A171798 a(n) = base-10 concatenation XYZ, where X = number of bits in binary expansion of n, Y = number of 0's, Z = number of 1's.

101, 211, 202, 321, 312, 312, 303, 431, 422, 422, 413, 422, 413, 413, 404, 541, 532, 532, 523, 532, 523, 523, 514, 532, 523, 523, 514, 523, 514, 514, 505, 651, 642, 642, 633, 642, 633, 633, 624, 642, 633, 633, 624, 633, 624, 624, 615, 642, 633, 633, 624

Offset: 1

Author

N. J. A. Sloane, Oct 15 2010, Oct 16 2010

Keywords

• nonn
• base
• easy

Comments

Start with n, repeatedly apply the map i -> a(i). Then every n converges to one of 1019, 1147, 1165 or 14311 (cf. A171813). Proof: this is true by direct calculation for n=1..2^14. For larger n, a(n) < n.

Examples

			14 = 1110 in base 2, so X=4, Y=1, Z=3, a(14)=413.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A171797, A171813, A171823, A171825.

Cf. A070939, A023416, A000120.

Programs

Haskell

a171798 n = read $ concatMap (show . ($ n))
                   [a070939, a023416, a000120] :: Integer
-- Reinhard Zumkeller, Feb 22 2012

Maple

# Maple code for trajectories of numbers from 1 to M:
F:=proc(n) local s,t1,t2; t1:=convert(n,base,2); t2:=nops(t1); s:=add(t1[i],i=1..t2);
parse(cat(t2,t2-s,s)); end;
M:=16384;
for n from 1 to M do t3:=F(n); sw:=-1;
for i from 1 to 10 do
if (t3 = 1147) or (t3 = 1165) or (t3 = 1019) or (t3 = 14311) then sw:=1; break; fi;
t3:=F(t3);
od;
if sw < 0 then lprint(n); fi;
od:
Contribution from R. J. Mathar, Oct 15 2010: (Start)
read("transforms") ; cat2 := proc(a,b) dgsb := max(1,ilog10(b)+1) ; a*10^dgsb+b ; end proc:
catL := proc(L) local a; a := op(1,L) ; for i from 2 to nops(L) do a := cat2(a,op(i,L)) ; end do; a; end proc:
A070939 := proc(n) max(1,ilog2(n)+1) ; end proc:
A171798 := proc(n) local n1,n3 ; n1 := A070939(n) ; n3 := wt(n) ; catL([n1,n1-n3,n3]) ; end proc:
seq(A171798(n),n=1..80) ; (End)

Mathematica

ans[n_]:=Module[{idn2=IntegerDigits[n,2]},FromDigits[{Length[idn2],Count[idn2,0],Count[idn2,1]}]]; Table[ans[i], {i, 50}] (* Harvey P. Dale, Nov 06 2010 *)

Python

def a(n):
    b = bin(n)[2:]
    z = b.count("0")
    return int(str(len(b)) + str(z) + str(len(b)-z))
print([a(n) for n in range(1, 52)]) # Michael S. Branicky, Mar 28 2022

Extensions

More terms from R. J. Mathar, Oct 15 2010