A356877 - OEIS

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

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).

LINKS

Table of n, a(n) for n=0..28.

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

EXAMPLE

-----------------------------------------------------------------------------

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

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 *)

PROG

(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)

CROSSREFS

Cf. A000120, A159918, A260986, A357750.