A334137 a(n) is the multiplicative inverse of A008514(n+1) modulo A008514(n).

0, 10, 19, 267, 338, 1868, 2069, 948, 7796, 5245, 22215, 17198, 3477, 43855, 21272, 95592, 60647, 186753, 135194, 45969, 263049, 139666, 467532, 301563, 55386, 559636, 241005, 948261, 543212, 1508854, 1001903, 318414, 1664590, 828391, 2587041, 1575400, 280143

Offset

0,2

Links

Daniel Hoyt, Table of n, a(n) for n = 0..10000

Example

For a(0), compute the inverse of 1^4 + 2^4 mod 0^4 + 1^4 which is 0 mod 1, since everything mod 1 is 0.
For a(1), compute the inverse of 2^4 + 3^4 mod 1^4 + 2^4. The inverse of 97 mod 17 (or 12 mod 17) is 10 mod 17 since 10*12 = 120 has remainder 1 mod 17.

Prog

(PARI)
f(n) = n^4 + (n+1)^4;
a(n) = lift(1/Mod(f(n+1), f(n)));
(Python 3)
import gmpy2
soc = [] # sum of 4d-centered cubes
a=0
b=1
for i in range(100):
    c = a**4 + b**4
    soc.append(c)
    a += 1
    b += 1
A334137 = []
for i in range(len(soc)-1):
    c = gmpy2.invert(soc[i+1], soc[i])
    A334137.append(int(c))
print(', '.join([str(x) for x in A334137]))

Crossrefs

Cf. A308215, A308217, A008514, A334121.

Sequence in context: A110368 A006050 A045646 * A212594 A255529 A023916

Adjacent sequences: A334134 A334135 A334136 * A334138 A334139 A334140

Keyword

nonn,easy

Author

Daniel Hoyt, Apr 15 2020

Status

approved