%I #12 Apr 18 2020 01:42:35
%S 0,10,19,267,338,1868,2069,948,7796,5245,22215,17198,3477,43855,21272,
%T 95592,60647,186753,135194,45969,263049,139666,467532,301563,55386,
%U 559636,241005,948261,543212,1508854,1001903,318414,1664590,828391,2587041,1575400,280143
%N a(n) is the multiplicative inverse of A008514(n+1) modulo A008514(n).
%H Daniel Hoyt, <a href="/A334137/b334137.txt">Table of n, a(n) for n = 0..10000</a>
%e 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.
%e 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.
%o (PARI)
%o f(n) = n^4 + (n+1)^4;
%o a(n) = lift(1/Mod(f(n+1), f(n)));
%o (Python 3)
%o import gmpy2
%o soc = [] # sum of 4d-centered cubes
%o a=0
%o b=1
%o for i in range(100):
%o c = a**4 + b**4
%o soc.append(c)
%o a += 1
%o b += 1
%o A334137 = []
%o for i in range(len(soc)-1):
%o c = gmpy2.invert(soc[i+1], soc[i])
%o A334137.append(int(c))
%o print(', '.join([str(x) for x in A334137]))
%Y Cf. A308215, A308217, A008514, A334121.
%K nonn,easy
%O 0,2
%A _Daniel Hoyt_, Apr 15 2020