cp's OEIS Frontend

A331548 15-adic integer x = ...2AA66B44A40E43797853AD13 satisfying x^5 = x; also x^3 = -x; (x^2)^3 = x^2 = A331550; (x^4)^2 = x^4 = A331549.

%I A331548 #31 May 02 2020 23:27:17
%S A331548 3,1,13,10,3,5,8,7,9,7,3,4,14,0,4,10,4,4,11,6,6,10,10,2,8,1,9,9,0,4,8,
%T A331548 3,10,11,5,9,11,0,8,0,10,9,2,6,0,8,11,5,8,5,7,1,6,10,5,12,14,0,0,6,10,
%U A331548 6,12,8,2,12,4,6,1,6,14,6,7,8,13,5,5,3,4,3,0
%N A331548 15-adic integer x = ...2AA66B44A40E43797853AD13 satisfying x^5 = x; also x^3 = -x; (x^2)^3 = x^2 = A331550; (x^4)^2 = x^4 = A331549.
%C A331548 The base-15 version of A120817.  A, B, C, D, and E are the standard notations for the hexadecimal digits 10, 11, 12, 13, and 14, respectively.
%C A331548 Conjecture: If k is the number of prime factors congruent to 1 (mod 4) of an integer n, then there are exactly k n-adic integers x satisfying x^5 = x, while not satisfying x^h = x for h = 2, 3, or 4. This does not count -x, which also satisfies, in each case. - _Patrick A. Thomas_, Mar 31 2020
%F A331548 x = 15-adic lim_{n->infinity} 3^(5^n).
%e A331548 x equals the limit of the (n+1) trailing digits of 3^(5^n):
%e A331548 3^(5^0) = (3), 3^(5^1) = 1(13), 3^(5^2) = 1708EB01(D13), ...
%e A331548 x   = ...2AA66B44A40E43797853AD13.
%e A331548 x^2 = ...65762C0520697E8CA1A31469 = A331550.
%e A331548 x^3 = ...C44883AA4AE0AB75769B41DC = -x.
%e A331548 x^4 = ...8978C2E9CE8570624D4BDA86 = A331549.
%e A331548 x^5 = ...2AA66B44A40E43797853AD13 = x.
%o A331548 (PARI) \\ after Paul D. Hanna's program in A120817
%o A331548 {a(n)=local(b=3, v=[]); for(k=1, n+1, b=b^5%15^k; v=concat(v, (15*b\15^k))); v[n+1]}
%o A331548 for(k=0,80,print1(a(k),", ")) \\ _Hugo Pfoertner_, Jan 26 2020
%o A331548 (PARI) (A331548_vec(n)=Vecrev(digits(lift(Mod(3,15^n)^5^(n-1)),15)))(99) \\ _M. F. Hasler_, Jan 26 2020
%Y A331548 Cf. A120817, A331549, A331550.
%K A331548 base,nonn
%O A331548 0,1
%A A331548 _Patrick A. Thomas_, Jan 20 2020