

A305461


The number of onedigit numbers, k, in base n such that k^2 and k^3 end in the same digit.


1



1, 2, 2, 3, 2, 4, 2, 3, 4, 4, 2, 6, 2, 4, 4, 5, 2, 8, 2, 6, 4, 4, 2, 6, 6, 4, 4, 6, 2, 8, 2, 5, 4, 4, 4, 12, 2, 4, 4, 6, 2, 8, 2, 6, 8, 4, 2, 10, 8, 12, 4, 6, 2, 8, 4, 6, 4, 4, 2, 12, 2, 4, 8, 9, 4, 8, 2, 6, 4, 8, 2, 12, 2, 4, 12, 6, 4, 8, 2, 10, 10, 4, 2, 12, 4
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OFFSET

1,2


COMMENTS

It appears that a(n) is equal to the number of factors of n, except when n cannot be divided by its multiplicative projection (A230542).  Ian Newman, Jun 01 2018
Number of solutions to x^3  x^2 == 0 (mod n).  Andrew Howroyd, Jul 22 2018


LINKS

Matthew Scroggs, Table of n, a(n) for n = 1..1000
Matthew Scroggs, Square and cube endings.


FORMULA

Multiplicative with a(p^e) = p^floor(e/2) + 1 for prime p.  Andrew Howroyd, Jul 22 2018


EXAMPLE

In base 4,
0^2 = 0, 0^3 = 0,
1^2 = 1, 1^3 = 1,
2^2 = 10, 2^3 = 20,
3^2 = 21, 3^3 = 123.
Three of these pairs have the same final digit, so a(4)=3.


MATHEMATICA

Table[Count[Range@ n, _?(PowerMod[#, 2, n] == PowerMod[#, 3, n] &)], {n, 85}] (* Michael De Vlieger, Jul 30 2018 *)


PROG

(Python) #
for base in range(1, 101):
....n = 0
....for j in range(base):
........if (j**2)%base == (j**3)%base:
............n += 1
....print(base, n)
(Haskell)
a305461 n = length $ filter (\i > (i^3  i^2) `mod` n == 0) [0..n1]
 Peter Kagey, Jun 10 2018
(PARI) a(n) = sum(k=0, n1, mk = Mod(k, n); mk^2 == mk^3); \\ Michel Marcus, Jul 03 2018
(PARI) a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); p^(e\2) + 1)} \\ Andrew Howroyd, Jul 22 2018


CROSSREFS

A034444 is the number of onedigit numbers, k, in base n such that k and k^2 end in the same digit.
Cf. A230542.
Sequence in context: A282446 A049599 A334762 * A043261 A157986 A025479
Adjacent sequences: A305458 A305459 A305460 * A305462 A305463 A305464


KEYWORD

nonn,mult


AUTHOR

Matthew Scroggs, Jun 01 2018


EXTENSIONS

a(1) inserted by Andrew Howroyd, Jul 22 2018


STATUS

approved



