A199238 -id:A199238 - cp's OEIS frontend

A199262 Smallest m such that A199238(m) = n.

1, 3, 11, 15, 59, 95, 223, 255, 1007, 1919, 4091, 6143, 16379, 28671, 61439, 65535, 261119, 516095, 1048571, 1966079, 4128767, 8380415, 16769023, 25165823, 67108799, 134184959, 268434431, 469762047, 1073741819, 2013265919, 4160749567, 4294967295, 17163091967

Offset: 0

Reinhard Zumkeller, Nov 04 2011

nonn

A199238(a(n)) = n and A199238(m) <> n for m < a(n).

Cf. A138791.

a199262 n = (fromJust $ elemIndex n a199238_list) + 1

a(24)-a(32) from Donovan Johnson, Nov 05 2011

A049445 Numbers k with the property that the number of 1's in binary expansion of k (see A000120) divides k.

Original entry on oeis.org

1, 2, 4, 6, 8, 10, 12, 16, 18, 20, 21, 24, 32, 34, 36, 40, 42, 48, 55, 60, 64, 66, 68, 69, 72, 80, 81, 84, 92, 96, 108, 110, 115, 116, 120, 126, 128, 130, 132, 136, 138, 144, 155, 156, 160, 162, 168, 172, 180, 184, 185, 192, 204, 205, 212, 216, 220, 222, 228

Offset: 1

N. J. A. Sloane

If instead of base 2 we take base 10, then we have the so-called Harshad or Niven numbers (i.e., positive integers divisible by the sum of their digits; A005349). - Emeric Deutsch, Apr 11 2007

A199238(a(n)) = 0. - Reinhard Zumkeller, Nov 04 2011

			20 is in the sequence because 20 is written 10100 in binary and 1 + 1 = 2, which divides 20.
21 is in the sequence because 21 is written 10101 in binary and 1 + 1 + 1 = 3, which divides 21.
22 is not in the sequence because 22 is written 10110 in binary 1 + 1 + 1 = 3, which does not divide 22.

Indranil Ghosh, Table of n, a(n) for n = 1..20000 (first 1000 terms from T. D. Noe)
Paul Dalenberg and Tom Edgar, Consecutive factorial base Niven numbers, Fibonacci Quarterly, Vol. 56, No. 2 (2018), pp. 163-166.
Jean-Marie De Koninck, Nicolas Doyon and Imre Kátai, On the counting function for the Niven numbers, Acta Arithmetica, Vol. 106, No. 3 (2003), pp. 265-275.

Cf. A000120, A005349, A199238.

a049445 n = a049445_list !! (n-1)
a049445_list = map (+ 1) $ elemIndices 0 a199238_list
-- Reinhard Zumkeller, Nov 04 2011

a:=proc(n) local n2: n2:=convert(n,base,2): if n mod add(n2[i],i=1..nops(n2)) = 0 then n else fi end: seq(a(n),n=1..300); # Emeric Deutsch, Apr 11 2007

binHarshadQ[n_] := Divisible[n, Count[IntegerDigits[n, 2], 1]]; Select[Range[228], binHarshadQ] (* Jean-François Alcover, Dec 01 2011 *)
Select[Range[300],Divisible[#,DigitCount[#,2,1]]&] (* Harvey P. Dale, Mar 20 2016 *)

for(n=1,1000,b=binary(n);l=length(b); if(n%sum(i=1,l, component(b,i))==0,print1(n,",")))

is_A049445(n)={n%norml2(binary(n))==0} \\ M. F. Hasler, Oct 09 2012

isok(n) = ! (n % hammingweight(n)); \\ Michel Marcus, Feb 10 2016

A049445 = [n for n in range(1,10**5) if not n % sum([int(d) for d in bin(n)[2:]])] # Chai Wah Wu, Aug 22 2014

{k : A000120(k) | k}. - R. J. Mathar, Mar 03 2008
a(n) seems to be asymptotic to c*n*log(n) where 0.7 < c < 0.8. - Benoit Cloitre, Jan 22 2003
Heuristically, c should be 1/(2*log(2)), since a random d-bit number should have probability approximately 2/d of being in the sequence. - Robert Israel, Aug 22 2014
{a(n)} = {k : A199238(k) = 0}. - M. F. Hasler, Oct 09 2012
De Koninck et al. (2003) proved that the number of base-b Niven numbers not exceeding x, N_b(x), is asymptotically equal to ((2*log(b)/(b-1)^2) * Sum_{j=1..b-1} gcd(j, b-1) + o(1)) * x/log(x). For b=2, N_2(n) ~ (2*log(2) + o(1)) * x/log(x). Therefore, the constant c mentioned above is indeed 1/(2*log(2)). - Amiram Eldar, Aug 16 2020

More terms from Michael Somos

Edited by N. J. A. Sloane, Oct 07 2005 and May 16 2008

A070635 a(n) = n mod (sum of digits of n).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 4, 3, 2, 1, 0, 9, 0, 0, 2, 3, 0, 4, 2, 0, 8, 7, 0, 3, 2, 3, 6, 3, 0, 7, 5, 3, 0, 1, 0, 1, 4, 0, 6, 3, 0, 10, 0, 3, 3, 5, 0, 5, 1, 9, 6, 3, 0, 5, 6, 0, 4, 10, 6, 2, 12, 9, 0, 7, 0, 3, 8, 3, 11, 7, 3, 15, 0, 0, 2, 6, 0, 7, 2, 12, 8

Offset: 1

Reinhard Zumkeller, May 13 2002

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

Cf. A007953.

Cf. A199238.

a070635 n = n `mod` (a007953 n)
-- Reinhard Zumkeller, Apr 07 2011

[n mod (&+Intseq(n)): n in [1..100]]; // Vincenzo Librandi, Dec 30 2015

Table[Mod[n,Total[IntegerDigits[n]]],{n,100}] (* Harvey P. Dale, Mar 11 2013 *)

a(n) = n % sumdigits(n); \\ Michel Marcus, Dec 30 2015

a(A005349(n)) = 0. - Reinhard Zumkeller, Mar 10 2008
A188641(n) = A000007(a(n)); a(A065877(n)) > 0. - Reinhard Zumkeller, Apr 07 2011
a(A138791(n)) = n and a(m) <> n for m < A138791(n). - Reinhard Zumkeller, Nov 07 2011

A161764 a(n) is the largest multiple of {the number of 1's in the binary representation of n} that is <= n.

Original entry on oeis.org

1, 2, 2, 4, 4, 6, 6, 8, 8, 10, 9, 12, 12, 12, 12, 16, 16, 18, 18, 20, 21, 21, 20, 24, 24, 24, 24, 27, 28, 28, 30, 32, 32, 34, 33, 36, 36, 36, 36, 40, 39, 42, 40, 42, 44, 44, 45, 48, 48, 48, 48, 51, 52, 52, 55, 54, 56, 56, 55, 60, 60, 60, 60, 64, 64, 66, 66, 68, 69, 69, 68, 72, 72

Offset: 1

Leroy Quet, Jun 18 2009

a(n) = n - A199238(n). - Reinhard Zumkeller, Nov 04 2011

			11 (decimal) in binary is 1011. There are three 1's. Because 9 is the largest multiple of 3 that is <= 11, then a(11) = 9.

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

Cf. A000120, A161765.

a161764 n = n - a199238 n  -- Reinhard Zumkeller, Nov 04 2011

a := proc (n) local n2, n1, j: n2 := convert(n, base, 2): n1 := add(n2[i], i = 1 .. nops(n2)): for j while j*n1 <= n do j*n1 end do end proc: seq(a(n), n = 1 .. 80); # Emeric Deutsch, Jun 26 2009

a(n)=local(B=binary(n),w=B*vector(#B,x,1)~);n-n%w \\ Hagen von Eitzen, Jun 22 2009

Extended by Hagen von Eitzen and Emeric Deutsch, Jun 26 2009

A265917 a(n) = floor(A070939(n)/A000120(n)) where A070939(n) is the binary length of n and A000120(n) is the binary weight of n.

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 1, 4, 2, 2, 1, 2, 1, 1, 1, 5, 2, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 6, 3, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 2, 1, 1, 1, 3, 2, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 7, 3, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 2, 1, 1, 1, 3, 2, 2, 1, 2, 1, 1

Offset: 1

Alex Ratushnyak, Dec 18 2015

1/a(n) gives a very rough approximation of the density of 1-bits in the binary representation (A007088) of n. This is 1 if more than half of the bits of n are 1. - Antti Karttunen, Dec 19 2015

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A000120, A007088, A070939, A135941, A199238, A265918.

Table[Floor[IntegerLength[n, 2]/Total@ IntegerDigits[n, 2]], {n, 120}] (* Michael De Vlieger, Dec 21 2015 *)

a(n) = #binary(n)\hammingweight(n); \\ Michel Marcus, Dec 19 2015

for n in range(1, 88):
    print(str((len(bin(n))-2) // bin(n).count('1')), end=',')

A265918 a(n) = A070939(n) mod A000120(n), where A070939(n) is the binary length of n and A000120(n) is the binary weight of n.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 2, 2, 1, 0, 0, 0, 2, 0, 2, 2, 1, 0, 2, 2, 1, 2, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 3, 2, 1, 1, 1, 3, 1, 3, 3

Offset: 1

Alex Ratushnyak, Dec 18 2015

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A000120, A070939, A135941, A199238, A265917.

Table[Mod[IntegerLength[n, 2], Total@ IntegerDigits[n, 2]], {n, 120}] (* Michael De Vlieger, Dec 21 2015 *)

a(n) = #binary(n) % hammingweight(n); \\ Michel Marcus, Dec 19 2015

for n in range(1, 88): print((len(bin(n))-2) % bin(n).count('1'), end=', ')

A358139 Numbers k > 0 sorted by k/A000120(k) in increasing order. A000120 is the binary weight of k. If k/A000120(k) yields equal values, the smaller k will appear first.

Original entry on oeis.org

1, 3, 2, 7, 5, 6, 11, 15, 4, 13, 9, 14, 10, 23, 12, 31, 19, 27, 21, 29, 22, 30, 8, 25, 17, 26, 18, 28, 47, 39, 20, 63, 43, 55, 45, 46, 35, 59, 24, 61, 37, 62, 38, 51, 53, 54, 41, 42, 57, 58, 44, 60, 79, 95, 16, 49, 33, 50, 34, 52, 87, 71, 36, 127, 91, 111, 93, 56

Offset: 1

Thomas Scheuerle, Oct 31 2022

A permutation of the positive integers.

This permutation satisfies a weak ordering: If b = a(c*d) and e = a(c) and f = a(d) then b > e and b > f with c,d > 1.

Thomas Scheuerle, a(1..25000) as scatter plot.
Thomas Scheuerle, a(1..1000) as scatter plot colored by binary weight and a(n)/A000120(a(n)) plotted as blue line.
Index entries for sequences that are permutations of the natural numbers

Cf. A000120, A049445, A161764, A199238, A235602, A348416.

f(x) = x/hammingweight(x);
cmpb(x, y) = my(hx=f(x), hy=f(y)); if (hx != hy, return(sign(hx-hy))); return(sign(x-y));
lista(nn) = Vec(vecsort([1..2*nn], cmpb, 1), nn); \\ Michel Marcus, Nov 05 2022

a(2^n) = 2^(n+1) - 1.

abs(a(n)-n) < n.

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A199262 Smallest m such that A199238(m) = n.

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A049445 Numbers k with the property that the number of 1's in binary expansion of k (see A000120) divides k.

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A070635 a(n) = n mod (sum of digits of n).

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A161764 a(n) is the largest multiple of {the number of 1's in the binary representation of n} that is <= n.

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A265917 a(n) = floor(A070939(n)/A000120(n)) where A070939(n) is the binary length of n and A000120(n) is the binary weight of n.

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A265918 a(n) = A070939(n) mod A000120(n), where A070939(n) is the binary length of n and A000120(n) is the binary weight of n.

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A358139 Numbers k > 0 sorted by k/A000120(k) in increasing order. A000120 is the binary weight of k. If k/A000120(k) yields equal values, the smaller k will appear first.

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