A054868 -id:A054868 - cp's OEIS frontend

A293168 Partial sums of A054868.

0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 13, 14, 16, 18, 19, 20, 21, 22, 24, 25, 27, 29, 30, 31, 33, 35, 36, 38, 39, 40, 42, 43, 44, 45, 47, 48, 50, 52, 53, 54, 56, 58, 59, 61, 62, 63, 65, 66, 68, 70, 71, 73, 74, 75, 77, 79, 80, 81, 83, 84, 86, 88, 90, 91, 92, 93, 95, 96, 98, 100, 101, 102, 104, 106

Offset: 0

N. J. A. Sloane, Oct 17 2017

Amiram Eldar, Table of n, a(n) for n = 0..10000
Richard Bellman and Harold N. Shapiro, On a problem in additive number theory, Annals Math., 49 (1948), 333-340.

This is the next term in the sequence A000120, A000788, A054868.

Accumulate[Table[DigitCount[DigitCount[n, 2, 1], 2, 1], {n, 0, 100}]] (* Amiram Eldar, Jul 20 2023 *)

A291317 A variation of the Josephus problem: a(n) is the surviving integer under the following elimination process. Arrange 1,2,3,...,n in a circle, increasing clockwise. Starting with i=1, at k-th stage, move k places clockwise and delete the current number.

Original entry on oeis.org

1, 1, 1, 3, 4, 3, 7, 7, 6, 10, 7, 12, 3, 10, 11, 7, 11, 1, 12, 6, 21, 1, 7, 12, 25, 3, 25, 28, 16, 26, 25, 6, 32, 19, 15, 21, 28, 3, 12, 21, 24, 13, 21, 36, 17, 45, 41, 45, 8, 40, 11, 6, 25, 41, 23, 4, 43, 52, 51, 57, 28, 21, 11, 47, 26, 29, 57, 51, 48, 56, 12

Offset: 1

Rémy Sigrist, Aug 22 2017

nonn

In the classical Josephus problem (A006257), one moves one place clockwise at each stage, and in the A054995 version, one moves two places clockwise at each stage; here, on the other hand, the number of moves is progressive, and the resulting sequence seems random.
No term belongs to A000096 (for the same reason that there are no even positive terms in A006257).
See also A128982 for another variation of the Josephus problem.
a(n) = 1 for n = 1, 2, 3, 18, 22, 171, 195, 234, 1262, 2136, ...
a(n) = n for n = 1, 7, 10, 12, 21, 25, 28, 235, 822, ...
More formally, for any function f over the natural numbers, let us define the function j_f with these rules: for any n > 0:
- let L = (1, 2, ..., n) be the list of the first n natural numbers,
- for k = 1 to n-1:
   - for i = 1 to f(k): move the first element of L to the end,
   - after these moves, discard the first element of L,
- j_f(n) = the remaining element in L.
In particular:
- j_A000012 = A006257,
- j_A007395 = A054995,
- and j_A000027 = a (this sequence),
- see also Links section for the scatterplots of j_f for certain classical or basic functions f.
We have the following properties:
- j_f(1) = 1,
- if f(1) = 1 mod 2 then j_f(2) = 1 else j_f(2) = 2,
- j_f(n) never equals k + Sum_{i=1..k} f(i),
- iterating j_f(n), j_f(j_f(n)), ... eventually leads to a fixed point,
- likely j_f = j_g iff f = g.

			The different stages for n=6 are (where ^ indicates the counting reference position):
- stage 1:  1^ 2  3  4  5  6
- stage 2:  1     3^ 4  5  6
- stage 3:  1     3  4     6^
- stage 4:  1     3        6^
- stage 5:        3^       6
- stage 6:        3^
Hence, a(6) = 3.

Rémy Sigrist, Table of n, a(n) for n = 1..5000
Rémy Sigrist, Scatterplot of the first 5000 terms of j_A000002 (A000002 = Kolakoski sequence).
Rémy Sigrist, Scatterplot of the first 5000 terms of j_A020639 (A020639(n) = least prime dividing n).
Rémy Sigrist, Scatterplot of the first 5000 terms of j_A006519 (A006519(n) = highest power of 2 dividing n).
Rémy Sigrist, Scatterplot of the first 5000 terms of j_A000120 (A000120(n) = Hamming weight of n).
Rémy Sigrist, Scatterplot of the first 5000 terms of j_A054868 (A054868(n) = sum of bits of sum of bits of n).
Index entries for sequences related to the Josephus Problem

Cf. A000002, A000012, A000027, A000096, A000120, A006519, A007395, A006257, A020639, A054868, A054995, A128982.

a(n) = my (l = List(vector(n,i,i)), i = 0); for (k = 1, n-1, i += k; my (p = i \ #l); listpop(l, 1 + (i % #l)); i -= p); return (l[1])

A089224 In binary representation: number of zeros of number of zeros of n.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 2, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 2, 2, 0, 2, 0, 0, 1, 2, 0, 0, 1, 0, 1, 1, 0, 2, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 2, 1, 2, 2, 0, 1, 2, 2, 0, 2, 0, 0, 1, 1, 2, 2, 0, 2, 0, 0, 1, 2, 0, 0, 1, 0, 1, 1, 0, 1, 2, 2, 0, 2, 0

Offset: 0

Reinhard Zumkeller, Dec 10 2003

			a(0) = 0; a(1) = 1; a(16) = 2; a(256) = 3; a(65536) = 4.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n.

Cf. A054868, A007088.

Cf. A023416.

a089224 = a023416 . a023416  -- Reinhard Zumkeller, Mar 31 2015

a:= n-> (z-> z(z(n)))(k-> `if`(k=0, 1, add(1-i, i=Bits[Split](k)))):
seq(a(n), n=0..100);  # Alois P. Heinz, Jul 04 2022

a[n_] := DigitCount[DigitCount[n, 2, 0], 2, 0]; Array[a, 100, 0] (* Amiram Eldar, Jul 24 2023 *)

def a(n): return bin(bin(n)[2:].count("0"))[2:].count("0")
print([a(n) for n in range(102)]) # Michael S. Branicky, Jul 04 2022

a(n) = A023416(A023416(n)).

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A293168 Partial sums of A054868.

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A291317 A variation of the Josephus problem: a(n) is the surviving integer under the following elimination process. Arrange 1,2,3,...,n in a circle, increasing clockwise. Starting with i=1, at k-th stage, move k places clockwise and delete the current number.

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A089224 In binary representation: number of zeros of number of zeros of n.

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