A356877 a(n) is the least number k such that (the binary weight of k) - (the binary weight of k^2) = n.

0, 23, 111, 479, 1471, 6015, 24319, 28415, 114175, 457727, 490495, 1964031, 6025215, 8122367, 32497663, 98549759, 132104191, 528449535, 1593769983, 1862205439, 7448952831, 25635323903, 29930291199, 119721689087, 411242070015, 479961546751, 514321285119, 2057287237631, 7687987265535

Offset: 0

Author

Karl-Heinz Hofmann, Oct 10 2022

Keywords

• nonn
• base

Comments

Note that the terms of A260986 with n > 1 can all be found here. Terms here that are not in A260986 have the property not to be a record value of the ratio (binary weight k) / (binary weight k^2).

Observation: The difference of two neighboring terms is a multiple of 2^(number of the ones after the last zero in binary expression of the smaller term).

Examples

			  -----------------------------------------------------------------------------
  n       k          k^2           binary k                          binary k^2
  -----------------------------------------------------------------------------
  0       0            0                  0                                   0
  1      23          529              10111                          1000010001
  2     111        12321            1101111                      11000000100001
  3     479       229441          111011111                  111000000001000001
  4    1471      2163841        10110111111              1000010000010010000001
  5    6015     36180225      1011101111111          10001010000001000100000001
  6   24319    591413761    101111011111111      100011010000000100001000000001
  7   28415    807412225    110111011111111      110000001000000010001000000001
  8  114175  13035930625  11011110111111111  1100001001000000001000010000000001
		

Links

• K. B. Stolarsky, The binary digits of a power, Proc. Amer. Math. Soc. 71 (1978), pp. 1-5.

Crossrefs

Cf. A000120, A159918, A260986, A357750.

Programs

Mathematica

a[0] = 0; a[n_] := a[n] = Module[{step = If[n == 1, 1, 2^Length[Split[IntegerDigits[a[n - 1], 2]][[-1]]]], k = a[n - 1]}, While[DigitCount[k, 2, 1] - DigitCount[k^2, 2, 1] != n, k += step]; k]; Array[a, 23, 0] (* Amiram Eldar, Oct 14 2022 *)

PARI

a(n) = my(k=0); while(hammingweight(k) - hammingweight(k^2) != n, k++); k; \\ Michel Marcus, Oct 14 2022

Python

A356877 = [0]
for n in range(1,29):
    s, k = -1, A356877[-1]
    while bin(A356877[-1])[s] == "1": s -= 1
    while bin(k)[2:].count("1")-bin(k**2)[2:].count("1") != n: k += 2**(abs(s)-1)
    A356877.append(k)
print(A356877)