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A385663 Binary numbers with an even number of 1s.

0, 11, 101, 110, 1001, 1010, 1100, 1111, 10001, 10010, 10100, 10111, 11000, 11011, 11101, 11110, 100001, 100010, 100100, 100111, 101000, 101011, 101101, 101110, 110000, 110011, 110101, 110110, 111001, 111010, 111100, 111111, 1000001, 1000010, 1000100, 1000111

Offset: 1

Matthew Schulz, Jul 06 2025

David Consiglio, Jr., Table of n, a(n) for n = 1..10001

Cf. A007088, A001969.

def a(n): return int(bin(((n-1)<<1)|((n-1).bit_count()&1))[2:])
print([a(n) for n in range(1, 37)]) # Michael S. Branicky, Jul 23 2025

a(n) = A007088(A001969(n)). - Michael S. Branicky, Jul 23 2025

More terms from David Consiglio, Jr., Jul 23 2025

Offset changed to 1 by Michael S. Branicky, Jul 23 2025

A306355 Numbers k such that the period of 1/k, or 0 if 1/k terminates, is strictly greater than the period of the decimal expansion of 1/m for all m < k.

Original entry on oeis.org

1, 3, 7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, 223, 229, 233, 257, 263, 269, 289, 313, 337, 361, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541, 571, 577, 593, 619, 647, 659, 701, 709, 727, 743, 811, 821, 823

Offset: 1

Matthew Schulz, Feb 09 2019

This sequence is infinite because 1/(10^k-1) has a period of k for all k, so the period can be arbitrarily large.

Are 1, 3, 289 and 361 the only terms that are not in A001913? - Robert Israel, Feb 10 2019

			7 is a term because 1/7 has a period of 6, which is greater than the periods of 1/m for m < 7.

Robert Israel, Table of n, a(n) for n = 1..10000
Project Euler, Reciprocal cycles: Problem 26
Eric Weisstein's World of Mathematics, Repeating Decimal

Cf. A051626, A007732.

Contains A001913.

count:= 1: A[1]:= 1: m:= 0:
for k from 0 to 100 do
  for d in [3,7,9,11] do
     x:= 10*k+d;
     p:= numtheory:-order(10,x);
     if p > m then
        m := p;
        count:= count+1;
        A[count]:= x
     fi
od od:
seq(A[i],i=1..count); # Robert Israel, Feb 10 2019

ResourceFunction["ProgressiveMaxPositions"]@
 Map[n |->
    First[RealDigits[n]] /. {{_, list_?ListQ} :> Length[list],
      list_?ListQ -> 0}][
  1/Range[1050]] (* Peter Cullen Burbery, Aug 05 2023 *)

RECORDS transform of A051626.

A287293 Golomb's sequence with powers of 2.

Original entry on oeis.org

2, 2, 4, 4, 8, 8, 8, 8, 16, 16, 16, 16, 32, 32, 32, 32, 32, 32, 32, 32, 64, 64, 64, 64, 64, 64, 64, 64, 128, 128, 128, 128, 128, 128, 128, 128, 256, 256, 256, 256, 256, 256, 256, 256, 512, 512, 512, 512, 512, 512, 512, 512, 512, 512, 512, 512, 512, 512, 512

Offset: 1

Matthew Schulz, May 22 2017

A Golomb-type sequence over the powers of 2 (A000079) instead of the integers.

			a(1) equals 2 so 2 appears twice. The next term is 4 because 2^2 is 4, and it appears twice because a(2)=4.

Rémy Sigrist, Table of n, a(n) for n = 1..10000 (first 108 terms from Matthew Schulz)

Like A001462, but instead of integers, uses powers of 2. The terms without repetition are A000079.

a = vector(59); a[1] = 2; for (p=1, oo, for (i=1, a[p], print1 (a[j++] = 2^p ", "); if (j==#a, break (2)))) \\ Rémy Sigrist, Dec 09 2018

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User: Matthew Schulz

A385663 Binary numbers with an even number of 1s.

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A306355 Numbers k such that the period of 1/k, or 0 if 1/k terminates, is strictly greater than the period of the decimal expansion of 1/m for all m < k.

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A287293 Golomb's sequence with powers of 2.

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