A011371 -id:A011371 - cp's OEIS frontend

A054896 a(n) = Sum_{k>0} floor(n/7^k).

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13

Offset: 0

Henry Bottomley, May 23 2000

nonn

Exponent of the highest power of 7 dividing n!.

			  a(10^0) = 0.
  a(10^1) = 1.
  a(10^3) = 16.
  a(10^3) = 164.
  a(10^4) = 1665.
  a(10^5) = 16662.
  a(10^6) = 166664.
  a(10^7) = 1666661.
  a(10^8) = 16666662.
  a(10^9) = 166666661

Hieronymus Fischer, Table of n, a(n) for n = 0..10000

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.
Cf. A053828, A054893, A054895, A054899, A067080, A098844, A132031, A214411.
Cf. A000142, A214411.

function A054896(n)
  if n eq 0 then return n;
  else return A054896(Floor(n/7)) + Floor(n/7);
  end if; return A054896;
end function;
[A054896(n): n in [0..100]]; // G. C. Greubel, Feb 09 2023

Table[t=0; p=7; While[s=Floor[n/p]; t=t+s; s>0, p *= 7]; t, {n,0,100}]

a(n)=(n-sumdigits(n, 7))\6 \\ Alan Michael Gómez Calderón, Oct 08 2024

def A054896(n):
    if (n==0): return 0
    else: return A054896(n//7) + (n//7)
[A054896(n) for n in range(101)] # G. C. Greubel, Feb 09 2023

a(n) = floor(n/7) + floor(n/49) + floor(n/343) + floor(n/2401) + ...
a(n) = (n - A053828(n))/6.
From Hieronymus Fischer, Aug 14 2007: (Start)
a(n) = a(floor(n/7)) + floor(n/7).
a(7*n) = n + a(n).
a(n*7^m) = a(n) + n*(7^m-1)/6.
a(k*7^m) = k*(7^m-1)/6, for 0 <= k < 7, m >= 0.
Asymptotic behavior:
a(n) = n/6 + O(log(n)).
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/6; equality holds for powers of 7.
a(n) >= (n-6)/6 - floor(log_7(n)); equality holds for n=7^m-1, m>0. -
lim inf (n/6 - a(n)) = 1/6, for n-->oo.
lim sup (n/6 - log_7(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_7(n)) = 0, for n-->oo.
G.f.: (1/(1-x))*Sum_{k > 0} x^(7^k)/(1-x^(7^k)). (End)
Partial sums of A214411. - R. J. Mathar, Jul 08 2021
a(n) = A214411(A000142(n)). - Michel Marcus, Oct 07 2024

Examples added by Hieronymus Fischer, Jun 06 2012

A219641 a(n) = n minus (number of 1's in Zeckendorf expansion of n).

Original entry on oeis.org

0, 0, 1, 2, 2, 4, 4, 5, 7, 7, 8, 9, 9, 12, 12, 13, 14, 14, 16, 16, 17, 20, 20, 21, 22, 22, 24, 24, 25, 27, 27, 28, 29, 29, 33, 33, 34, 35, 35, 37, 37, 38, 40, 40, 41, 42, 42, 45, 45, 46, 47, 47, 49, 49, 50, 54, 54, 55, 56, 56, 58, 58, 59, 61, 61, 62, 63, 63, 66

Offset: 0

Antti Karttunen, Nov 24 2012

nonn

See A014417 for the Fibonacci number system representation, also known as Zeckendorf expansion.

Antti Karttunen, Table of n, a(n) for n = 0..10000
Paul Baird-Smith, Alyssa Epstein, Kristen Flint, and Steven J. Miller, The Zeckendorf Game, arXiv:1809.04881 [math.NT], 2018.

Cf. A007895, A014417. A022342 gives the positions of records, resulting the same sequence with duplicates removed: A219640. A035336 gives the positions of values that occur only once: A219639. Cf. also A219637, A219642. Analogous sequence for binary system: A011371, for factorial number system: A219651.

zeck = DigitCount[Select[Range[0, 500], BitAnd[#, 2*#] == 0&], 2, 1];
Range[0, Length[zeck]-1] - zeck (* Jean-François Alcover, Jan 25 2018 *)

from sympy import fibonacci
def a(n):
    k=0
    x=0
    while n>0:
        k=0
        while fibonacci(k)<=n: k+=1
        x+=10**(k - 3)
        n-=fibonacci(k - 1)
    return str(x).count("1")
print([n - a(n) for n in range(101)]) # Indranil Ghosh, Jun 09 2017

Scheme
```
(define (A219641 n) (- n (A007895 n)))
```

a(n) = n - A007895(n).

A295296 Numbers n such that the sum of their divisors + the number of ones in their binary expansion = 2n; numbers for which A000203(n) + A000120(n) = 2n.

Original entry on oeis.org

1, 2, 3, 4, 8, 10, 16, 32, 64, 128, 136, 256, 315, 512, 1024, 2048, 4096, 8192, 16384, 32768, 32896, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 2147516416

Offset: 1

Antti Karttunen, Nov 26 2017

Numbers n such that their binary weight is equal to their deficiency.
Numbers n such that A000120(n) = A033879(n), or equally A000203(n) = A005187(n), or equally A001065(n) = A011371(n).
2^(2^k-1) * (2^(2^k) + 1) is in the sequence if and only if (2^(2^k) + 1) is a (Fermat) prime (A019434) which as of today is known to be the case for 0 <= k <= 4 giving the terms 3, 10, 136, 32896 and 2147516416. - David A. Corneth, Nov 26 2017
It would be nice to know whether 315 is the only term that is neither in A191363 nor a power of two.
Any term that is either a square or twice a square (in A028982) must be odious (in A000069), and vice versa.
If there's an odd term below 10^30 besides 315 then it must be divisible by a prime >= 23. - David A. Corneth, Nov 27 2017
221753180448460815 is odd and also a term of this sequence. - Alexander Violette, Dec 24 2020

			A000203(315) = 1 + 3 + 5 + 7 + 9 + 15 + 21 + 35 + 45 + 63 + 105 + 315 = 624. 2*315 - 624 = 6, and when 315 is written in binary, 100111011, we see that it has six 1-bits. Thus 315 is included in the sequence.

Max Alekseyev, Table of n, a(n) for n = 1..70 (first 52 terms from Giovanni Resta)
Index entries for sequences related to binary expansion of n
Index entries for sequences related to sigma(n)

Positions of zeros in A294898 and A294899.
Subsequence of A005100 and also of A295298.
Subsequences: A000079, A191363 (the five known terms).
Cf. A000120, A000203, A001065, A005187, A011371, A019434, A033879, A141548.

Select[Range[2^20], DivisorSigma[1, #] + DigitCount[#, 2, 1] == 2 # &] (* Michael De Vlieger, Nov 26 2017 *)

for(n=1, oo, if(sigma(n)+hammingweight(n) == 2*n, print1(n, ", ")));

Terms a(35) and beyond from Giovanni Resta, Feb 27 2020

A317932 Denominators of certain "Dirichlet Square Root" sequences: a(n) = A046644(n)/(2^A007949(n)).

Original entry on oeis.org

1, 2, 1, 8, 2, 2, 2, 16, 2, 4, 2, 8, 2, 4, 2, 128, 2, 4, 2, 16, 2, 4, 2, 16, 8, 4, 2, 16, 2, 4, 2, 256, 2, 4, 4, 16, 2, 4, 2, 32, 2, 4, 2, 16, 4, 4, 2, 128, 8, 16, 2, 16, 2, 4, 4, 32, 2, 4, 2, 16, 2, 4, 4, 1024, 4, 4, 2, 16, 2, 8, 2, 32, 2, 4, 8, 16, 4, 4, 2, 256, 8, 4, 2, 16, 4, 4, 2, 32, 2, 8, 4, 16, 2, 4, 4, 256, 2, 16, 4, 64, 2, 4, 2, 32, 4

Offset: 1

Antti Karttunen, Aug 11 2018

These are denominators for rational valued sequences that are obtained as "Dirichlet Square Roots" of sequences b that satisfy the condition b(3) = 2, and b(p) = odd number for any other primes p. For example, A064989, A065769 and A234840. - Antti Karttunen, Aug 31 2018

The original definition was: Denominators of the rational valued sequence whose Dirichlet convolution with itself yields A002487, Stern's Diatomic sequence. However, this definition depends on the conjecture given in A261179.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Index entries for sequences related to Stern's sequences

Cf. A317930, A318319, A318669 (some of the numerator sequences), A317931 (conjectured, for A002487).
Cf. A305439 (the 2-adic valuation), A318666.
Cf. also A046644, A261179, A299150, A237649, A299150, A317928, A317839, A317843, A317934, A318319.

\\ Original program, based on conjectural formula:
A002487(n) = { my(a=1, b=0); while(n>0, if(bitand(n, 1), b+=a, a+=b); n>>=1); (b); }; \\ From A002487
A317931perA317932(n) = if(1==n,n,(A002487(n)-sumdiv(n,d,if((d>1)&&(dA317931perA317932(d)*A317931perA317932(n/d),0)))/2);
A317932(n) = denominator(A317931perA317932(n));

\\ New fast program implementing the new definition:
A005187(n) = { my(s=n); while(n>>=1, s+=n); s; };
A046644(n) = factorback(apply(e -> 2^A005187(e),factor(n)[,2]));
A317932(n) = (A046644(n)/2^valuation(n,3)); \\ Antti Karttunen, Aug 31 2018

A011371(n) = (A005187(n)-n);
A317932(n) = { my(f = factor(n), m=1); for(i=1, #f~, if(3 == f[i,1], m *= 2^(A011371(f[i,2])), m *= 2^A005187(f[i,2]))); (m); }; \\ Antti Karttunen, Sep 03 2018

a(n) = A046644(n)/A318666(n) = 2^A305439(n).
a(n) = denominator of f(n), where f(1) = 1, f(n) = (1/2) * (b(n) - Sum_{d|n, d>1, d 1, where b can be A064989, A065769 or A234840 for example, conjecturally also A002487.
Multiplicative with a(3^e) = 2^A011371(e), a(p^e) = 2^A005187(e) for any other primes. - Antti Karttunen, Sep 03 2018

Definition changed, the original (now conjectured alternative definition) moved to the comments section by Antti Karttunen, Aug 31 2018

Keyword:mult added by Antti Karttunen, Sep 03 2018

A102730 Number of factorials contained in the binary representation of n!

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 5, 6, 7, 6, 7, 6, 7, 6, 6, 6, 7, 6, 6, 6, 7, 6, 7, 8, 6, 7, 6, 7, 6, 7, 7, 7, 8, 7, 7, 7, 6, 8, 7, 7, 7, 7, 7, 8, 7, 7, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 7, 7, 7, 7, 7, 7, 7, 7, 8, 7, 7, 7, 7, 7, 7, 8, 7, 7, 8, 7, 7, 7, 7, 7, 7, 8, 7, 7, 7, 7, 8, 7, 7, 7, 7, 8, 7, 7, 8, 8, 7, 7, 7, 8, 8, 7, 8, 7, 7

Offset: 0

Reinhard Zumkeller, Feb 07 2005

Conjecture: the sequence is bounded.
I conjecture the contrary: for every k, there exists n with a(n) > k. See A103670. - Charles R Greathouse IV, Aug 21 2011
For n > 0: A103670(n) = smallest m such that a(m) = n.
A103671(n) = smallest m such that the binary representation of n! does not contain m!.
A103672(n) = greatest m less than n such that the binary representation of n! contains m!.

			n = 6: 6! = 720 -> '1011010000' contains a(6) = 5 factorials: 0! = 1 -> '1', 1! = 1 -> '1', 2! = 2 -> '10', 3! = 6 -> '110' and 6! itself, but not 4! = 24-> '11000' and 5! = 120 -> '1111000'.

Charles R Greathouse IV, Table of n, a(n) for n = 0..1000
Index entries for sequences related to binary expansion of n.
Index entries for sequences related to factorial numbers.

Cf. A036603, A007088, A000142, A011371, A093684, A103673, A103676, A103677, A103674, A103678, A103679, A103675, A103680, A103681.

a[n_] := Sum[Boole[StringContainsQ[IntegerString[n!, 2], IntegerString[k!, 2]]], {k, 0, n}]; Array[a, 100, 0] (* Amiram Eldar, Apr 03 2025 *)

contains(v,u)=for(i=0,#v-#u,for(j=1,#u,if(v[i+j]!=u[j],next(2)));return(1));0
a(n)=my(v=binary(n--!));sum(i=0,n-1,contains(v,binary(i!)))+1 \\ Charles R Greathouse IV, Aug 21 2011

A054895 a(n) = Sum_{k>0} floor(n/6^k).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 16, 16, 16

Offset: 0

Henry Bottomley, May 23 2000

nonn

Different from the highest power of 6 dividing n! (cf. A054861). - Hieronymus Fischer, Aug 14 2007

Partial sums of A122841. - Hieronymus Fischer, Jun 06 2012

			  a(10^0) = 0.
  a(10^1) = 1.
  a(10^2) = 18.
  a(10^3) = 197.
  a(10^4) = 1997.
  a(10^5) = 19996.
  a(10^6) = 199995.
  a(10^7) = 1999995.
  a(10^8) = 19999994.
  a(10^9) = 199999993.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.

Cf. A053827, A054861, A054899, A067080, A098844, A122841, A132030.

a054895 n = a054895_list !! n
a054895_list = scanl (+) 0 a122841_list
-- Reinhard Zumkeller, Nov 10 2013

function A054895(n)
  if n eq 0 then return n;
  else return A054895(Floor(n/6)) + Floor(n/6);
  end if; return A054895;
end function;
[A054895(n): n in [0..100]]; // G. C. Greubel, Feb 09 2023

Table[t=0; p=6; While[s=Floor[n/p]; t=t+s; s>0, p *= 6]; t, {n,0,100}]

def A054895(n):
    if (n==0): return 0
    else: return A054895(n//6) + (n//6)
[A054895(n) for n in range(104)] # G. C. Greubel, Feb 09 2023

a(n) = floor(n/6) + floor(n/36) + floor(n/216) + floor(n/1296) + ...
a(n) = (n - A053827(n))/5.
From Hieronymus Fischer, Aug 14 2007: (Start)
a(n) = a(floor(n/6)) + floor(n/6).
a(6*n) = n + a(n).
a(n*6^m) = n*(6^m-1)/5 + a(n).
a(k*6^m) = k*(6^m-1)/5, for 0 <= k < 6, m >= 0.
Asymptotic behavior:
a(n) = (n/5) + O(log(n)).
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/5; equality holds for powers of 6.
a(n) >= ((n-5)/5) - floor(log_6(n)); equality holds for n=6^m-1, m>0.
lim inf (n/5 - a(n)) = 1/5, for n-->oo.
lim sup (n/5 - log_6(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_6(n)) = 0, for n-->oo.
G.f.: (1/(1-x))*Sum_{k > 0} x^(6^k)/(1-x^(6^k)). (End)

An incorrect formula was deleted by N. J. A. Sloane, Nov 18 2008

Examples added by Hieronymus Fischer, Jun 06 2012

A218788 a(n) = A014486-index for the n-th tendril of infinite beanstalk (A213730(n)), with the "lesser numbers to the right side" construction.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 3, 0, 0, 0, 1, 3, 0, 4, 0, 0, 1, 0, 0, 1, 3, 0, 4, 0, 0, 1, 4, 0, 0, 2, 0, 3, 1, 0, 0, 0, 1, 3, 0, 4, 0, 0, 1, 4, 0, 0, 2, 0, 3, 1, 0, 4, 0, 0, 2, 0, 37, 0, 0, 110, 0, 3, 0, 1, 0, 0, 1, 3, 0, 4, 0, 0, 1, 4, 0, 0, 2, 0, 3, 1, 0, 4, 0, 0

Offset: 1

Antti Karttunen, Nov 11 2012

nonn

"Tendrils" of the beanstalk are the finite side-trees sprouting from its infinite trunk (see A179016) at the numbers given by A213730.

			A213730(9)=22, and from that branches 24 and 25 (as both A011371(24)=A011371(25)=22) and while 24 is a leaf (in A055938) the other branch 25 further branches to two leaves (as both A011371(28)=A011371(29)=25).
When we construct a binary tree from this in such a fashion that the larger numbers go to the left, we obtain:
..........
29...28...
..\./.....
...25..24.
....\./...
.....22...
..........
and the binary tree
.......
.\./...
..*....
...\./.
....*..
.......
is located as A014486(3) in the normal encoding order of binary trees, thus a(9)=3.

A. Karttunen, Table of n, a(n) for n = 1..8727
A. Karttunen, Illustration of how binary trees (the second rightmost column) are encoded by A014486

These are the mirror-images of binary trees given in A218787, i.e. a(n) = A057163(A218787(n)). A218786 gives the sizes of these trees. Cf. A072764, A218610, A218611.

A218616 The infinite trunk of beanstalk (A179016) with reversed subsections.

Original entry on oeis.org

0, 1, 3, 7, 4, 15, 11, 8, 31, 26, 23, 19, 16, 63, 57, 53, 49, 46, 42, 39, 35, 32, 127, 120, 116, 112, 109, 104, 101, 97, 94, 89, 85, 81, 78, 74, 71, 67, 64, 255, 247, 240, 236, 231, 225, 221, 215, 209, 205, 200, 197, 193, 190, 184, 180, 176, 173, 168, 165, 161

Offset: 0

Antti Karttunen, Nov 10 2012

This can be viewed as an irregular table: after the initial zero on row 0, start each row n with (2^n)-1 and subtract repeatedly the number of 1-bits to get successive terms, until the number that has already been listed (which is always (2^(n-1))-1) is encountered, which is not listed second time, but instead, the current row is finished and the next row starts with (2^(n+1))-1, with the same process repeated.
This contains the terms in the infinite trunk of beanstalk (A179016) listed in partially reversed manner: after the initial zero each subsequence lists A213709(n) successive terms  from A179016, descending from (2^n)-1 downwards, usually down to 2^(n-1) (conjectured to indeed be a power of 2 in each case, apart from 2 itself missing from the beginning of the sequence).
Currently A179016 and many of the derived sequences are much easier and somewhat faster to compute with the help of this sequence, especially if the program computes any other required values incrementally in the same loop.

			After zero, we start with (2^1)-1 = 1, subtract A000120(1)=1 from it, resulting 1-1=0 (which is of the form (2^0)-1, thus not listed second time), instead, start the next row with (2^2)-1 = 3, subtract A000120(3)=2 from it, resulting 3-2=1, which has been already encountered, thus start the next row with (2^3)-1 = 7, subtract A000120(7)=3 from it, resulting 7-3=4, which is listed after 7, then 4-A000120(4)=4-1=3, which is of the form (2^k)-1 and already encountered, thus start the next row with (2^4)-1 = 15, etc. This results an irregular table which begins as:
0; 1; 3; 7, 4; 15, 11, 8; 31, 26, 23, 19, 16; 63, 57, ...
After zero, each row n is A213709(n-1) elements long.

Antti Karttunen, Rows 0..16, flattened

a(n) = A179016(A218602(n)). The rows are the initial portions of every (2^n)-1:th row in A218254.

a(0)=0, a(1)=1, and for n > 1, if n = A213710(A213711(n)-1) then a(n) = (2^A213711(n)) - 1, and in other cases, a(n) = A011371(a(n-1)).

Alternatively: For n < 4, a(n) = (2^n)-1, and for n >= 4, a(n) = A004755(A004755(A011371(a(n-1)))) if A011371(a(n-1))+1 is power of 2, otherwise just A011371(a(n-1)).

A325618 Numbers m such that there exists an integer partition of m whose reciprocal factorial sum is 1.

Original entry on oeis.org

1, 4, 11, 18, 24, 31, 37, 44, 45, 50, 52, 57, 58, 65, 66, 70, 71, 73, 76, 78, 79, 83, 86, 87, 89, 91, 92, 94, 96, 97, 99, 100, 102, 104, 107, 108, 109, 110, 112, 113, 114, 115, 117, 118, 119, 120, 121, 122, 123, 125, 126, 127, 128, 130, 131

Offset: 1

Gus Wiseman, May 13 2019

nonn

The reciprocal factorial sum of an integer partition (y_1,...,y_k) is 1/y_1! + ... + 1/y_k!.

Conjecture: 137 is the greatest integer not in this sequence. - Charlie Neder, May 14 2019

			The sequence of terms together with an integer partition of each whose reciprocal factorial sum is 1 begins:
   1: (1)
   4: (2,2)
  11: (3,3,3,2)
  18: (3,3,3,3,3,3)
  24: (4,4,4,4,3,3,2)
  31: (4,4,4,4,3,3,3,3,3)
  37: (4,4,4,4,4,4,4,4,3,2)
  44: (4,4,4,4,4,4,4,4,3,3,3,3)
  45: (5,5,5,5,5,4,4,4,3,3,2)
  50: (4,4,4,4,4,4,4,4,4,4,4,4,2)

Charlie Neder, Table of n, a(n) for n = 1..1000

Factorial numbers: A000142, A002982, A007489, A011371, A022559, A064986, A115627, A284605, A325616.

Reciprocal factorial sum: A002966, A051908, A058360, A316854, A316855, A325619, A325620, A325621, A325622, A325623, A325624.

a(11)-a(55) from Charlie Neder, May 14 2019

A090616 Exponent of highest power of 4 dividing n!.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 5, 5, 7, 7, 8, 8, 9, 9, 9, 9, 11, 11, 11, 11, 12, 12, 13, 13, 15, 15, 16, 16, 17, 17, 17, 17, 19, 19, 19, 19, 20, 20, 21, 21, 23, 23, 23, 23, 24, 24, 25, 25, 26, 26, 27, 27, 28, 28, 28, 28, 31, 31, 32, 32, 33, 33, 33, 33, 35, 35, 35, 35, 36

Offset: 0

Henry Bottomley, Dec 06 2003

nonn

			a(6)=2 since 6! = 720 = 4^2 * 45.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

Cf. A011371, A054861, A027868, A054896, A235127.

IntegerExponent[Range[0, 100]!, 4] (* Vincenzo Librandi, Mar 10 2013 *)

a(n) = valuation( n!, 4 ); /* Joerg Arndt, Mar 10 2013 */
(Python 3.10+)
def A090616(n): return (n-n.bit_count())>>1 # Chai Wah Wu, Jul 09 2022

a(n) = A090622(n, 4) = floor(A011371(n)/2) = floor((floor(n/2) + floor(n/4) + floor(n/8) + floor(n/16) + ...)/2).

a(n) = A235127(n!). - R. J. Mathar, Jul 08 2021

cp's OEIS Frontend

A054896 a(n) = Sum_{k>0} floor(n/7^k).

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A219641 a(n) = n minus (number of 1's in Zeckendorf expansion of n).

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A295296 Numbers n such that the sum of their divisors + the number of ones in their binary expansion = 2n; numbers for which A000203(n) + A000120(n) = 2n.

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A317932 Denominators of certain "Dirichlet Square Root" sequences: a(n) = A046644(n)/(2^A007949(n)).

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A102730 Number of factorials contained in the binary representation of n!

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A054895 a(n) = Sum_{k>0} floor(n/6^k).

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A218788 a(n) = A014486-index for the n-th tendril of infinite beanstalk (A213730(n)), with the "lesser numbers to the right side" construction.