

A336520


Primes in Pi: a(n) is the smallest prime factor of A090897(n) that does not appear in earlier terms of A090897, or 1, if no such factor exists.


2



3, 2, 53, 379, 58979, 161923, 2643383, 1746893, 6971, 5, 17, 1499, 11, 1555077581737, 297707, 4733, 37, 126541, 2130276389911155737, 1429, 71971, 383, 61, 1559, 29, 193, 12073, 698543, 157, 20289606809, 23687, 1249, 59, 2393, 251, 101, 15827173, 82351, 661
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OFFSET

1,1


COMMENTS

Inspired by a comment of Mario Cortés in A090897, who suggests that 1 might not appear in this sequence.
Differs from A336519 for n = 4, 16, 73, 83, 90, ....
a(n) is not 1 for the first 2000 terms. We can prove that a(n) has a prime factor p that does not divide LCM(A090897(1), ..., A090897(n1)) without using prime number factorization. The method is explained in the link below.  David A. Corneth, Aug 22 2020


LINKS



EXAMPLE

[ 1] 3, {3} > 3;
[ 2] 14, {2, 7} > 2;
[ 3] 159, {3, 53} > 53;
[ 4] 2653, {7, 379} > 379;
[ 5] 58979, {58979} > 58979;
[ 6] 323846, {2, 161923} > 161923;
[ 7] 2643383, {2643383} > 2643383;
[ 8] 27950288, {2, 1746893} > 1746893;
[ 9] 419716939, {6971, 60209} > 6971;
[10] 9375105820, {2, 5, 1163, 403057} > 5.


PROG

(SageMath)
def Select(item, Selected):
return next((x for x in item if not (x in Selected)), 1)
def PiPart(n):
return floor(pi * 10^(n * (n + 1) // 2  1)) % 10^n
def A336520List(len):
prev = []; ret = []
for n in range(1, len + 1):
p = prime_factors(PiPart(n))
ret.append(Select(p, prev))
prev.extend(p)
return ret
print(A336520List(39))
def LcmPiPart(n):
return lcm([PiPart(n) for n in (1..n)])
def is_an_prime(n):
lcmpi = LcmPiPart(n  1)
lm, m = 1, PiPart(n)
while lm != m:
lm, m = m, lcm(lcmpi, m) // lcmpi
return m > 1


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



