A137727 Final digit of prime(n)*prime(n+1).

6, 5, 5, 7, 3, 1, 3, 7, 7, 9, 7, 7, 3, 1, 1, 7, 9, 7, 7, 3, 7, 7, 7, 3, 7, 3, 1, 3, 7, 1, 7, 7, 3, 1, 9, 7, 1, 1, 1, 7, 9, 1, 3, 1, 3, 9, 3, 1, 3, 7, 7, 9, 1, 7, 1, 7, 9, 7, 7, 3, 9, 1, 7, 3, 1, 7, 7, 9, 3, 7, 7, 3, 1, 7, 7, 7, 3, 7, 9, 1, 9, 1, 3, 7, 7, 7, 3, 7, 3, 1, 3, 3, 7, 9, 7, 7, 9, 3, 3, 7, 9, 1, 7, 9, 7

Offset: 1

Author

Alexander Adamchuk, Feb 08 2008

Keywords

• nonn
• base

Comments

a(n) is 1, 3, 7, or 9 for n > 3. I conjecture that 1 and 9 appear 17/66 of the time and 3 and 7 appear 8/33 of the time. - Charles R Greathouse IV, Jan 03 2013

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A006094 (Products of 2 successive primes), A007652 (Final digit of prime(n)), A010879 (final digit of n), A110923 (final two digits of prime(n) (with leading zero omitted)), A137728 (second digit from the end of product of first n primes).

Programs

Mathematica

Table[ Mod[ Prime[n]*Prime[n+1], 10 ], {n,1,1000} ]
Mod[Times@@@Partition[Prime[Range[110]],2,1],10] (* Harvey P. Dale, Oct 05 2014 *)

PARI

a(n)=prime(n)*prime(n+1)%10 \\ Charles R Greathouse IV, Dec 29 2012

Python

from sympy import prime
def a(n): return (prime(n)*prime(n+1))%10
print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Jun 05 2021

Python

# much faster alternate for initial segment of sequence
from sympy import nextprime
def aupton(terms):
    p1, p2, alst = 2, 3, []
    while len(alst) < terms:
        p1, p2, alst = p2, nextprime(p2), alst + [(p1*p2)%10]
    return alst
print(aupton(105)) # Michael S. Branicky, Jun 05 2021

Formula

a(n) = A010879(A006094(n)). - Felix Fröhlich, Jun 05 2021