A344851 a(n) = (n^2) mod (2^A070939(n)).

0, 1, 0, 1, 0, 1, 4, 1, 0, 1, 4, 9, 0, 9, 4, 1, 0, 1, 4, 9, 16, 25, 4, 17, 0, 17, 4, 25, 16, 9, 4, 1, 0, 1, 4, 9, 16, 25, 36, 49, 0, 17, 36, 57, 16, 41, 4, 33, 0, 33, 4, 41, 16, 57, 36, 17, 0, 49, 36, 25, 16, 9, 4, 1, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100

Offset

0,7

Comments

Informally, if n has w binary digits, a(n) is obtained by keeping the w final binary digits of n^2.

For n > 0, a(n) is the final digit of n^2 in base A062383(n).

This sequence has interesting graphical features (see illustration in Links section).

Links

Rémy Sigrist, Table of n, a(n) for n = 0..8192

John S. McCaskill and Peter R.Wills, On permutations derived from integer powers x^n, arXiv:1907.01890 [math.NT], 2019.

Rémy Sigrist, Scatterplot of the sequence for n = 2^18..2^19

Formula

a(n) = 0 iff n = 0 or n > 1 and n belongs to A116882.

a(n) = 1 iff n belongs to A086341.

a(2^k + m) = a(2^(k+1)-m) for any k > 0 and m = 0..2^k.

Example

For n = 42:
- A070939(42) = 6,
- a(42) = (42^2) mod (2^6) = 1764 mod 64 = 36.

Mathematica

{0}~Join~Table[Mod[n^2, 2^(1+Floor@Log2@n)], {n, 100}] (* Giorgos Kalogeropoulos, Jun 02 2021 *)

Prog

(PARI) a(n) = (n^2) % 2^#binary(n)
(Python)
def a(n): return (n**2) % (2**n.bit_length())
print([a(n) for n in range(75)]) # Michael S. Branicky, May 30 2021

Crossrefs

Cf. A000290, A048152, A062383, A070939, A086341, A116882, A316347 (decimal analog).

Sequence in context: A070432 A170989 A290457 * A346202 A253004 A253011

Adjacent sequences: A344848 A344849 A344850 * A344852 A344853 A344854

Keyword

nonn,base,easy

Author

Rémy Sigrist, May 30 2021

Status

approved