A245196 -id:A245196 - cp's OEIS frontend

A038374 Length of longest contiguous block of 1's in binary expansion of n.

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

Offset: 1

Jeffrey Shallit

			a(157) = 3 because 157 in base 2 is 10011101 and longest contiguous block of 1's is of length 3.
May be arranged into blocks of lengths 1, 2, 4, 8, 16, ...:
1,
1, 2,
1, 1, 2, 3,
1, 1, 1, 2, 2, 2, 3, 4,
1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2, 2, 3, 3, 4, 5,
1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 5, 6,
... - _N. J. A. Sloane_, Jul 25 2014

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to binary expansion of n

Cf. A087117, A090000, A090001, A090002, A090003, A090050, A245196.

import Data.List (unfoldr, group)
a038374 = maximum . map length . filter ((== 1) . head) . group .
   unfoldr (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2)
-- Reinhard Zumkeller, May 01 2012

A038374 := proc(n) local nshft,thisr,resul; nshft := n ; resul :=0 ; thisr :=0 ; while nshft > 0 do if nshft mod 2 <> 0 then thisr := thisr+1 ; else resul := max(resul,thisr) ; thisr := 0 ; fi ; nshft := floor(nshft/2) ; od ; resul := max(resul,thisr) ; RETURN(resul) ; end : for n from 1 to 80 do printf("%d,",A038374(n)) ; od : # R. J. Mathar, Jun 15 2006

Table[Max[Length/@DeleteCases[Split[IntegerDigits[n,2]],?(MemberQ[ #,0] &)]],{n,120}] (* _Harvey P. Dale, Jun 10 2013 *)

a(n)=if (n==0, return (0)); n>>=valuation(n, 2); if(n<2, return(n)); my(e=valuation(n+1, 2)); max(e, a(n>>e)) \\ Charles R Greathouse IV, Jan 12 2014; edited by Michel Marcus, Apr 14 2019

from itertools import groupby
def a(n): return max(len(list(g)) for k, g in groupby(bin(n)[1:]) if k=='1')
print([a(n) for n in range(1, 91)]) # Michael S. Branicky, Jul 04 2022

a(n) >= A089309(n). a(n) >= A089310(n). a(2^i)=1. a(2^i-1)=i. - R. J. Mathar, Jun 15 2006

May be defined by the recurrence given in A245196, taking G(n)=n+1 (n>=0) and m=1. - N. J. A. Sloane, Jul 25 2014

A245180 A160239(n)/8.

Original entry on oeis.org

1, 1, 3, 1, 8, 3, 14, 1, 8, 8, 24, 3, 24, 14, 52, 1, 8, 8, 24, 8, 64, 24, 112, 3, 24, 24, 72, 14, 112, 52, 216, 1, 8, 8, 24, 8, 64, 24, 112, 8, 64, 64, 192, 24, 192, 112, 416, 3, 24, 24, 72, 24, 192, 72, 336, 14, 112, 112, 336, 52, 416, 216, 848, 1, 8, 8, 24, 8, 64, 24, 112, 8, 64, 64, 192

Offset: 1

N. J. A. Sloane, Jul 16 2014

			The entries may be arranged into blocks of sizes 1,2,4,8,...:
B_0: 1,
B_1: 1, 3,
B_2: 1, 8, 3, 14,
B_3: 1, 8, 8, 24, 3, 24, 14, 52,
B_4: 1, 8, 8, 24, 8, 64, 24, 112, 3, 24, 24, 72, 14, 112, 52, 216,
B_5: 1, 8, 8, 24, 8, 64, 24, 112, 8, 64, 64, 192, 24, 192, 112, 416, 3, 24, 24, 72, 24, 192, 72, 336, 14, 112, 112, 336, 52, 416, 216, 848,
...
The first half of each block is equal to 1 followed by 8 times an initial segment of the sequence itself.
The next quarter of each block consists of 3 times (1 followed by 8 times an initial segment of the sequence itself).
The next one-eighth of each block consists of 14 times (1 followed by 8 times an initial segment of the sequence itself).
And so on, the successive multipliers 1,3,14,52,... being given by A083424.
Also, the final quarter of any block consists of the twice the last half of the previous block added to eight times the full block before that.
Consider for example the 4th block,
[1, 8, 8, 24, 8, 64, 24, 112; 3, 24, 24, 72; 14, 112, 52, 216].
This is [1 8*(1,1,3,1,8,3,14); 3*(1 8*(1,1,3)); 2*(3,24,14,52)+8*(1,8,3,14)].
The final entries in the blocks give A083424.
See also the formula section.
.
From _Omar E. Pol_, Mar 18 2015: (Start)
Also, the sequence can be written as an irregular tetrahedron T(s,r,k) as shown below:
1;
..
1;
3;
........
1,    8;
3;
14;
................
1,    8,  8, 24;
3,   24;
14;
52;
..................................
1,    8,  8, 24,  8,  64, 24, 112;
3,   24, 24, 72;
14, 112;
52;
216;
.....................................................................
1,    8,  8, 24,  8,  64, 24, 112, 8, 64, 64, 192, 24, 192, 112, 416;
3,   24, 24, 72, 24, 192, 72, 336;
14, 112,112,336;
52, 416;
216;
848;
...
Note that T(s,r,k) = T(s+1,r,k).
(End)

N. J. A. Sloane, Table of n, a(n) for n = 1..16383
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.], which is also available at arXiv:1004.3036v2

Cf. A160239, A083424, A245190, A245195, A245196, A245562, A245540 (partial sums).

See A245181 for the numbers that appear.

a245180 = flip div 8 . a160239  -- Reinhard Zumkeller, Feb 13 2015

R:=proc(n) option remember;
if n=1 then 1
elif (n mod 2) = 0 then R(n/2)
elif (n mod 4) = 3 then 2*R((n-1)/2)+R(n-2)
else 8*R((n-1)/4); fi; end;
[seq(R(n),n=1..200)];

R[n_] := R[n] = Which[n == 1, 1, Mod[n, 2] == 0, R[n/2], Mod[n, 4] == 3, 2*R[(n - 1)/2] + R[n - 2], True, 8*R[(n - 1)/4] ];
Array[R, 200] (* Jean-François Alcover, Nov 16 2017, translated from Maple *)

The following is a fairly simple explicit formula for a(n) as a function of n: a(n) = 8^(r-1) * Product_{lengths i of runs of 1 in binary expansion of n} R(i), where r is the number of runs of 1 in the binary expansion of n and R(i) = A083424(i-1) = (5*4^(i-1)+(-2)^(i-1))/6. Note that row i of the table in A245562 lists the lengths of runs of 1 in binary expansion of i. Example: n = 27 = 11011 in binary, there are two runs each of length 2, so r=1, R(2) = A083424(1) = 3, and so a(27) = 8^1*3*3 = 72. - N. J. A. Sloane, Aug 10 2014
Many 2-D cellular automata studied in the Toothpick paper (Applegate et al.) have a recursive formula for the general term in a typical block of 2^k terms (see Equations 2, 4, 5, 9, 10 12, 38, 39 of that paper). An analogous formula for the present sequence is the following.
Consider the block B_{k-1} containing terms a(2^(k-1)), a(2^(k-1)+1), ..., a(2^k-1). It is convenient to index the terms working backwards from the next, 2^k-th, term. For n in the range 2^(k-1) <= n < 2^k, write n = 2^k-2^r+j, with 0 <= r <= k-1 and 0 <= j < 2^(r-1), and j=0 if r=0. Then
(if j=0) a(2^k-2^r) = f(k-r-1),
(if j>0) a(2^k-2^r+j) = 8*f(k-r-1)*a(j),
where f(i) = A083424(i) = (5*4^i+(-2)^i)/6.
For example, here is block B_4, consisting of terms a(16)=a(31), so k=5:
n:   16 17 18 19 20 21 22 23 24 25 26 27 28  29 30 31
a(n): 1  8  8 24  8 64 24 112 3 24 24 72 14 112 52 216
r:    4  4  4  4  4  4  4  4  3  3  3  3  2   2  1  0
j:    0  1  2  3  4  5  6  7  0  1  2  3  0   1  0  0
Then we have a(24) = a(32-8) = f(5-3-1) = f(1) = 3, illustrating the first equation, and a(21) = a(32-16+5) = 8*f(0)*a(5) = 8*1*8 = 64, illustrating the second equation.
See A245196 for a list of sequences produced by this type of recurrence.

A245195 a(n) = 2^A014081(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 2, 2, 2, 4, 2, 2, 4, 8, 4, 4, 4, 8, 8, 8, 16, 32, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16, 2, 2, 2, 4, 2

Offset: 0

N. J. A. Sloane, Jul 24 2014

This sequence provides a bridge between A245180 (and, presumably, A160239) and A014081.
See A245196 for more about this class of sequences.
Run length transform of A011782: 1,1,2,4,8,16,32,64,... - Chai Wah Wu, Oct 19 2016

Chai Wah Wu and Robert Israel, Table of n, a(n) for n = 0..10000
Robert Israel, Proof that A277560 is the same as A245195 [This will be modified to reflect the fact that the two sequences have now been merged]
Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166 [math.CO], 2016.

Cf. A014081, A245180, A160239, A048896, A245196, A005725, A005940, A011782, A106737, A162510.

# This Maple program applies more generally to a sequence where the recurrence across a block is as follows. The parameters to be set are the sequence G(0), G(1), G(2), ... (the final terms in the blocks), and the multiplier m.
# For n in the range 2^(k-1) <= n < 2^k, write n = 2^k-2^r+j, with 0 <= r <= k-1 and 0 <= j < 2^(r-1), and j=0 if r=0. Then
# (if j=0) a(2^k-2^r) = G(k-r-1),
# (if j>0) a(2^k-2^r+j) = m*G(k-r-1)*a(j).
# Since Maple gives its lists an offset of 1, it is necessary to add 1 to the arguments of G.
# For the present sequence, G(n)=2^n and m=1.
G:=[seq(2^n,n=0..30)];
m:=1;
f:=proc(n) option remember; global m,G; local k,r,j,np;
if n <= 2 then G[0+1] elif n=3 then G[1+1]
elif n=4 then G[0+1] elif n=5 then m*G[0+1] elif n=6 then G[1+1] elif n=7 then G[2+1]
else
   k:=1+floor(log[2](n)); np:=2^k-n;
   if np=1 then r:=0; j:=0; else r:=1+floor(log[2](np-1)); j:=2^r-np; fi;
   if j=0 then G[k-r-1+1]; else m*G[k-r-1+1]*f(j); fi;
fi;
end;
[seq(f(n),n=1..520)]:
# Setting G(n) = A083424(n) and m = 8 gives A245180. Setting G(n) = 2^n and m = 2 gives A048896.
A245195:=n->add(binomial(n,2*k)*binomial(n,k) mod 2, k=0..floor(n/2)): seq(A245195(n), n=0..200); # Wesley Ivan Hurt, Nov 01 2016

Table[Sum[Mod[Binomial[n, 2 k] Binomial[n, k], 2], {k, 0, n}], {n, 0, 85}] (* Michael De Vlieger, Oct 21 2016 *)

a(n) = 2^hammingweight(bitand(n, n>>1)) \\ Charles R Greathouse IV, Jul 16 2016

a(n) = sum(k=0, n, binomial(n, 2*k)*binomial(n,k) % 2); \\ Michel Marcus, Oct 21 2016

from _future_ import division
def A277560(n):
    return sum(int(not (~n & 2*k) | (~n & k)) for k in range(n//2+1))

def A245195(n): return 1<<(n&(n>>1)).bit_count() # Chai Wah Wu, Feb 11 2023

The entries may be arranged into blocks of sizes 1,2,4,8,...:
B_0: 1,
B_1: 1, 2,
B_2: 1, 1, 2, 4,
B_3: 1, 1, 1, 2, 2, 2, 4, 8,
B_4: 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16,
B_5: 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 2, 2, 2, 4, 2, 2, 4, 8, 4, 4, 4, 8, 8, 8, 16, 32,
...
Consider the block B_{k-1} containing terms a(2^(k-1)), a(2^(k-1)+1), ..., a(2^k-1). It is convenient to index the terms working backwards from the next, 2^k-th, term. For n in the range 2^(k-1) <= n < 2^k, write n = 2^k-2^r+j, with 0 <= r <= k-1 and 0 <= j < 2^(r-1), and j=0 if r=0. Then
(if j=0) a(2^k-2^r) = 2^(k-r-1),
(if j>0) a(2^k-2^r+j) = 2^(k-r-1)*a(j).
a(n) = A162510(A005940(1+n)). - Antti Karttunen, Oct 29 2016
From Robert Israel, Nov 02 2016: (Start)
a(2*k)   = a(k).
a(4*k+1) = a(k).
a(4*k+3) = 2*a(2*k+1).
G.f. g(x) satisfies g(x) = x + (2*x+1)*g(x^2) - x*g(x^4). (End)
Also, a(n) = Sum_{k=0..floor(n/2)} ((binomial(n,2k)*binomial(n,k)) mod 2). - Chai Wah Wu, Oct 19 2016 and Robert Israel, Nov 04 2016. For proof, see the article by Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166, or the Robert Israel link.

Changed offset to 0, merged former entry A277560 from Chai Wah Wu (Oct 19 2016) with this sequence. - N. J. A. Sloane, Nov 05 2016

A245541 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=(k-r)(k-r+1)/2, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=((k-r)(k-r+1)/2)*a(j).

Original entry on oeis.org

1, 1, 3, 1, 1, 3, 6, 1, 1, 1, 3, 3, 3, 6, 10, 1, 1, 1, 3, 1, 1, 3, 6, 3, 3, 3, 9, 6, 6, 10, 15, 1, 1, 1, 3, 1, 1, 3, 6, 1, 1, 1, 3, 3, 3, 6, 10, 3, 3, 3, 9, 3, 3, 9, 18, 6, 6, 6, 18, 10, 10, 15, 21, 1, 1, 1, 3, 1, 1, 3, 6, 1, 1, 1, 3, 3, 3, 6, 10, 1, 1, 1, 3, 1, 1, 3, 6, 3, 3, 3, 9, 6, 6, 10, 15

Offset: 1

N. J. A. Sloane, Jul 26 2014

See A245196 for a list of other sequences produced by this type of recurrence.

It follows from the definition that the final entries in the blocks are triangular numbers.

			Arranged into blocks:
1,
1, 3,
1, 1, 3, 6,
1, 1, 1, 3, 3, 3, 6, 10,
1, 1, 1, 3, 1, 1, 3, 6, 3, 3, 3, 9, 6, 6, 10, 15,
1, 1, 1, 3, 1, 1, 3, 6, 1, 1, 1, 3, 3, 3, 6, 10, 3, 3, 3, 9, 3, 3, 9, 18, 6, 6, 6, 18, 10, 10, 15, 21,
...

Cf. A245196, A245547.

G:=[seq(n,n=0..30)];
m:=1;
f:=proc(n) option remember; global m,G; local k,r,j,np;
   k:=1+floor(log[2](n)); np:=2^k-n;
   if np=1 then r:=0; j:=0; else r:=1+floor(log[2](np-1)); j:=2^r-np; fi;
   if j=0 then G[k-r]; else m*G[k-r]*f(j); fi;
end;
[seq(f(n),n=1..120)];

A245547 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=(k-r)(k-r+1)/2, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=(k-r)(k-r+1)*a(j).

Original entry on oeis.org

1, 1, 3, 1, 2, 3, 6, 1, 2, 2, 6, 3, 6, 6, 10, 1, 2, 2, 6, 2, 4, 6, 12, 3, 6, 6, 18, 6, 12, 10, 15, 1, 2, 2, 6, 2, 4, 6, 12, 2, 4, 4, 12, 6, 12, 12, 20, 3, 6, 6, 18, 6, 12, 18, 36, 6, 12, 12, 36, 10, 20, 15, 21, 1, 2, 2, 6, 2, 4, 6, 12, 2, 4, 4, 12, 6, 12, 12, 20, 2, 4, 4, 12, 4, 8, 12, 24, 6

Offset: 1

N. J. A. Sloane, Jul 27 2014

See A245196 for a list of other sequences produced by this type of recurrence.

It follows from the definition that the final entries in the blocks are triangular numbers.

			Arranged into blocks:
1,
1, 3,
1, 2, 3, 6,
1, 2, 2, 6, 3, 6, 6, 10,
1, 2, 2, 6, 2, 4, 6, 12, 3, 6, 6, 18, 6, 12, 10, 15,
1, 2, 2, 6, 2, 4, 6, 12, 2, 4, 4, 12, 6, 12, 12, 20, 3, 6, 6, 18, 6, 12, 18, 36, 6, 12, 12, 36, 10, 20, 15, 21,
...

Cf. A245196, A245541.

G:=[seq(n,n=0..30)];
m:=2;
f:=proc(n) option remember; global m,G; local k,r,j,np;
k:=1+floor(log[2](n)); np:=2^k-n;
if np=1 then r:=0; j:=0; else r:=1+floor(log[2](np-1)); j:=2^r-np; fi;
if j=0 then G[k-r]; else m*G[k-r]*f(j); fi;
end;
[seq(f(n),n=1..120)];

A245536 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=k-r-1, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=(k-r-1)*a(j).

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 0, 1, 0, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 2, 0, 3, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 2, 2, 0, 0, 2, 3, 0, 4, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 2

Offset: 1

N. J. A. Sloane, Jul 25 2014

nonn

Defined by the recurrence given in A245196, taking G(n)=n (n>=0) and m=1.

Changing G from [0,1,2,3,4,...] to [1,2,3,4,5,6,...] produces A038374.

Cf. A245196, A038374.

G:=[seq(n,n=0..30)];
m:=1;
f:=proc(n) option remember; global m,G; local k,r,j,np;
   k:=1+floor(log[2](n)); np:=2^k-n;
   if np=1 then r:=0; j:=0; else r:=1+floor(log[2](np-1)); j:=2^r-np; fi;
   if j=0 then G[k-r-1+1]; else m*G[k-r-1+1]*f(j); fi;
end;
[seq(f(n),n=1..120)];

A245537 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=A083424(k-r-1), or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=A083424(k-r-1)*a(j).

Original entry on oeis.org

1, 1, 3, 1, 1, 3, 14, 1, 1, 1, 3, 3, 3, 14, 52, 1, 1, 1, 3, 1, 1, 3, 14, 3, 3, 3, 9, 14, 14, 52, 216, 1, 1, 1, 3, 1, 1, 3, 14, 1, 1, 1, 3, 3, 3, 14, 52, 3, 3, 3, 9, 3, 3, 9, 42, 14, 14, 14, 42, 52, 52, 216, 848, 1, 1, 1, 3, 1, 1, 3, 14, 1, 1, 1, 3, 3, 3, 14, 52

Offset: 1

N. J. A. Sloane, Jul 26 2014

Similar to A245180, except the multiplier 8 in that recurrence is set here to be 1.

See A245196 for a list of other sequences produced by this type of recurrence.

			Arranged into blocks:
1,
1, 3,
1, 1, 3, 14,
1, 1, 1, 3, 3, 3, 14, 52,
1, 1, 1, 3, 1, 1, 3, 14, 3, 3, 3, 9, 14, 14, 52, 216,
1, 1, 1, 3, 1, 1, 3, 14, 1, 1, 1, 3, 3, 3, 14, 52, 3, 3, 3, 9, 3, 3, 9, 42, 14, 14, 14, 42, 52, 52, 216, 848,
...

Cf. A083424, A245180, A245196.

A245538 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=A083424(k-r-1), or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=2A083424(k-r-1)a(j).

Original entry on oeis.org

1, 1, 3, 1, 2, 3, 14, 1, 2, 2, 6, 3, 6, 14, 52, 1, 2, 2, 6, 2, 4, 6, 28, 3, 6, 6, 18, 14, 28, 52, 216, 1, 2, 2, 6, 2, 4, 6, 28, 2, 4, 4, 12, 6, 12, 28, 104, 3, 6, 6, 18, 6, 12, 18, 84, 14, 28, 28, 84, 52, 104, 216, 848, 1, 2, 2, 6, 2, 4, 6, 28, 2, 4, 4, 12, 6, 12, 28, 104

Offset: 1

N. J. A. Sloane, Jul 26 2014

Similar to A245180, except the multiplier 8 in that recurrence is set here to be 2.

See A245196 for a list of other sequences produced by this type of recurrence.

			Arranged into blocks:
1,
1, 3,
1, 2, 3, 14,
1, 2, 2, 6, 3, 6, 14, 52,
1, 2, 2, 6, 2, 4, 6, 28, 3, 6, 6, 18, 14, 28, 52, 216,
1, 2, 2, 6, 2, 4, 6, 28, 2, 4, 4, 12, 6, 12, 28, 104, 3, 6, 6, 18, 6, 12, 18, 84, 14, 28, 28, 84, 52, 104, 216, 848,
...

Cf. A245196, A245180.

A245539 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=A083424(k-r-1), or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=4A083424(k-r-1)a(j).

Original entry on oeis.org

1, 1, 3, 1, 4, 3, 14, 1, 4, 4, 12, 3, 12, 14, 52, 1, 4, 4, 12, 4, 16, 12, 56, 3, 12, 12, 36, 14, 56, 52, 216, 1, 4, 4, 12, 4, 16, 12, 56, 4, 16, 16, 48, 12, 48, 56, 208, 3, 12, 12, 36, 12, 48, 36, 168, 14, 56, 56, 168, 52, 208, 216, 848

Offset: 1

N. J. A. Sloane, Jul 26 2014

Similar to A245180, except the multiplier 8 in that recurrence is set here to be 4.

See A245196 for a list of other sequences produced by this type of recurrence.

			Arranged into blocks:
1,
1, 3,
1, 4, 3, 14,
1, 4, 4, 12, 3, 12, 14, 52,
1, 4, 4, 12, 4, 16, 12, 56, 3, 12, 12, 36, 14, 56, 52, 216,
1, 4, 4, 12, 4, 16, 12, 56, 4, 16, 16, 48, 12, 48, 56, 208, 3, 12, 12, 36, 12, 48, 36, 168, 14, 56, 56, 168, 52, 208, 216, 848,
...

Cf. A245196, A245180.

cp's OEIS Frontend

A038374 Length of longest contiguous block of 1's in binary expansion of n.

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A245180 A160239(n)/8.

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A245195 a(n) = 2^A014081(n).

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A245541 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=(k-r)*(k-r+1)/2, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=((k-r)*(k-r+1)/2)*a(j).

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A245547 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=(k-r)*(k-r+1)/2, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=(k-r)*(k-r+1)*a(j).

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A245536 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=k-r-1, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=(k-r-1)*a(j).

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A245537 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=A083424(k-r-1), or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=A083424(k-r-1)*a(j).

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A245538 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=A083424(k-r-1), or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=2*A083424(k-r-1)*a(j).

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A245541 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=(k-r)(k-r+1)/2, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=((k-r)(k-r+1)/2)*a(j).

A245547 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=(k-r)(k-r+1)/2, or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=(k-r)(k-r+1)*a(j).

A245538 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=A083424(k-r-1), or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=2A083424(k-r-1)a(j).

A245539 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=A083424(k-r-1), or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=4A083424(k-r-1)a(j).