A011371 -id:A011371 - cp's OEIS frontend

A073642 Replace 2^k in the binary representation of n with k (i.e., if n = 2^b + 2^c + 2^d + ... then a(n) = b + c + d + ...).

0, 0, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 4, 4, 5, 5, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 10, 10, 5, 5, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 10, 10, 11, 11, 9, 9, 10, 10, 11, 11, 12, 12, 12, 12, 13, 13, 14, 14, 15, 15, 6, 6, 7, 7, 8, 8, 9, 9, 9, 9, 10, 10, 11, 11, 12, 12, 10, 10, 11, 11, 12, 12, 13, 13

Offset: 0

Benoit Cloitre, Aug 29 2002

For n >= 1, a(n) is the n-th row sum of the irregular triangle A133457. - Vladimir Shevelev, Dec 14 2015

For n >= 0, 2^a(n) is the number of partitions of n whose dimension (given by the hook-length formula) is an odd integer. See the MacDonald reference. - Arvind Ayyer, May 12 2016

			9 = 2^3 + 2^0, hence a(9) = 3 + 0 = 3;
25 = 2^4 + 2^3 + 2^0, hence a(25) = 4 + 3 + 0 = 7.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Arvind Ayyer, Amritanshu Prasad and Steven Spallone, Odd partitions in Young's lattice, arXiv:1601.01776 [math.CO], 2016.
Ian G. MacDonald, On the degrees of the irreducible representations of symmetric groups, Bulletin of the London Mathematical Society, 3(2):189-192, 1971.

Cf. A059867, A029931, A087135, A000009, A087136, A272090.

a073642 = sum . zipWith (*) [0..] . a030308_row
-- Reinhard Zumkeller, Jun 01 2013

A073642 := proc(n)
        local bdgs ;
        bdgs := convert(n,base,2) ;
        add( op(i,bdgs)*(i-1), i=1..nops(bdgs)) ;
end proc: # R. J. Mathar, Nov 17 2011

Total[Flatten[Position[Rest[Reverse[IntegerDigits[#, 2]]], 1]]] & /@ Range[0, 87] (* Jayanta Basu, Jul 03 2013 *)

a(n)=sum(i=1,length(binary(n)), component(binary(n),i)*(length(binary(n))-i))

a(n) = my(b=binary(n)); b*-[-#b+1..0]~; \\ Ruud H.G. van Tol, Oct 17 2023

def A073642(n):
    a, i = 0, 0
    while n > 0:
        a, n, i = a+(n%2)*i, n//2, i+1
    return a
print([A073642(n) for n in range(30)]) # A.H.M. Smeets, Aug 17 2019

a(n) = log_2(A059867(n)).
It seems that for n > 10 a(n) < n/(2*log(n)) and that Sum_{k=1..n} a(k) is asymptotic to C*n*log(n)^2 with 1/2 > C > 0.47.
a(1)=0, a(2n) = a(n) + e1(n), a(2n+1) = a(2n), where e1(n) = A000120(n). - Ralf Stephan, Jun 19 2003
If n = 2^log_2(n) then a(n) = log_2(n); otherwise, a(n) = log_2(n) + a(n-2^log_2(n)), where log_2=A000523. a(2*n+1) = a(2*n), as the least significant bit of n does not contribute to a(n). - Reinhard Zumkeller, Aug 17 2003, edited by A.H.M. Smeets, Aug 17 2019
a(n) = A029931(floor(n/2)). - Franklin T. Adams-Watters, Oct 22 2011
a(n) = Sum_{k=0..A070939(n)-1} k*A030308(n,k). - Reinhard Zumkeller, Jun 01 2013
Conjecture: a(n) = (3*A011371(n) - Sum_{k=1..n} A007814(k)^2)/2 for n > 0. - Velin Yanev, Sep 09 2017
G.f.: (1/(1 - x)) * Sum_{k>=1} k*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Aug 17 2019
From A.H.M. Smeets, Aug 17 2019: (Start)
floor(log_2(n)) <= a(n) <= floor(log_2(n+2)*(log_2(n+2)-1)/2), n > 0.
Lower bound: floor(log_2(n)) = a(n) for n = 2^m or n = 2^m+1, m >= 0.
Upper bound: a(n) = floor(log_2(n+2)*(log_2(n+2)-1)/2) for n = 2^m-2 or n = 2^m-1, m >= 0. (End)
From Aayush Soni, Feb 12 2022: (Start)
For k < 2^n, a((2^n)+k) + a((2^n)-k-1) = n*(n+1)/2.
Proof: Any (n+1)-bit number 111...11_2 can only be split into two numbers 2^n + k and 2^n - k - 1 which never share any bits in common. Since a(111...11_2) = 0+1+2+...+n, this implies the stated formula. (End)

a(0)=0 and offset corrected by Philippe Deléham, Apr 20 2009

A219651 a(n) = n minus (sum of digits in factorial base expansion of n).

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 5, 5, 6, 6, 7, 7, 10, 10, 11, 11, 12, 12, 15, 15, 16, 16, 17, 17, 23, 23, 24, 24, 25, 25, 28, 28, 29, 29, 30, 30, 33, 33, 34, 34, 35, 35, 38, 38, 39, 39, 40, 40, 46, 46, 47, 47, 48, 48, 51, 51, 52, 52, 53, 53, 56, 56, 57, 57, 58, 58, 61, 61

Offset: 0

Antti Karttunen, Nov 25 2012

See A007623 for the factorial base number system representation.

A. Karttunen, Table of n, a(n) for n = 0..10080

Bisection: A219650. Analogous sequence for binary system: A011371, for Zeckendorf expansion: A219641.

Cf. A007623, A034968, A219652-A219655, A219656, A219659, A219666.

(* First run program for A007623 to define factBaseIntDs *) Table[n - Plus@@factBaseIntDs[n], {n, 0, 99}] (* Alonso del Arte, Nov 25 2012 *)

from itertools import count
def A219651(n):
    c, f = 0, 1
    for i in count(2):
        f *= i
        if f>n:
            break
        c += (i-1)*(n//f)
    return c # Chai Wah Wu, Oct 11 2024

Scheme
```
(define (A219651 n) (- n (A034968 n)))
```

a(n) = n - A034968(n).

A255131 n minus the least number of squares that add up to n: a(n) = n - A002828(n).

Original entry on oeis.org

0, 0, 0, 0, 3, 3, 3, 3, 6, 8, 8, 8, 9, 11, 11, 11, 15, 15, 16, 16, 18, 18, 19, 19, 21, 24, 24, 24, 24, 27, 27, 27, 30, 30, 32, 32, 35, 35, 35, 35, 38, 39, 39, 40, 41, 43, 43, 43, 45, 48, 48, 48, 50, 51, 51, 51, 53, 54, 56, 56, 56, 59, 59, 59, 63, 63, 63, 64, 66, 66, 67, 67, 70, 71, 72, 72, 73, 74, 75, 75, 78, 80, 80, 80, 81

Offset: 0

Antti Karttunen, Feb 24 2015

nonn

The associated beanstalk-sequence starts from a(0) as: 0, 3, 6, 8, 11, 15, 16, 18, 21, ... (A276573).

			a(0) = 0, because no squares are needed for an empty sum, and 0 - 0 = 0.
a(3) = 0, because 3 cannot be represented as a sum of less than three squares (1+1+1), and 3 - 3 = 0.
a(4) = 3, because 4 can be represented as a sum of just one square (namely 4 itself), and 4 - 1 = 3.

Robert Israel, Table of n, a(n) for n = 0..10000
Discussion on the SeqFan mailing list

Subsequence: A005563.
Cf. A000290, A002828, A062535, A260731, A260732, A260733, A260734, A262689, A276573.
Cf. also A011371, A236840, A260740.

f:= proc(n) local F, x;
   if issqr(n) then return n-1 fi;
   if nops(select(t -> t[1] mod 4 = 3 and t[2]::odd, ifactors(n)[2])) = 0 then return n-2 fi;
   x:= n/4^floor(padic:-ordp(n, 2)/2);
   if x mod 8 = 7 then n-4 else n-3 fi
end proc:
f(0):= 0:
map(f, [$0..100]); # Robert Israel, Mar 27 2018

{0}~Join~Table[n - (If[First@ # > 0, 1, Length[First@ Split@ #] + 1] &@ SquaresR[Range@ 4, n]), {n, 84}] (* Michael De Vlieger, Sep 08 2016, after Harvey P. Dale at A002828 *)

a(n) = n - A002828(n).

a(n) = A260740(n) + A062535(n).

A299150 Denominators of the positive solution to n = Sum_{d|n} a(d) * a(n/d).

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 2, 2, 8, 2, 2, 4, 2, 2, 4, 8, 2, 8, 2, 4, 4, 2, 2, 4, 8, 2, 16, 4, 2, 4, 2, 8, 4, 2, 4, 16, 2, 2, 4, 4, 2, 4, 2, 4, 16, 2, 2, 16, 8, 8, 4, 4, 2, 16, 4, 4, 4, 2, 2, 8, 2, 2, 16, 16, 4, 4, 2, 4, 4, 4, 2, 16, 2, 2, 16, 4, 4, 4, 2, 16, 128, 2, 2

Offset: 1

Gus Wiseman, Feb 03 2018

			Sequence begins: 1, 1, 3/2, 3/2, 5/2, 3/2, 7/2, 5/2, 27/8, 5/2, 11/2, 9/4, 13/2, 7/2.

Antti Karttunen, Table of n, a(n) for n = 1..65537 (first 1000 terms from Andrew Howroyd)

Cf. A000010, A003958, A005187, A006519, A007431, A011371, A018804, A023900, A029935, A037445, A046643, A046644, A165825, A257098, A298971, A299119, A299149 (numerators), A299151, A317848, A317929, A317932, A318314, A318512, A318653, A318681.

Cf. A318440 (the 2-adic valuation).

nn=50;
sys=Table[n==Sum[a[d]*a[n/d],{d,Divisors[n]}],{n,nn}];
Denominator[Array[a,nn]/.Solve[sys,Array[a,nn]][[2]]]
f[p_, e_] := 2^((1 + Mod[p, 2])*e - DigitCount[e, 2, 1]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Apr 28 2023 *)

a(n)={my(v=factor(n)[,2]); denominator(n*prod(i=1, #v, my(e=v[i]); binomial(2*e, e)/4^e))} \\ Andrew Howroyd, Aug 09 2018

A299150(n) = { my(f = factor(n), m=1); for(i=1, #f~, m *= 2^(((1+(f[i,1]%2))*f[i,2]) - hammingweight(f[i,2]))); (m); }; \\ Antti Karttunen, Sep 03 2018

for(n=1, 100, print1(denominator(direuler(p=2, n, 1/(1-p*X)^(1/2))[n]), ", ")) \\ Vaclav Kotesovec, May 08 2025

a(n) = denominator(n*A317848(n)/A165825(n)) = A165825(n)/(A037445(n) * A006519(n)). - Andrew Howroyd, Aug 09 2018
a(n) = A046644(n)/A006519(n). - Andrew Howroyd and Antti Karttunen, Aug 30 2018
From Antti Karttunen, Sep 03 2018: (Start)
a(n) = 2^A318440(n).
Multiplicative with a(2^e) = 2^A011371(e), a(p^e) = 2^A005187(e) for odd primes p.
Multiplicative with a(p^e) = 2^(((1+A000035(p))*e)-A000120(e)) for all primes p.
(End)

Keyword:mult added by Andrew Howroyd, Aug 09 2018

A236840 n minus number of runs in the binary expansion of n: a(n) = n - A005811(n).

Original entry on oeis.org

0, 0, 0, 2, 2, 2, 4, 6, 6, 6, 6, 8, 10, 10, 12, 14, 14, 14, 14, 16, 16, 16, 18, 20, 22, 22, 22, 24, 26, 26, 28, 30, 30, 30, 30, 32, 32, 32, 34, 36, 36, 36, 36, 38, 40, 40, 42, 44, 46, 46, 46, 48, 48, 48, 50, 52, 54, 54, 54, 56, 58, 58, 60, 62, 62, 62, 62, 64, 64, 64

Offset: 0

Antti Karttunen, Apr 18 2014

All terms are even. Used by the "number-of-runs beanstalk" sequence A255056 and many of its associated sequences.

Antti Karttunen, Table of n, a(n) for n = 0..8192

Cf. A091067 (the positions of records), A106836 (run lengths).
Cf. A255070 (terms divided by 2).
Cf. A005811, A000120, A003188.
Cf. A255056, A255066, A255061, A255062, A255071, A255072, A255058, A255327.
Other subtracting maps: A011371, A178503, A219641, A219651, A227190, A227191, A237449, A244320, A244234, A255131.

A236840 := proc(n) local i, b; if n=0 then 0 else b := convert(n, base, 2); select(i -> (b[i-1]<>b[i]), [$2..nops(b)]); n-1-nops(%) fi end: seq(A236840(i), i=0..69); # Peter Luschny, Apr 19 2014

a[n_] := n - Length@ Split[IntegerDigits[n, 2]]; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Jul 16 2023 *)

Scheme
```
(define (A236840 n)  (- n (A005811 n)))
```

a(n) = n - A005811(n) = n - A000120(A003188(n)).

a(n) = 2*A255070(n).

A325508 Product of primes indexed by the prime exponents of n!.

Original entry on oeis.org

1, 1, 2, 4, 10, 20, 42, 84, 204, 476, 798, 1596, 3828, 7656, 12276, 24180, 36660, 73320, 120840, 241680, 389424, 785680, 1294440, 2588880, 3848880, 7147920, 11264760, 15926040, 26057304, 52114608, 74421648, 148843296, 187159392, 340949280, 527531760, 926505360

Offset: 0

Gus Wiseman, May 08 2019

nonn

The prime indices of a(n) are the signature of n!, which is row n of A115627.

			We have 7! = 2^4 * 3^2 * 5^1 * 7^1, so a(7) = prime(4)*prime(2)*prime(1)*prime(1) = 84.
The sequence of terms together with their prime indices begins:
          1: {}
          1: {}
          2: {1}
          4: {1,1}
         10: {1,3}
         20: {1,1,3}
         42: {1,2,4}
         84: {1,1,2,4}
        204: {1,1,2,7}
        476: {1,1,4,7}
        798: {1,2,4,8}
       1596: {1,1,2,4,8}
       3828: {1,1,2,5,10}
       7656: {1,1,1,2,5,10}
      12276: {1,1,2,2,5,11}
      24180: {1,1,2,3,6,11}
      36660: {1,1,2,3,6,15}
      73320: {1,1,1,2,3,6,15}
     120840: {1,1,1,2,3,8,16}
     241680: {1,1,1,1,2,3,8,16}

Cf. A000142, A011371, A022559, A056171, A071626, A076934, A115627, A118914, A122111, A135291, A181819, A307035, A323014, A325272, A325273, A325276, A325509.

Table[Times@@Prime/@Last/@If[(n!)==1,{},FactorInteger[n!]],{n,0,30}]

a(n) = A181819(n!).
A001221(a(n)) = A071626(n).
A001222(a(n)) = A000720(n).
A056239(a(n)) = A022559(n).
A003963(a(n)) = A135291(n).
A061395(a(n)) = A011371(n).
A007814(a(n)) = A056171(n).
a(n) = A122111(A307035(n)). - Antti Karttunen, Nov 19 2019

A076934 Smallest integer of the form n/k!.

Original entry on oeis.org

1, 1, 3, 2, 5, 1, 7, 4, 9, 5, 11, 2, 13, 7, 15, 8, 17, 3, 19, 10, 21, 11, 23, 1, 25, 13, 27, 14, 29, 5, 31, 16, 33, 17, 35, 6, 37, 19, 39, 20, 41, 7, 43, 22, 45, 23, 47, 2, 49, 25, 51, 26, 53, 9, 55, 28, 57, 29, 59, 10, 61, 31, 63, 32, 65, 11, 67, 34, 69, 35, 71

Offset: 1

Amarnath Murthy, Oct 19 2002

Equivalently, n divided by the largest factorial divisor of n.
Also, the smallest r such that n/r is a factorial number.
Positions of 1's are the factorial numbers A000142. Is every positive integer in this sequence? - Gus Wiseman, May 15 2019
Let m = A055874(n), the largest integer such that 1,2,...,m divides n. Then a(n*m!) = n since m+1 does not divide n, showing that every integer is part of the sequence. - Etienne Dupuis, Sep 19 2020

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A000142, A011371, A022559, A055874, A055881, A071626, A111701, A115627, A135291, A229020.

Cf. A325272, A325273, A325508, A325509, A325543, A325544.

Table[n/Max@@Intersection[Divisors[n],Array[Factorial,n]],{n,100}] (* Gus Wiseman, May 15 2019 *)
a[n_] := Module[{k=1}, While[Divisible[n, k!], k++]; n/(k-1)!]; Array[a, 100] (* Amiram Eldar, Dec 25 2023 *)

first(n) = {my(res = [1..n]); for(i = 2, oo, k = i!; if(k <= n, for(j = 1, n\k, res[j*k] = j ) , return(res) ) ) } \\ David A. Corneth, Sep 19 2020

From Amiram Eldar, Dec 25 2023: (Start)
a(n) = n/A055881(n)!.
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = BesselI(2, 2) = 0.688948... (A229020). (End)

More terms from David A. Corneth, Sep 19 2020

A090622 Square array read by antidiagonals of highest power of k dividing n! (with n,k>1).

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 0, 1, 3, 0, 0, 1, 1, 4, 0, 1, 0, 1, 2, 4, 0, 0, 1, 1, 2, 2, 7, 0, 0, 0, 1, 1, 2, 2, 7, 0, 0, 1, 0, 2, 1, 3, 4, 8, 0, 0, 0, 1, 0, 2, 1, 3, 4, 8, 0, 0, 0, 0, 1, 1, 2, 1, 4, 4, 10, 0, 0, 0, 1, 1, 1, 1, 4, 2, 4, 5, 10, 0, 0, 1, 0, 1, 1, 2, 1, 4, 2, 5, 5, 11, 0, 0, 0, 1, 0, 1, 1, 2, 1, 4, 2, 5, 5, 11

Offset: 2

Henry Bottomley, Dec 06 2003

			Square array starts:
1, 0, 0, 0, 0, 0, 0, ...
1, 1, 0, 0, 1, 0, 0, ...
3, 1, 1, 0, 1, 0, 1, ...
3, 1, 1, 1, 1, 0, 1, ...
4, 2, 2, 1, 2, 0, 1, ...
4, 2, 2, 1, 2, 1, 1, ...
7, 2, 3, 1, 2, 1, 2, ...

Alois P. Heinz, Antidiagonals n = 2..142, flattened

Columns include A011371, A054861, A090616, A027868, A054861, A054896, A090617, A090618, A027868, A064458, A090619, A090620, A054896, A027868, A090621.
Cf. A090623, A090624, A115627.
Diagonal gives A011776. - Alois P. Heinz, Oct 04 2012

f:= proc(n, p) local c, k; c, k:= 0, p;
       while n>=k do c:= c+iquo(n, k); k:= k*p od; c
    end:
T:= (n, k)-> min(seq(iquo(f(n, i[1]), i[2]), i=ifactors(k)[2])):
seq(seq(T(n, 2+d-n), n=2..d), d=2..20);  # Alois P. Heinz, Oct 04 2012

f[n_, p_] := Module[{c = 0, k = p}, While[n >= k , c = c + Quotient[n, k]; k = k*p ]; c ]; t[n_, k_] := Min[ Table[ Quotient[f[n, i[[1]]], i[[2]]], {i, FactorInteger[k]}]]; Table[ Table[t[n, 2 + d - n], {n, 2, d}], {d, 2, 20}] // Flatten (* Jean-François Alcover, Oct 03 2013, translated from Alois P. Heinz's Maple program *)

For k=p prime: T(n,p) = [n/p] + [n/p^2] + [n/p^3] + .... For k = p^m a prime power: T(n,p^m) = [T(n,p)/m]. For k = b*c with b and c coprime: T(n,a*b) = min(T(n,a), T(n,b)). T(n,k) is close to, but below, n/A090624(k).

A136695 Final nonzero digit of n! in base 8.

Original entry on oeis.org

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

Offset: 0

Carl R. White, Jan 16 2008

			6! = 720 decimal = 1320 octal, so a(6) = 2.

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

Cf. A000142, A136690, A136691, A136692, A136693, A136694, A136696, A008904, A136697, A136698, A136699, A136700, A136701, A136702.

Cf. A011371, A049606.

P:= 1: E:= 0: a[0]:= 1:
for n from 1 to 100 do
  v:= padic:-ordp(n,2);
  P:= P*(n/2^v) mod 8;
  E:= E + v;
  if E mod 3 = 0 then a[n]:= P
  elif E mod 3 = 1 then a[n]:= 2*(P mod 4)
  else a[n]:= 4
  fi
od:
seq(a[n],n=0..100); # Robert Israel, Sep 26 2018

Table[IntegerDigits[FromDigits[Reverse[IntegerDigits[n!,8]]]][[1]],{n,0,100}] (* Harvey P. Dale, Nov 29 2024 *)

From Robert Israel, Sep 26 2018: (Start)
If A011371(n) == 0 (mod 3) then a(n) = A049606(n) mod 8.
If A011371(n) == 1 (mod 3) then a(n) = 2*(A049606(n) mod 4).
If A011371(n) == 2 (mod 3) then a(n) = 4. (End)

A218787 a(n) = A014486-index for the n-th tendril of infinite beanstalk (A213730(n)), with the "lesser numbers to the left side" construction.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 2, 0, 8, 0, 0, 1, 0, 0, 1, 2, 0, 8, 0, 0, 1, 8, 0, 0, 3, 0, 2, 1, 0, 0, 0, 1, 2, 0, 8, 0, 0, 1, 8, 0, 0, 3, 0, 2, 1, 0, 8, 0, 0, 3, 0, 60, 0, 0, 172, 0, 2, 0, 1, 0, 0, 1, 2, 0, 8, 0, 0, 1, 8, 0, 0, 3, 0, 2, 1, 0, 8, 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 lesser numbers go to the left, we obtain:
...........
...28...29.
.....\./...
..24..25...
...\ /.....
....22.....
...........
and the binary tree
........
...\./..
....*...
.\./....
..*.....
........
is located as A014486(2) in the normal encoding order of binary trees, thus a(9)=2.

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 A218788, i.e. a(n) = A057163(A218788(n)). A218786 gives the sizes of these trees. Cf. A072764, A218609, A218611.

cp's OEIS Frontend

A073642 Replace 2^k in the binary representation of n with k (i.e., if n = 2^b + 2^c + 2^d + ... then a(n) = b + c + d + ...).

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A219651 a(n) = n minus (sum of digits in factorial base expansion of n).

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A255131 n minus the least number of squares that add up to n: a(n) = n - A002828(n).

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A299150 Denominators of the positive solution to n = Sum_{d|n} a(d) * a(n/d).

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A236840 n minus number of runs in the binary expansion of n: a(n) = n - A005811(n).

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A325508 Product of primes indexed by the prime exponents of n!.

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A076934 Smallest integer of the form n/k!.

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