Gage Schacher - cp's OEIS frontend

User: Gage Schacher

Gage Schacher has authored 2 sequences.

A336953 In binary representation, rotate the digits of n right n places.

0, 1, 2, 3, 2, 3, 6, 7, 8, 12, 10, 7, 12, 14, 11, 15, 8, 12, 10, 7, 20, 26, 21, 30, 17, 25, 13, 30, 19, 27, 30, 31, 8, 12, 10, 7, 36, 50, 41, 60, 34, 19, 42, 53, 11, 45, 58, 31, 48, 56, 44, 30, 19, 43, 54, 59, 14, 15, 43, 55, 60, 62, 47, 63, 32, 48, 40, 28

Offset: 0

Gage Schacher, Aug 08 2020

On the graph, there are a series of larger and larger parallelograms joined together by a straight line on y=x where n is unchanged, mostly in the case where n is a multiple of the bit length of n. In addition to the main line that cuts through the graph, each parallelogram has the same few sloped lines in its borders.

			a(3) = a('11') = '11' = 3;
a(4) = a('100') = '010' = '10' = 2;
a(5) = a('101') = '011' = '11' = 3;

Rémy Sigrist, Table of n, a(n) for n = 0..16384

Cf. A007088, A038572 (rotated one binary place to the right).

Cf. A366139 (rotate left), A366140 (fixed points).

Array[FromDigits[RotateRight[IntegerDigits[#, 2], #], 2] &, 68, 0] (* Michael De Vlieger, Oct 05 2020 *)

a(n) = my(d=binary(n)); for (k=1, n, d = concat(d[#d], d[1..#d-1])); fromdigits(d, 2); \\ Michel Marcus, Aug 09 2020

def A336953(n):
    if n == 0: return 0
    l, m = -(n%n.bit_length()), bin(n)[2:]
    return int(m[l:]+m[:l],2) # Chai Wah Wu, Jan 22 2023

A336823 a(1)=1; thereafter a(n) = a(n-1) / wt(n-1) if wt(n-1) divides a(n-1), otherwise a(n) = a(n-1) * wt(n-1) where wt(n) is the binary weight of n.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 2, 6, 6, 3, 6, 2, 1, 3, 1, 4, 4, 2, 1, 3, 6, 2, 6, 24, 12, 4, 12, 3, 1, 4, 1, 5, 5, 10, 5, 15, 30, 10, 30, 120, 60, 20, 60, 15, 5, 20, 5, 1, 2, 6, 2, 8, 24, 6, 24, 120, 40, 10, 40, 8, 2, 10, 2, 12, 12, 6, 3, 1, 2, 6, 2, 8, 4, 12, 4, 1, 3, 12

Offset: 1

Gage Schacher, Aug 04 2020

			a(1)=1 is given, and has 1, the binary weight of 1, as a factor so a(2) =1/1=1;
a(2)=1 which has 1, the binary weight of 2, as a factor so a(3) =1/1=1;
a(3)=1 which does not have 2, the binary weight of 3, as a factor so a(4) =1*2=2.

Gage Schacher, Table of n, a(n) for n = 1..10000

Cf. A000120 (binary weight), A326889.

a[1] = 1; a[n_] := a[n] = If[Divisible[a[n-1], (w = DigitCount[n-1, 2, 1])], a[n-1] / w, a[n-1] * w]; Array[a, 100] (* Amiram Eldar, Aug 05 2020 *)

a(n) = if (n==1, 1, my(h=hammingweight(n-1), last=a(n-1)); if (last%h, last*h, last/h)); \\ Michel Marcus, Aug 05 2020

a(n) = a(n-1)/A000120(n-1) iff A000120(n-1) is a factor of a(n-1), otherwise a(n) = a(n-1)*A000120(n-1).

a(1)=1 added to definition. - N. J. A. Sloane, Sep 21 2020

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User: Gage Schacher

A336953 In binary representation, rotate the digits of n right n places.

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