A361130 -id:A361130 - cp's OEIS frontend

A361128 Let b = A360519; let Lg = gcd(b(n-1),b(n)), Rg = gcd(b(n),b(n+1)); let L(n) = prod_{primes p|Lg} p-part of b(n), R(n) = prod_{primes p|Rg} p-part of b(n), M(n) = b(n)/(L(n)*R(n)); sequence gives L(n).

1, 2, 5, 7, 3, 4, 5, 11, 9, 2, 7, 11, 3, 5, 2, 11, 13, 3, 4, 7, 13, 5, 2, 17, 7, 9, 2, 13, 17, 3, 2, 19, 5, 3, 4, 11, 17, 25, 2, 23, 3, 19, 4, 13, 3, 5, 2, 29, 3, 31, 8, 7, 3, 37, 4, 17, 3, 41, 16, 5, 23, 7, 12, 5, 29, 49, 2, 3, 43, 25, 2, 3, 47, 5, 8, 3, 7, 19, 2, 27, 5, 31

Offset: 2

Scott R. Shannon, Rémy Sigrist, and N. J. A. Sloane, Mar 09 2023

nonn

The p-part of a number k is the highest power of p that divides k. For example, the 2-part of 24 is 8, the 3-part is 3.
One can think of A360519 as a chain of circles, each circle linked to its neighbors to the left and the right. The n-th term b(n) = A360519(n) is a product L(n)*M(n)*R(n), where L(n) is the part of b(n) sharing primes with the term to the left, R(n) the part sharing primes with the term to the right, and M(n) is the rest of b(n).
By definition of A360519, the set of primes in L(n) is disjoint from the primes in R(n).

N. J. A. Sloane, Table of n, a(n) for n = 2..20000

Cf. A360519, A361118, A361129, A361130.

# Suppose bW is a list of the terms of A360519.
# Then f3(bW[n-1], bW[n], bW[n+1]); returns [L(n), M(n), R(n)] where:
with(numtheory);
f3:=proc(a,b,c)
local lefta,midb,rightc,i,p,pa,pc,ta,tb,tc,t1,t2;
ta:=a; tb:=b; tc:=c;
# left
t1:=igcd(a,b);
t2:=factorset(t1);
t2:=convert(t2,list);
lefta:=1;
for i from 1 to nops(t2) do
p:=t2[i];
while (tb mod p) = 0 do lefta:=lefta*p; tb:=tb/p; od;
od:
# right
t1:=igcd(b,c);
t2:=factorset(t1);
t2:=convert(t2,list);
rightc:=1;
for i from 1 to nops(t2) do
p:=t2[i];
while (tb mod p) = 0 do rightc:=rightc*p; tb:=tb/p; od;
od:
# middle
midb:=b/(lefta*rightc);
[lefta,midb,rightc];
end; # N. J. A. Sloane, Mar 09 2023

A361129 Let b = A360519; let Lg = gcd(b(n-1),b(n)), Rg = gcd(b(n),b(n+1)); let L(n) = prod_{primes p|Lg} p-part of b(n), R(n) = prod_{primes p|Rg} p-part of b(n), M(n) = b(n)/(L(n)*R(n)); sequence gives M(n).

Original entry on oeis.org

3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 11, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 5, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1

Offset: 2

Scott R. Shannon, Rémy Sigrist, and N. J. A. Sloane, Mar 09 2023

nonn

The p-part of a number k is the highest power of p that divides k. For example, the 2-part of 24 is 8, the 3-part is 3.
Since so many of the initial terms are 1, we show more than the usual number of terms in the DATA section.
Conjecture: All terms are odd, and every odd number eventually appears.

N. J. A. Sloane, Table of n, a(n) for n = 2..20000

Cf. A360519, A361118, A361128, A361130.

cp's OEIS Frontend

A361128 Let b = A360519; let Lg = gcd(b(n-1),b(n)), Rg = gcd(b(n),b(n+1)); let L(n) = prod_{primes p|Lg} p-part of b(n), R(n) = prod_{primes p|Rg} p-part of b(n), M(n) = b(n)/(L(n)*R(n)); sequence gives L(n).

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