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

Author

Benoit Cloitre, Aug 29 2002

Keywords

• easy
• nonn

Comments

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

Examples

			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.
		

Links

• 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.

Crossrefs

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

Programs

Haskell

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

Maple

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

Mathematica

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

PARI

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

PARI

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

Python

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

Formula

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)

Extensions

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