A227192 Sum of the partial sums of the run lengths of binary expansion of n, when starting scanning from the least significant end; Row sums of A227188 and A227738.

1, 3, 2, 5, 6, 4, 3, 7, 8, 10, 9, 6, 7, 5, 4, 9, 10, 12, 11, 14, 15, 13, 12, 8, 9, 11, 10, 7, 8, 6, 5, 11, 12, 14, 13, 16, 17, 15, 14, 18, 19, 21, 20, 17, 18, 16, 15, 10, 11, 13, 12, 15, 16, 14, 13, 9, 10, 12, 11, 8, 9, 7, 6, 13, 14, 16, 15, 18, 19, 17, 16

Offset

1,2

Comments

Equivalently, sum of bit-indices in binary expansion of n (counted from the right hand end, with the least significant bit having bit-index 0) of the positions where a bit differs from its immediate right-hand neighbor, counted up to the first leading zero.

a(0) could be 0 or 1, depending on how the binary expansion of zero is conceived, thus its value is left unspecified here.

From Jason Kimberley, Feb 22 2022: (Start)

Also, the total length of string movement required to display the binary expansion of n by the positions of the vanes of vertical blinds (starting with all 0).

The transitions from 0000 to 1011 are:

0001, 0011, 0111, 1111;

1110, 1100, 1000;

The transitions from 0000 to 1101 are:

0001, 0011, 0111, 1111;

1110, 1100;

1101. (End)

Links

Antti Karttunen, Table of n, a(n) for n = 1..8192

Index entries for sequences related to binary expansion of n

Formula

a(n) = Sum_{i=0..A005811(n)-1} A227188(n,i). [Row sums of A227188]

Equivalently, for n>=1, a(n) = Sum_{i=(A173318(n-1)+1)..A173318(n)} A227738(i). [Row sums of A227738]

a(n) = A227183(n) + A000217(A005811(n)-1). [Alternative definition]

a(n) = A029931(A003188(n)).

Recurrence: a(2n) = a(n) + 2*A069010(n), a(2n+1) = a(2n) +1 or -1, according to if n is even or odd. - Ralf Stephan, Sep 04 2013

Example

For 11, whose binary expansion is "1011", the run lengths, when starting scanning from the right, are: [2,1,1]. Their partial sums are [2,2+1,2+1+1] = [2,3,4] which sum to total 9, thus a(11)=9. Equivalently, the zero-based positions (counted from the right) where bits change from one to zero or vice versa in "...01011" are 2, 3, 4 and 2+3+4 = 9.
For 13, whose binary expansion is "1101", the run lengths similarly scanned are [1,1,2]. Their partial sums are [1,1+1,1+1+2] = [1,2,4] which sum to total 7, thus a(13)=7. Equivalently, the positions where bits change in "...01101" are 1, 2, 4 and  1+2+4 = 7.

Mathematica

Table[Tr[FoldList[Plus, 0, Length /@ Split[Reverse[IntegerDigits[n, 2]]]] ], {n, 71}] - Wouter Meeussen, Aug 22 2013

Prog

(Scheme)
(define (A227192 n) (let loop ((i (- (A005811 n) 1)) (s 0)) (cond ((< i 0) s) (else (loop (- i 1) (+ s (A227188bi n i))))))) ;; This version sums the nonzero terms of the n-th row of table A227188.
(define (A227192v2 n) (+ (A227183 n) (A000217 (- (A005811 n) 1)))) ;; Another variant.
(define (A227192v3 n) (add A227738 (+ 1 (A173318 (- n 1))) (A173318 n))) ;; This sums terms of table A227738.
;; With the help of this function that implements Sum_{i=lowlim..uplim} intfun(i)
(define (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (1+ i) (+ res (intfun i)))))))
(Python)
def A227192(n):
  '''Sum of the partial sums of the run lengths of binary expansion of n, starting from the least significant end.'''
  s = 0
  b = n%2
  i = 0
  while (n != 0):
    n >>= 1
    i += 1
    if((n%2) != b):
      b = n%2
      s += i
  return(s)
(PARI) a(n)=local(b, s, t); b=binary(n); s=#b; t=b[#b]; forstep(i=#b-1, 1, -1, if(b[i]!=t, s=s+#b-i; t=!t)); s /* Ralf Stephan, Sep 04 2013 */
(Ruby)
def a(n)
  k = n.to_s(2).scan(/((\d)\2*)/)
  k.each_index.map { |i| (i + 1) * k[i][0].size }.reduce(:+)
end # Peter Kagey, Aug 06 2015

Crossrefs

Cf. A005811, A227183. Row sums of A227188 and A227738.

Sequence in context: A198755 A134237 A341521 * A360260 A099889 A115511

Adjacent sequences: A227189 A227190 A227191 * A227193 A227194 A227195

Keyword

nonn,base,look

Author

Antti Karttunen, Jul 06 2013

Status

approved