A171813 -id:A171813 - cp's OEIS frontend

A171797 A modified Sisyphus function: a(n) = concatenation of (number of digits in n) (number of even digits) (number of odd digits).

110, 101, 110, 101, 110, 101, 110, 101, 110, 101, 211, 202, 211, 202, 211, 202, 211, 202, 211, 202, 220, 211, 220, 211, 220, 211, 220, 211, 220, 211, 211, 202, 211, 202, 211, 202, 211, 202, 211, 202, 220, 211, 220, 211, 220, 211, 220, 211, 220, 211, 211, 202

Offset: 0

N. J. A. Sloane, Oct 15 2010

Start with n, repeatedly apply the map i -> a(i). Then every number converges to 312. - Eric Angelini and Alexandre Wajnberg, Oct 15 2010

			11 has 2 digits, both odd, so a(11) = 202.
12 has 2 digits, one even and one odd, so a(12)=211. Then a(211) = 312.

M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Reinhard Zumkeller, Table of n, a(n) for n = 0..25000

Cf. A073053 (Sisyphus), A171798, A171813, A055642, A196563, A196564, A308002, A308003 (another version).

A100961 gives steps to reach 312.

a171797 n = read $ concatMap (show . ($ n))
                   [a055642, a196563, a196564] :: Integer
-- Reinhard Zumkeller, Feb 22 2012, Oct 15 2010

nevenDgs := proc(n) local a, d; a := 0 ; for d in convert(n,base,10) do if type(d,'even') then a :=a +1 ; end if; end do; a ; end proc:
cat2 := proc(a,b) local ndigsb; ndigsb := max(ilog10(b)+1,1) ; a*10^ndigsb+b ; end:
catL := proc(L) local a, i; a := op(1,L) ; for i from 2 to nops(L) do a := cat2(a,op(i,L)) ; end do; a; end proc:
A055642 := proc(n) max(1,ilog10(n)+1) ; end proc:
A171797 := proc(n) local n1,n2 ; n1 := A055642(n) ; n2 := nevenDgs(n) ; catL([n1,n2,n1-n2]) ; end proc:
seq(A171797(n),n=1..80) ; # R. J. Mathar, Oct 15 2010 and Oct 18 2010

def a(n):
    s = str(n); e = sum(d in "02468" for d in s)
    return int("".join(map(str, (len(s), e, len(s)-e))))
print([a(n) for n in range(52)]) # Michael S. Branicky, Jun 15 2021

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

a(0) added by N. J. A. Sloane, May 12 2019

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.

Original entry on oeis.org

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

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

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.

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

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

Cf. A171797, A171813, A171823, A171825.

Cf. A070939, A023416, A000120.

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

# 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)

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 *)

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

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

A308003 A modified Sisyphus function: a(n) = concatenation of (number of even digits in n) (number of digits in n) (number of odd digits in n).

Original entry on oeis.org

110, 11, 110, 11, 110, 11, 110, 11, 110, 11, 121, 22, 121, 22, 121, 22, 121, 22, 121, 22, 220, 121, 220, 121, 220, 121, 220, 121, 220, 121, 121, 22, 121, 22, 121, 22, 121, 22, 121, 22, 220, 121, 220, 121, 220, 121, 220, 121, 220, 121, 121, 22, 121, 22, 121

Offset: 0

N. J. A. Sloane, May 12 2019

If we start with n and repeatedly apply the map i -> a(i), we eventually reach 132 (see A073054).

			11 has 2 digits, both odd, so a(11)=22 (leading zeros are omitted).
12 has 2 digits, one even and one odd, so a(12)=121. Then a(121) = 132.

M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

N. J. A. Sloane, Table of n, a(n) for n = 0..28000

Cf. A073053 (Sisyphus), A171797, A171798, A171813, A055642, A196563, A196564, A308002.

A073054 gives steps to reach 132.

# Maple code based on R. J. Mathar's code for A171797:
nevenDgs := proc(n) local a, d; a := 0 ; for d in convert(n,base,10) do if type(d,'even') then a :=a +1 ; end if; end do; a ; end proc:
cat2 := proc(a,b) local ndigsb; ndigsb := max(ilog10(b)+1,1) ; a*10^ndigsb+b ; end:
catL := proc(L) local a, i; a := op(1,L) ; for i from 2 to nops(L) do a := cat2(a,op(i,L)) ; end do; a; end proc:
A055642 := proc(n) max(1,ilog10(n)+1) ; end proc:
A308003 := proc(n) local n1,n2 ; n1 := A055642(n) ; n2 := nevenDgs(n) ; catL([n2,n1,n1-n2]) ; end proc:
seq(A308003(n),n=0..80) ;

def a(n):
    s = str(n)
    e = sum(1 for c in s if c in "02468")
    return int(str(e) + str(len(s)) + str(len(s)-e))
print([a(n) for n in range(55)]) # Michael S. Branicky, Mar 29 2022

A350709 Modified Sisyphus function of order 3: a(n) is the concatenation of (number of digits of n)(number digits of n congruent to 0 modulo 3)(number of digits of n congruent to 1 modulo 3)(number of digits of n congruent to 2 modulo 3).

Original entry on oeis.org

1100, 1010, 1001, 1100, 1010, 1001, 1100, 1010, 1001, 1100, 2110, 2020, 2011, 2110, 2020, 2011, 2110, 2020, 2011, 2110, 2101, 2011, 2002, 2101, 2011, 2002, 2101, 2011, 2002, 2101, 2200, 2110, 2101, 2200, 2110, 2101, 2200

Offset: 0

Matthew E. Coppenbarger, Mar 28 2022

If we start with n and repeatedly apply the map i -> a(i), we eventually get the cycle {4031, 4112, 4220}

			11 has two digits, both congruent to 1 modulo 3, so a(11) = 2020.
a(20) = 2101.
a(30) = 2200.
a(1111123567) = 10262.

M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Cf. A073053 (Sisyphus), A171797, A171798, A171813, A055642, A196563, A196564, A308002.

def a(n):
    d, m = list(map(int, str(n))), [0, 0, 0]
    for di in d: m[di%3] += 1
    return int(str(len(d)) + "".join(map(str, m)))
print([a(n) for n in range(37)]) # Michael S. Branicky, Mar 28 2022

A171823 a(n) = base-2 concatenation XYZ, where X = number of bits in binary expansion of n, Y = number of 0's, Z = number of 1's.

Original entry on oeis.org

101, 1011, 10010, 11101, 11110, 11110, 11011, 100111, 1001010, 1001010, 100111, 1001010, 100111, 100111, 1000100, 1011001, 1011110, 1011110, 1011011, 1011110, 1011011, 1011011, 1011100, 1011110, 1011011, 1011011, 1011100, 1011011, 1011100, 1011100

Offset: 1

N. J. A. Sloane, Oct 16 2010

			14 = 1110 in base 2, so X = 4 = 100, Y = 1, Z = 3 = 11, a(14) = 100.1.11 = 100111.

Cf. A171825, A171798, A171813.

F:=proc(n) local t1,t2,t2b,n1,n1b,n0,n0b,t3,t4;
t1:=convert(n,base,2);
t2:=nops(t1);
t2b:=convert(t2,base,2);
n1:=add(t1[i],i=1..t2);
n1b:=convert(n1,base,2);
n0:=t2-n1;
n0b:=convert(n0,base,2);
t3:=[
seq(t2b[nops(t2b)+1-i],i=1..nops(t2b)),
seq(n0b[nops(n0b)+1-i],i=1..nops(n0b)),
seq(n1b[nops(n1b)+1-i],i=1..nops(n1b))
];
t4:="";
for i from 1 to nops(t3) do t4:=cat(t4,t3[i]); od:
parse(t4);
end;

a[n_] := IntegerDigits[{z, y} = DigitCount[n, 2]; {y+z, y, z}, 2] // Flatten // FromDigits; Array[a, 30] (* Jean-François Alcover, Oct 20 2016 *)

A308005 A modified Sisyphus function: a(n) = concatenation of (number of odd digits in n) (number of digits in n) (number of even digits in n).

Original entry on oeis.org

11, 110, 11, 110, 11, 110, 11, 110, 11, 110, 121, 220, 121, 220, 121, 220, 121, 220, 121, 220, 22, 121, 22, 121, 22, 121, 22, 121, 22, 121, 121, 220, 121, 220, 121, 220, 121, 220, 121, 220, 22, 121, 22, 121, 22, 121, 22, 121, 22, 121, 121, 220, 121, 220, 121, 220, 121, 220, 121, 220, 22, 121, 22, 121, 22

Offset: 0

N. J. A. Sloane, May 12 2019

If we start with n and repeatedly apply the map i -> a(i), it appears that we eventually reach one of the two fixed points 22 or 231, or enter the two-cycle (33, 220). Are there any other possibilities? This is in contrast to the behavior of the closely related A308003.

			11 has 2 digits, both odd, so a(11)=220.
12 has 2 digits, one even and one odd, so a(12)=121. Then a(121) = 231, a fixed point.
22 has two digits, both even, so 22 -> 22, another fixed point  (leading zeros are omitted).

M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A073053 (Sisyphus), A171797, A171798, A171813, A055642, A196563, A196564, A308002, A308003.

Maple code based on R. J. Mathar's code for A171797:
nevenDgs := proc(n) local a, d; a := 0 ; for d in convert(n, base, 10) do if type(d, 'even') then a :=a +1 ; end if; end do; a ; end proc:
cat2 := proc(a, b) local ndigsb; ndigsb := max(ilog10(b)+1, 1) ; a*10^ndigsb+b ; end:
catL := proc(L) local a, i; a := op(1, L) ; for i from 2 to nops(L) do a := cat2(a, op(i, L)) ; end do; a; end proc:
A055642 := proc(n) max(1, ilog10(n)+1) ; end proc:
A308005 := proc(n) local n1, n2 ; n1 := A055642(n) ; n2 := nevenDgs(n) ; catL([n1-n2, n1, n2]) ; end proc:
[seq(A308005(n), n=0..80)];

A375208 Modified Sisyphus function of order 5.

Original entry on oeis.org

110000, 101000, 100100, 100010, 100001, 110000, 101000, 100100, 100010, 100001, 211000, 202000, 201100, 201010, 201001, 211000, 202000, 201100, 201010, 201001, 210100, 201100, 200200, 200110, 200101, 210100, 201100, 200200, 200110, 200101, 210010, 201010, 200110, 200020, 200011, 210010, 201010, 200110, 200020, 200011, 210001, 201001, 200101, 200011, 200002

Offset: 0

Matt Coppenbarger, Oct 16 2024

a(n) is the concatenation of the number of digits in n with number of digits of n congruent to k modulo 5 for each k from 0 to 4 in turn. See Example.

If we start with n and repeatedly apply the map i -> a(i), we eventually get the cycle {613200, 622110}.

			11 has two digits, both congruent to 1 modulo 5, so a(11) = 202000.
a(20) = 210100.
a(30) = 210010.
a(2527200000) = 1060400.

M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Cf. A073053 (Sisyphus), A171797, A171798, A171813, A055642, A196563, A196564, A308002, A350709.

a:= n-> (l-> parse(cat(nops(l), seq(add(`if`(irem(i, 5)=k
          , 1, 0), i=l), k=0..4))))(convert(n, base, 10)):
seq(a(n), n=0..44);  # Alois P. Heinz, Oct 23 2024

# based on Michael S. Branicky in A350709
def a(n, order=5):
    d, m = list(map(int, str(n))), [0]*order
    for di in d: m[di%order] += 1
    return int(str(len(d)) + "".join(map(str, m)))
print([a(n) for n in range(37)])

from collections import Counter
def A375208(n):
    s = str(n)
    c = Counter(int(d)%5 for d in s)
    return int(str(len(s))+''.join(str(c[i]) for i in range(5))) # Chai Wah Wu, Nov 26 2024

A352751 Modified Sisyphus function of order 4: a(n) is the concatenation of (number of digits of n)(number digits of n congruent to 0 modulo 4)(number of digits of n congruent to 1 modulo 4)(number of digits of n congruent to 2 modulo 4)(number of digits of n congruent to 3 modulo 4).

Original entry on oeis.org

11000, 10100, 10010, 10001, 11000, 10100, 10010, 10001, 11000, 10100, 21100, 20200, 20110, 20101, 21100, 20200, 20110, 20101, 21100, 20200, 21010, 20110, 20020, 20011, 21010, 20110, 20020, 20011, 21010, 20110, 21001, 20101, 20011, 20002, 21001, 20101, 20011, 20002, 21001, 20101, 22000, 21100, 21010

Offset: 0

Matthew E. Coppenbarger, Apr 01 2022

If we start with n and repeatedly apply the map i -> a(i), we eventually get one of three cycles: {51220}, {50410, 52111, 53200}, or {51301}

			11 has two digits, both congruent to 1 modulo 4, so a(11) = 20200.
a(20) = 21010.
a(30) = 21001.
a(1111123567) = 100622.

M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.

Cf. A073053 (Sisyphus), A171797, A171798, A171813, A055642, A196563, A196564, A308002, A350709.

def a(n, order=4):
    d, m = list(map(int, str(n))), [0]*order
    for di in d: m[di%order] += 1
    return int(str(len(d)) + "".join(map(str, m)))
print([a(n) for n in range(37)]) # Michael S. Branicky, Apr 01 2022

cp's OEIS Frontend

A171797 A modified Sisyphus function: a(n) = concatenation of (number of digits in n) (number of even digits) (number of odd digits).

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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.

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A308003 A modified Sisyphus function: a(n) = concatenation of (number of even digits in n) (number of digits in n) (number of odd digits in n).

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A350709 Modified Sisyphus function of order 3: a(n) is the concatenation of (number of digits of n)(number digits of n congruent to 0 modulo 3)(number of digits of n congruent to 1 modulo 3)(number of digits of n congruent to 2 modulo 3).

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A171823 a(n) = base-2 concatenation XYZ, where X = number of bits in binary expansion of n, Y = number of 0's, Z = number of 1's.

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A308005 A modified Sisyphus function: a(n) = concatenation of (number of odd digits in n) (number of digits in n) (number of even digits in n).

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A375208 Modified Sisyphus function of order 5.

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