A373100 -id:A373100 - cp's OEIS frontend

A373098 Last digit of n*2^n.

0, 2, 8, 4, 4, 0, 4, 6, 8, 8, 0, 8, 2, 6, 6, 0, 6, 4, 2, 2, 0, 2, 8, 4, 4, 0, 4, 6, 8, 8, 0, 8, 2, 6, 6, 0, 6, 4, 2, 2, 0, 2, 8, 4, 4, 0, 4, 6, 8, 8, 0, 8, 2, 6, 6, 0, 6, 4, 2, 2, 0, 2, 8, 4, 4, 0, 4, 6, 8, 8, 0, 8, 2, 6, 6, 0, 6, 4, 2, 2, 0, 2, 8, 4, 4, 0, 4, 6, 8, 8, 0, 8, 2, 6, 6, 0, 6, 4, 2, 2

Offset: 0

Javier Rodríguez Ríos, May 23 2024

Periodic with period 20: [0,2,8,4,4,0,4,6,8,8,0,8,2,6,6,0,6,4,2,2].

Index entries for linear recurrences with constant coefficients, signature (0,1,0,-1,1,1,-1,-1,1,0,-1,0,1).

Cf. A010879, A036289.

Cf. A373099, A373100.

a:= n-> n*2&^n mod 10:
seq(a(n), n=0..100);

Mathematica
```
a[n_] := Mod[n*2^n, 10]
```

a(n,b=10) = lift(n*Mod(2,b)^n) \\ Hugo Pfoertner, May 24 2024

Python
```
def a(n):
    return (n * 2**n) % 10
```

a(n) = A010879(A036289(n)).
a(n) = A373099(n) - 1.
a(n) = A373100(n) + 1 (n >= 1).
a(n) = 0 if n mod 5 = 0,
       2 if n mod 20 = {1, 12, 18, 19},
       8 if n mod 20 = {2,  8,  9, 11},
       4 if n mod 20 = {3,  4,  6, 17},
       6 if n mod 20 = {7, 13, 14, 16}.

A373099 Last digit of n*2^n + 1.

Original entry on oeis.org

1, 3, 9, 5, 5, 1, 5, 7, 9, 9, 1, 9, 3, 7, 7, 1, 7, 5, 3, 3, 1, 3, 9, 5, 5, 1, 5, 7, 9, 9, 1, 9, 3, 7, 7, 1, 7, 5, 3, 3, 1, 3, 9, 5, 5, 1, 5, 7, 9, 9, 1, 9, 3, 7, 7, 1, 7, 5, 3, 3, 1, 3, 9, 5, 5, 1, 5, 7, 9, 9, 1, 9, 3, 7, 7, 1, 7, 5, 3, 3

Offset: 0

Javier Rodríguez Ríos, May 23 2024

This is a cyclic sequence of 20 numbers, using only 1,3,5,7 and 9 (4 times each).

James Cullen, Question 15897, Educational Times, Vol. 58 (December 1905), p. 534.
Richard K. Guy (2004), Unsolved Problems in Number Theory (3rd ed.), New York: Springer Verlag, pp. section B20, ISBN 0-387-20860-7.

Wikipedia, Cullen number.
Index entries for linear recurrences with constant coefficients, signature (0,1,0,-1,1,1,-1,-1,1,0,-1,0,1).

Cf. A010879, A002064.

Cf. A373098, A373100.

lastDigit := proc(n)
    return (n * 2^n + 1) mod 10;
end proc:
# Example usage
minN := 1; maxN := 10;
lastDigits := [seq(lastDigit(n), n = minN .. maxN)];
print(lastDigits);

lastDigit[n_] := Mod[n * 2^n + 1, 10]
(* Example usage *)
minN = 1; maxN = 10;
lastDigits = Table[lastDigit[n], {n, minN, maxN}]
Print[lastDigits]

PARI
```
a(n) = lift(Mod(n*2^n + 1, 10))
```

def last_digit(n):
    return (n * 2**n + 1) % 10
# Example usage
min_n, max_n = 1, 10
last_digits = [last_digit(n) for n in range(min_n, max_n + 1)]
print(last_digits)

a(n) = A010879(A002064(n)).
From Chai Wah Wu, Jul 06 2024: (Start)
a(n) = a(n-2) - a(n-4) + a(n-5) + a(n-6) - a(n-7) - a(n-8) + a(n-9) - a(n-11) + a(n-13) for n > 12.
G.f.: (-3*x^12 - 3*x^11 - 2*x^10 - 4*x^9 + x^8 - 5*x^6 + 2*x^5 + 3*x^4 - 2*x^3 - 8*x^2 - 3*x - 1)/((x - 1)*(x^4 + x^3 + x^2 + x + 1)*(x^8 - x^6 + x^4 - x^2 + 1)). (End)

cp's OEIS Frontend

A373098 Last digit of n*2^n.

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