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A286877
One of the two successive approximations up to 17^n for 17-adic integer sqrt(-1). Here the 4 (mod 17) case (except for n=0).
13
0, 4, 38, 2928, 27493, 1029745, 23747457, 313398285, 3596107669, 94280954402, 450044583893, 28673959190179, 28673959190179, 3524407382568745, 13428985415474682, 13428985415474682, 42949774758062711577, 91610966633729580058, 6709533061724423693474
OFFSET
0,2
COMMENTS
x = ...GC5A24,
x^2 = ...GGGGGG = -1.
FORMULA
a(0) = 0 and a(1) = 4, a(n) = a(n-1) + 2 * (a(n-1)^2 + 1) mod 17^n for n > 1.
a(n) == L(17^n,4) (mod 17^n) == (2 + sqrt(5))^(17^n) + (2 - sqrt(5))^(17^n) (mod 17^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Dec 02 2022
EXAMPLE
a(1) = ( 4)_17 = 4,
a(2) = ( 24)_17 = 38,
a(3) = ( A24)_17 = 2928,
a(4) = (5A24)_17 = 27493.
PROG
(Ruby)
def A(k, m, n)
ary = [0]
a, mod = k, m
n.times{
b = a % mod
ary << b
a = b ** m
mod *= m
}
ary
end
def A286877(n)
A(4, 17, n)
end
p A286877(100)
(Python)
def A(k, m, n):
ary=[0]
a, mod = k, m
for i in range(n):
b=a%mod
ary.append(b)
a=b**m
mod*=m
return ary
def a286877(n):
return A(4, 17, n)
print(a286877(100)) # Indranil Ghosh, Aug 03 2017
(PARI) a(n) = truncate(sqrt(-1+O(17^n))); \\ Michel Marcus, Aug 04 2017
CROSSREFS
The two successive approximations up to p^n for p-adic integer sqrt(-1): A048898 and A048899 (p=5), A286840 and A286841 (p=13), this sequence and A286878 (p=17).
Sequence in context: A131591 A030259 A218710 * A373826 A018860 A016484
KEYWORD
nonn,easy
AUTHOR
Seiichi Manyama, Aug 02 2017
STATUS
approved