login
A373099
Last digit of n*2^n + 1.
2
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
0,2
COMMENTS
This is a cyclic sequence of 20 numbers, using only 1,3,5,7 and 9 (4 times each).
REFERENCES
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.
FORMULA
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)
MAPLE
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);
MATHEMATICA
lastDigit[n_] := Mod[n * 2^n + 1, 10]
(* Example usage *)
minN = 1; maxN = 10;
lastDigits = Table[lastDigit[n], {n, minN, maxN}]
Print[lastDigits]
PROG
(Python)
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)
(PARI) a(n) = lift(Mod(n*2^n + 1, 10))
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
STATUS
approved