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
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.
Programs
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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
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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)
Comments