV2 #bbcbasic
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@geoffl #bbcmicrobot MO.0:V.22,7:N%=1E4:@%=5:P.'"Numbers that are the sum of two squares"
F.X%=0TOSQRN%:S%=X%*X%:F.Y%=0TOX%:A%=S%+Y%*Y%:IFA%<N%A%?H.=TRUE:N.,EL.N.X%
F.L%=0TON%-1:IFL%?H. P.L%;
N.
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I ran @rheolism's program and got this.
Source: https://bbcmic.ro/?t=bfOfM #bbcbasic
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@rheolism
[#]bbcmicrobot MO.0:V.22,4:N%=1E4:@%=5:P.'"Numbers that are the sum of two squares (by two or more different methods)."
F.X%=0TOSQRN%:S%=X%*X%:F.Y%=0TOX%:A%=S%+Y%*Y%:IFA%<N%A%?H.=-NOTA%?H.:N.,EL.N.X%
F.L%=0TON%-1:IFL%?H.>1P.L%;
N.
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I ran @geoffl's program and got this.
Source: https://bbcmic.ro/?t=bg1bM #bbcbasic
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[#]bbcmicrobot MO.0:V.22,7:w=!-52:F.O=0TO2S.2:P%=@%:[OPTO
SED
.l
LDA H.,X
BEQn
CLD
STX81
LDA#32
JSRw
TAX
LDA80
LSRA:LSRA:LSRA:LSRA
BEQz0
ORA#48:JSRw:LDX#48
BNEZ0
.z0:TXA:JSRw:.Z0
LDA80
AND#15
BEQz1
ORA#48:JSRw
BNEZ1
.z1:TXA:JSRw:.Z1
TYA
LSRA:LSRA:LSRA:LSRA
BEQz2
ORA#48:JSRw:LDX#48
BNEZ2
.z2:TXA:JSRw:.Z2
TYA
AND#15
ORA#48:JSRw
LDX81
SED
.n
TYA
CLC
ADC#1
TAY
BCCc
LDA80
ADC#0
BCSz
STA80
.c
INX
BNEl
INCl+2
BPLl
.z
CLD
RTS
]:N.
N%=1E4:P.'"Numbers that are the sum of two squares"
F.U%=0TOSQRN%:S%=U%*U%:F.V%=0TOU%:A%=S%+V%*V%:IFA%<N%A%?H.=TRUE:N.,EL.N.U%
CA.@%
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I ran @rheolism's program and got this.
Source: https://bbcmic.ro/?t=bg3jW #bbcbasic
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@bbcmicrobot The assembler printing loop increments a BCD counter as well as the binary counter for checking the array of flags so printing the value out is much simpler as each decimal digit is its own nibble. Handily (for doing this at least) INC/INX/etc ignore BCD mode so we only need to turn off BCD mode to call into the OS to write out characters.
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[#]bbcmicrobot MO.0:V.-233;96,144,32,64,240;0;22,6:N%=12288:@%=0:F.U%=1TOSQRN%:S%=U%*U%:IFS%?H. T%=S%?H.:P.TAB(H%)S%?(H.-T%*T%)"每+"T%"每="U%"每";:H%=(H%+13)MOD39:IF0EL.S%?H.=U%:F.V%=1TOU%:A%=S%+V%*V%:IFA%<N%A%?H.=U%:N.,EL.N.U%:P.
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I ran @rheolism's program and got this.
Source: https://bbcmic.ro/?t=bg8tg #bbcbasic
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[#]bbcmicrobot
MO.6:VDU-233;96,144,32,64,240;0;
P."Galloping Primitive Pythagorean Triples"'
W=0:X=0:Y=0:Z=1
FORI=0TO22
A=W*Z
B=2XY
C=WY+XZ
IFA P.;A"每+"B"每="C"每"
W=X:X=Y:Y=Z:Z=X+Y
N.
@rheolism
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I ran @geoffl's program and got this.
Source: https://bbcmic.ro/?t=bl5mH #bbcbasic
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[#]bbcmicrobot @geoffl MO.6:?584=9:V.-233;96,144,32,64,240;0;:P."Galloping Primitive Pythagorean Triples"':@%=0:Z%=1:F.I=0TO23:W%=X%:X%=Y%:Y%=Z%:Z%=X%+Y%:IFW%P.W%Z%"每+"2X%*Y%"每="W%*Y%+X%*Z%"每":N.EL.N.
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I ran @rheolism's program and got this.
Source: https://bbcmic.ro/?t=bl79K #bbcbasic
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text/gemini