Displaying 1-10 of 12 results found.
10-adic integer x=...92160195896736500120813568 satisfying x^5 = x; also x^3 = -x = A120817; (x^2)^3 = x^2 = A091664; (x^4)^2 = x^4 = A018248.
+20
10
8, 6, 5, 3, 1, 8, 0, 2, 1, 0, 0, 5, 6, 3, 7, 6, 9, 8, 5, 9, 1, 0, 6, 1, 2, 9, 5, 9, 6, 4, 4, 3, 8, 5, 7, 7, 8, 5, 5, 8, 4, 5, 7, 6, 9, 6, 4, 4, 5, 9, 6, 6, 7, 7, 6, 7, 4, 0, 5, 3, 0, 6, 1, 6, 0, 4, 7, 3, 1, 3, 9, 0, 4, 2, 7, 9, 0, 8, 5, 3, 5, 6, 3, 5, 0, 3, 6, 6, 6, 9, 1, 7, 9, 6, 6, 4, 1, 1, 6, 5, 9, 5, 6, 4, 4
FORMULA
x = 10-adic lim_{n->oo} 8^(5^n).
EXAMPLE
x equals the limit of the (n+1) trailing digits of 8^(5^n):
8^(5^0)=(8), 8^(5^1)=327(68), 8^(5^2)=37778931862957161709(568), ...
x=...06160350476776695446967548558775834469592160195896736500120813568.
x^2=...0557423423230896109004106619977392256259918212890624 ( A091664). x^3=...3304553032451441224165530407839804103263499879186432 ( A120817). x^4=...9442576576769103890995893380022607743740081787109376 ( A018248). x^5=...6695446967548558775834469592160195896736500120813568 = x.
PROG
(PARI) {a(n)=local(b=8, v=[]); for(k=1, n+1, b=b^5%10^k; v=concat(v, (10*b\10^k))); v[n+1]}
Hypotenuse numbers (squares are sums of 2 nonzero squares).
+10
74
5, 10, 13, 15, 17, 20, 25, 26, 29, 30, 34, 35, 37, 39, 40, 41, 45, 50, 51, 52, 53, 55, 58, 60, 61, 65, 68, 70, 73, 74, 75, 78, 80, 82, 85, 87, 89, 90, 91, 95, 97, 100, 101, 102, 104, 105, 106, 109, 110, 111, 113, 115, 116, 117, 119, 120, 122, 123, 125, 130, 135, 136, 137, 140
COMMENTS
Circumradius R of the triangles such that the area, the sides and R are integers. - Michel Lagneau, Mar 03 2012 The 2 squares summing to a(n)^2 cannot be equal because sqrt(2) is not rational. - Jean-Christophe Hervé, Nov 10 2013 Closed under multiplication. The primitive elements are those with exactly one prime divisor of the form 4k + 1 with multiplicity one, which are also those for which there exists a unique integer triangle = A084645. - Jean-Christophe Hervé, Nov 11 2013 a(n) are numbers whose square is the mean of two distinct nonzero squares. This creates 1-to-1 mapping between a Pythagorean triple and a "Mean" triple. If the Pythagorean triple is written, abnormally, as {j, k, h} where j^2 +(j+k)^2 = h^2, and h = a(n), then the corresponding "Mean" triple with the same h is {k, 2j, h} where (k^2 + (k+2j)^2)/2 = h^2. For example for h = 5, the Pythagorean triple is {3, 1, 5} and the Mean triple is {1, 6, 5}. - Richard R. Forberg, Mar 01 2015 Integral side lengths of rhombuses with integral diagonals p and q (therefore also with integral areas A because A = pq/2 is some multiple of 24). No such rhombuses are squares. - Rick L. Shepherd, Apr 09 2017 Conjecture: these are bases n in which exists an n-adic integer x satisfying x^5 = x, and 5 is the smallest k>1 such that x^k =x (so x^2, x^3 and x^4 are not x). Example: the 10-adic integer x = ...499879186432 ( A120817) satisfies x^5 = x, and x^2, x^3, and x^4 are not x, so 10 is in this sequence. See also A120817, A210850 and A331548. - Patrick A. Thomas, Mar 01 2020 Didactic comment: When students solve a quadratic equation a*x^2 + b*x + c = 0 (a, b, c: integers) with the solution formula, they often make the mistake of calculating b^2 + 4*a*c instead of b^2 - 4*a*c (especially if a or c is negative). If the root then turns out to be an integer, they feel safe. This sequence lists the absolute values of b for which this error can happen. Reasoning: With p^2 = b^2 - 4*a*c and q^2 = b^2 + 4*a*c it follows by addition immediately that p^2 + q^2 = 2*b^2. If 4*a*c < 0, let p = x + y and q = x - y. If 4*a*c > 0, let p = x - y and q = x + y. In both cases follows that y^2 + x^2 = b^2. So every Pythagorean triple gives an absolute value of b for which this error can occur. Example: From (y, x, b) = (3, 4, 5) follows (q^2, b^2, p^2) = (1, 25, 49) or (p^2, b^2, q^2) = (1, 25, 49) with abs(4*a*c) = 24. - Felix Huber, Jul 22 2023 Conjecture: Numbers m such that the limit: Limit_{s->1} zeta(s)*Sum_{k=1..m} [k|m]* A008683(k)*(i^k)/(k^(s - 1)) exists, which is equivalent to numbers m such that abs(Sum_{k=1..m} [k|m]* A008683(k)*(i^k)) = 0. - Mats Granvik, Jul 06 2024
REFERENCES
Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 98-104.
MAPLE
isA009003 := proc(n)
local p;
for p in numtheory[factorset](n) do
if modp(p, 4) = 1 then
return true;
end if;
end do:
false;
end proc:
for n from 1 to 200 do
if isA009003(n) then
printf("%d, ", n) ;
end if;
MATHEMATICA
Select[Range[200], Length[PowersRepresentations[#^2, 2, 2]] > 1 &] (* Alonso del Arte, Feb 11 2014 *)
PROG
(PARI) list(lim)=my(v=List(), u=vectorsmall(lim\=1)); forprimestep(p=5, lim, 4, forstep(n=p, lim, p, u[n]=1)); for(i=5, lim, if(u[i], listput(v, i))); u=0; Vec(v) \\ Charles R Greathouse IV, Jan 13 2022 (Haskell)
import Data.List (findIndices)
a009003 n = a009003_list !! (n-1)
a009003_list = map (+ 1) $ findIndices (> 0) a005089_list
(Python)
from itertools import count, islice
from sympy import primefactors
def A009003_gen(): # generator of terms return filter(lambda n:any(map(lambda p: p % 4 == 1, primefactors(n))), count(1))
CROSSREFS
Complement of A004144. Primes in this sequence give A002144. Same as A146984 (integer contraharmonic means) as sets - see Pahikkala 2010, Theorem 5.
Convergence speed of m^^m, where m = A067251(n) and n >= 2. a(n) = f(m, m) - f(m, m - 1), where f(x, y) corresponds to the maximum value of k, such that x^^y == x^^(y + 1) (mod 10^k).
+10
30
0, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 2, 1, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 6, 1, 1, 3, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 5, 1, 1
COMMENTS
It is possible to anticipate the convergence speed of a^^m, where ^^ indicates tetration or hyper-4 (e.g., 3^^4=3^(3^(3^3))), simply looking at the congruence (mod 25) of m. In fact, assuming m > 2, a(n) = 1 for any m == 2, 3, 4, 6, 8, 9, 11, 12, 13, 14, 16, 17, 19, 21, 22, 23 (mod 25), and a(n) >= 2 otherwise.
It follows that 32/45 = 71.11% of the a(n) assume unitary value.
You can also obtain an arbitrary high convergence speed, such as taking the beautiful base b = 999...99 (9_9_9... n times), which gives a(n) = len(b), for any len(b) > 1. Thus, 99...9^^m == 99...9^^(m + 1) (mod m*10^len(b)), as proved by Ripà in "La strana coda della serie n^n^...^n", pages 25-26. In fact, m = 99...9 == 24 (mod 25) and a(m=24) > 1.
Knowing the "constant congruence speed" of a given base (a.k.a. the convergence speed of the base m, assuming m > 2) is very useful in order to calculate the exact number of stable digits of all its tetrations of height b > 1. As an example, let us consider all the a(n) such that n is congruent to 4 (mod 9) (i.e., all the tetration bases belonging to the congruence class 5 (mod 10)). Then, the exact number of stable digits (#S(m, b)) of any tetration m^^b (i.e., the number of its last "frozen" digits) such that m is congruent to 5 (mod 10), for any b >= 3, can automatically be calculated by simply knowing that (under the stated constraint) the congruence speed of the m corresponds to the 2-adic valuation of (m^2 - 1) minus 1. Thus, let k = 1, 2, 3, ..., and we have that
If m = 20*k - 5, then #S(m, b > 2) = b*(v_2(m^2 - 1) - 1) + 1 = b*(v_2(m + 1) + 1);
If m = 20*k + 5, then #S(m, b > 2) = (b + 1)*(v_2(m^2 - 1) - 1) = (b + 1)*(v_2(m - 1));
If m = 5, then #S(m, 1) = 1, #S(m, 2) = 4, #S(m, b > 2) = 8 + 2*(b - 3).
(End)
For any n > 2, the value of a(n) depends on the congruence modulo 18 of n, since the constant congruence speed of m arises from the 14 nontrivial solutions of the fundamental equation y^5 = y in the (commutative) ring of decadic integers (e.g., y = -1 = ...9999 is a solution of y^5 = y, so it originates the law a(n) = min(v_2(m + 1), v_5(m + 1)) concerning every n belonging to the congruence class 0 modulo 18, as stated in the "Formula" section of the present sequence).
(End)
REFERENCES
Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. ISBN 978-88-6178-789-6
FORMULA
Let n > 2. For any integer c >= 0, if n is an element of the set {5,7,14,17,22,23,24,29,32,39,41,45,46}, then a(n+45*c) >= 2; whereas a(n) = 1 otherwise. - Marco Ripà, Sep 28 2018 If n == 5 (mod 9), then a(n) = v_2(a(n)^2 - 1) - 1, where v_2(x) indicates the 2-adic valuation of x. - Marco Ripà, Dec 19 2021 If n == 1 (mod 18) and n<>1, then a(n) = min(v_2(m - 1), v_5(m - 1)) (i.e., 1 plus the number of trailing zeros, if any, next to the rightmost digit of m);
if n == 10 (mod 18), then a(n) = min(v_2(m + 1), v_5(m - 1));
if n == {2,8}(mod 9) and n<>2, then a(n) = v_5(m^2 + 1);
if n == {3,7}(mod 18), then a(n) = min(v_2(m + 1), v_5(n^2 + 1));
if n == {12,16}(mod 18), then a(n) = min(v_2(m - 1), v_5(n^2 + 1));
if n == 4 (mod 9), then a(n) = v_5(m + 1);
if n == 5 (mod 18), then a(n) = v_2(m - 1);
if n == 14 (mod 18), then a(n) = v_2(m + 1);
if n == 6 (mod 9), then a(n) = v_5(m - 1);
if n == 9 (mod 18), then a(n) = min(v_2(m - 1), v_5(m + 1));
if n == 0 (mod 18), then a(n) = min(v_2(m + 1), v_5(m + 1)) (i.e., number of digits of the rightmost repunit "9's" of m); where v_2(x) and v_5(x) indicates the 2-adic valuation of (x) and the 5-adic valuation of (x), respectively. - Marco Ripà, Feb 17 2022
EXAMPLE
For m = 25, a(23) = 3 implies that 25^^(25+i) freezes 3*i "new" rightmost digits (i >= 0).
PROG
(PARI) \\ uses reducetower.gp from links
f2(x, y) = my(k=0); while(reducetower(x, 10^k, y) == reducetower(x, 10^k, y+1), k++); k;
f1(n) = polcoef(x*(x+1)*(x^4-x^3+x^2-x+1)*(x^4+x^3+x^2+x+1) / ((x-1)^2*(x^2+x+1)*(x^6+x^3+1)) + O(x^(n+1)), n, x); \\ A067251a(n) = my(m=f1(n)); f2(m, m) - f2(m, m-1);
lista(nn) = {for (n=2, nn, print1(a(n), ", "); ); } \\ Michel Marcus, Jan 27 2021
CROSSREFS
Cf. A067251, A317824, A317903, A349425, A370211, A370775, A371129, A371671, A371720, A372490, A373387. Cf. A000007, A018247, A018248, A063006, A091661, A091663, A091664, A120817, A120818, A290372, A290373, A290374, A290375.
The 10-adic integer x = ...1787109376 satisfies x^2 = x.
+10
27
6, 7, 3, 9, 0, 1, 7, 8, 7, 1, 8, 0, 0, 4, 7, 3, 4, 7, 7, 0, 6, 2, 2, 0, 0, 8, 3, 3, 9, 8, 5, 9, 9, 0, 9, 8, 3, 0, 1, 9, 6, 7, 6, 7, 5, 6, 7, 5, 2, 4, 4, 9, 9, 9, 8, 8, 1, 6, 3, 1, 9, 1, 4, 0, 9, 4, 3, 3, 8, 7, 3, 9, 9, 0, 1, 0, 9, 4, 1, 6, 0, 7, 9, 1, 0, 3, 8, 1, 9, 8, 0, 8, 6, 2, 9, 9, 6, 4, 0, 6, 9, 0, 6, 3, 7, 5, 3, 2
COMMENTS
The 10-adic numbers a and b defined in A018247 and this sequence satisfy a^2=a, b^2=b, a+b=1, ab=0. - Michael Somos
REFERENCES
W. W. R. Ball, Mathematical Recreations & Essays, N.Y. Macmillan Co, 1947.
R. Cuculière, Jeux Mathématiques, in Pour la Science, No. 6 (1986), 10-15.
V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.
M. Kraitchik, Sphinx, 1935, p. 1.
A. M. Robert, A Course in p-adic Analysis, Springer, 2000; see pp. 63, 419.
LINKS
V. deGuerre and R. A. Fairbairn, Automorphic numbers, Jnl. Rec. Math., 1 (No. 3, 1968), 173-179.
FORMULA
x = r^4 where r=...1441224165530407839804103263499879186432 ( A120817). x = 10-adic limit_{n->oo} 6^(5^n). - Paul D. Hanna, Jul 06 2006 For n >= 2, the final n+1 digits of either 2^(10^n), 4^(10^n) or 6^(10^n), when read from right to left, give the first n+1 entries in the sequence. - Peter Bala, Nov 05 2022
EXAMPLE
x equals the limit of the (n+1) trailing digits of 6^(5^n):
6^(5^0)=(6), 6^(5^1)=77(76), 6^(5^2)=28430288029929701(376), ...
x = ...9442576576769103890995893380022607743740081787109376.
Trailing digits of 2^(10^n), 4^(10^n) and 6^(10^n) for n = 5:
2^(10^5) = ...9883(109376);
4^(10^5) = ...7979(109376);
6^(10^5) = ...4155(109376). (End)
MAPLE
a := proc (n) option remember; if n = 1 then 2 else irem(a(n-1)^10, 10^n) end if; end proc:
# display the digits of a(100) from right to left
S := convert(a(100), string):
with(ListTools):
the_List := [seq(parse(S[i]), i = 1..length(S))]:
MATHEMATICA
b = {6}; g[n_] := Block[{k = 0, c}, While[c = FromDigits[Prepend[b, k]]; Mod[c^2, 10^n] != c, k++ ]; b = Prepend[b, k]]; Do[ g[n], {n, 2, 105}]; Reverse[b]
With[{n = 150}, Reverse[IntegerDigits[PowerMod[16, 5^n, 10^n]]]] (* IWABUCHI Yu(u)ki, Feb 16 2024 *)
PROG
(PARI) {a(n)=local(b=6, v=[]); for(k=1, n+1, b=b^5%10^k; v=concat(v, (10*b\10^k))); v[n+1]} \\ Paul D. Hanna, Jul 06 2006 (PARI) Vecrev(digits(lift(chinese(Mod(0, 2^100), Mod(1, 5^100))))) \\ Seiichi Manyama, Aug 07 2019
CROSSREFS
A016090 gives associated automorphic numbers.
AUTHOR
Yoshihide Tamori (yo(AT)salk.edu)
10-adic integer x=.....06619977392256259918212890624 satisfying x^3 = x.
+10
17
4, 2, 6, 0, 9, 8, 2, 1, 2, 8, 1, 9, 9, 5, 2, 6, 5, 2, 2, 9, 3, 7, 7, 9, 9, 1, 6, 6, 0, 1, 4, 0, 0, 9, 0, 1, 6, 9, 8, 0, 3, 2, 3, 2, 4, 3, 2, 4, 7, 5, 5, 0, 0, 0, 1, 1, 8, 3, 6, 8, 0, 8, 5, 9, 0, 5, 6, 6, 1, 2, 6, 0, 0, 9, 8, 9, 0, 5, 8, 3, 9, 2, 0, 8, 9, 6, 1, 8, 0, 1, 9, 1, 3, 7, 0, 0, 3, 5, 9, 3
COMMENTS
Let a,b be integers defined in A018247, A018248 satisfying a^2=a, b^2=b, obviously a^3=a, b^3=b; let c,d,e,f be integers defined in A091661, A063006, A091663, A091664; then c^3=c, d^3=d, e^3=e, f^3=f, c+d=1, a+e=1, b+f=1, b+c=a, d+f=e, a+f=c, a=f+1, b=e+1, cd=-1, af=-1, gh=-1 where -1=.....999999999.
FORMULA
x = r^2 where r = ...1441224165530407839804103263499879186432 ( A120817). x = 10-adic lim_{n->oo} 4^(5^n). - Paul D. Hanna, Jul 06 2006
EXAMPLE
x equals the limit of the (n+1) trailing digits of 4^(5^n):
4^(5^0) = (4), 4^(5^1) = 10(24), 4^(5^2) = 1125899906842(624), ...
x = ...0557423423230896109004106619977392256259918212890624.
MATHEMATICA
To calculate c, d, e, f use Mathematica algorithms for a, b and equations: c=a-b, d=1-c, e=b-1, f=a-1.
PROG
(PARI) {a(n)=local(b=4, v=[]); for(k=1, n+1, b=b^5%10^k; v=concat(v, (10*b\10^k))); v[n+1]} \\ Paul D. Hanna, Jul 06 2006 (PARI) ( A091664_vec(n)=Vecrev(digits(lift(chinese(Mod(0, 2^n), Mod(-1, 5^n))))))(99) \\ M. F. Hasler, Jan 26 2020
AUTHOR
Edoardo Gueglio (egueglio(AT)yahoo.it), Jan 28 2004
10-adic integer x = ...5807 satisfying x^5 = x.
+10
8
7, 0, 8, 5, 9, 2, 6, 6, 6, 1, 8, 5, 3, 0, 0, 7, 4, 8, 1, 1, 4, 2, 6, 8, 7, 8, 7, 3, 2, 4, 1, 6, 1, 5, 1, 1, 5, 4, 5, 0, 2, 2, 9, 0, 6, 9, 2, 1, 7, 4, 7, 2, 2, 2, 2, 1, 7, 5, 8, 7, 8, 5, 2, 4, 8, 0, 6, 9, 6, 4, 4, 8, 5, 8, 3, 0, 8, 6, 5, 2, 5, 0, 6, 6, 9, 9, 1, 5
FORMULA
p = A120817 = ...186432, q = A018247 = ...890625, x = p - q = ...295807.
EXAMPLE
7^5 - 7 == 0 mod 10,
7^5 - 7 == 0 mod 10^2,
807^5 - 807 == 0 mod 10^3,
5807^5 - 5807 == 0 mod 10^4.
2^(5^0) - 5^(2^0) == 7 mod 10,
2^(5^1) - 5^(2^1) == 7 mod 10^2,
2^(5^2) - 5^(2^2) == 807 mod 10^3,
2^(5^3) - 5^(2^3) == 5807 mod 10^4. (End)
PROG
(Ruby)
def P(n)
s1, s2 = 2, 8
n.times{|i|
m = 10 ** (i + 1)
(0..9).each{|j|
k1, k2 = j * m + s1, (9 - j) * m + s2
if (k1 ** 5 - k1) % (m * 10) == 0 && (k2 ** 5 - k2) % (m * 10) == 0
s1, s2 = k1, k2
break
end
}
}
s1
end
def Q(s, n)
n.times{|i|
m = 10 ** (i + 1)
(0..9).each{|j|
k = j * m + s
if (k ** 2 - k) % (m * 10) == 0
s = k
break
end
}
}
s
end
str = (10 ** (n + 1) + P(n) - Q(5, n)).to_s.reverse
(0..n).map{|i| str[i].to_i}
end
10-adic integer x = ...2943 satisfying x^5 = x.
+10
8
3, 4, 9, 2, 2, 9, 7, 0, 9, 1, 8, 5, 6, 7, 4, 0, 4, 6, 3, 0, 8, 2, 8, 1, 2, 7, 9, 2, 6, 3, 0, 3, 8, 6, 6, 6, 2, 6, 6, 7, 1, 3, 4, 4, 5, 3, 2, 0, 8, 3, 1, 6, 7, 7, 5, 6, 6, 6, 8, 4, 9, 7, 5, 6, 9, 8, 0, 7, 9, 0, 3, 0, 4, 3, 8, 9, 9, 2, 7, 9, 5, 3, 3, 7, 0, 6, 4, 8
FORMULA
p = A120818 = ...813568, q = A018247 = ...890625, x = p - q = ...922943.
EXAMPLE
3^5 - 3 == 0 mod 10,
43^5 - 43 == 0 mod 10^2,
943^5 - 943 == 0 mod 10^3,
2943^5 - 2943 == 0 mod 10^4.
8^(5^0) - 5^(2^0) == 3 mod 10,
8^(5^1) - 5^(2^1) == 43 mod 10^2,
8^(5^2) - 5^(2^2) == 943 mod 10^3,
8^(5^3) - 5^(2^3) == 2943 mod 10^4. (End)
PROG
(Ruby)
def P(n)
s1, s2 = 2, 8
n.times{|i|
m = 10 ** (i + 1)
(0..9).each{|j|
k1, k2 = j * m + s1, (9 - j) * m + s2
if (k1 ** 5 - k1) % (m * 10) == 0 && (k2 ** 5 - k2) % (m * 10) == 0
s1, s2 = k1, k2
break
end
}
}
s2
end
def Q(s, n)
n.times{|i|
m = 10 ** (i + 1)
(0..9).each{|j|
k = j * m + s
if (k ** 2 - k) % (m * 10) == 0
s = k
break
end
}
}
s
end
str = (10 ** (n + 1) + P(n) - Q(5, n)).to_s.reverse
(0..n).map{|i| str[i].to_i}
end
10-adic integer x = ...7057 satisfying x^5 = x.
+10
8
7, 5, 0, 7, 7, 0, 2, 9, 0, 8, 1, 4, 3, 2, 5, 9, 5, 3, 6, 9, 1, 7, 1, 8, 7, 2, 0, 7, 3, 6, 9, 6, 1, 3, 3, 3, 7, 3, 3, 2, 8, 6, 5, 5, 4, 6, 7, 9, 1, 6, 8, 3, 2, 2, 4, 3, 3, 3, 1, 5, 0, 2, 4, 3, 0, 1, 9, 2, 0, 9, 6, 9, 5, 6, 1, 0, 0, 7, 2, 0, 4, 6, 6, 2, 9, 3, 5, 1
FORMULA
p = A120817 = ...186432, q = A018247 = ...890625, x = p + q = ...077057.
EXAMPLE
7^5 - 7 == 0 mod 10,
57^5 - 57 == 0 mod 10^2,
57^5 - 57 == 0 mod 10^3,
7057^5 - 7057 == 0 mod 10^4.
2^(5^0) + 5^(2^0) == 7 mod 10,
2^(5^1) + 5^(2^1) == 57 mod 10^2,
2^(5^2) + 5^(2^2) == 57 mod 10^3,
2^(5^3) + 5^(2^3) == 7057 mod 10^4. (End)
PROG
(Ruby)
def P(n)
s1, s2 = 2, 8
n.times{|i|
m = 10 ** (i + 1)
(0..9).each{|j|
k1, k2 = j * m + s1, (9 - j) * m + s2
if (k1 ** 5 - k1) % (m * 10) == 0 && (k2 ** 5 - k2) % (m * 10) == 0
s1, s2 = k1, k2
break
end
}
}
s1
end
def Q(s, n)
n.times{|i|
m = 10 ** (i + 1)
(0..9).each{|j|
k = j * m + s
if (k ** 2 - k) % (m * 10) == 0
s = k
break
end
}
}
s
end
str = (P(n) + Q(5, n)).to_s.reverse
(0..n).map{|i| str[i].to_i}
end
10-adic integer x = ...4193 satisfying x^5 = x.
+10
8
3, 9, 1, 4, 0, 7, 3, 3, 3, 8, 1, 4, 6, 9, 9, 2, 5, 1, 8, 8, 5, 7, 3, 1, 2, 1, 2, 6, 7, 5, 8, 3, 8, 4, 8, 8, 4, 5, 4, 9, 7, 7, 0, 9, 3, 0, 7, 8, 2, 5, 2, 7, 7, 7, 7, 8, 2, 4, 1, 2, 1, 4, 7, 5, 1, 9, 3, 0, 3, 5, 5, 1, 4, 1, 6, 9, 1, 3, 4, 7, 4, 9, 3, 3, 0, 0, 8, 4
FORMULA
p = A120818 = ...813568, q = A018247 = ...890625, x = p + q = ...704193.
EXAMPLE
3^5 - 3 == 0 mod 10,
93^5 - 93 == 0 mod 10^2,
193^5 - 193 == 0 mod 10^3,
4193^5 - 4193 == 0 mod 10^4.
8^(5^0) + 5^(2^0) == 3 mod 10,
8^(5^1) + 5^(2^1) == 93 mod 10^2,
8^(5^2) + 5^(2^2) == 193 mod 10^3,
8^(5^3) + 5^(2^3) == 4193 mod 10^4. (End)
PROG
(Ruby)
def P(n)
s1, s2 = 2, 8
n.times{|i|
m = 10 ** (i + 1)
(0..9).each{|j|
k1, k2 = j * m + s1, (9 - j) * m + s2
if (k1 ** 5 - k1) % (m * 10) == 0 && (k2 ** 5 - k2) % (m * 10) == 0
s1, s2 = k1, k2
break
end
}
}
s2
end
def Q(s, n)
n.times{|i|
m = 10 ** (i + 1)
(0..9).each{|j|
k = j * m + s
if (k ** 2 - k) % (m * 10) == 0
s = k
break
end
}
}
s
end
str = (P(n) + Q(5, n)).to_s.reverse
(0..n).map{|i| str[i].to_i}
end
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.
+10
4
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, 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, 6, 12, 8, 2, 12, 4, 6, 1, 6, 14, 6, 7, 8, 13, 5, 5, 3, 4, 3, 0
COMMENTS
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. 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
FORMULA
x = 15-adic lim_{n->infinity} 3^(5^n).
EXAMPLE
x equals the limit of the (n+1) trailing digits of 3^(5^n):
3^(5^0) = (3), 3^(5^1) = 1(13), 3^(5^2) = 1708EB01(D13), ...
x = ...2AA66B44A40E43797853AD13.
x^2 = ...65762C0520697E8CA1A31469 = A331550. x^3 = ...C44883AA4AE0AB75769B41DC = -x.
x^4 = ...8978C2E9CE8570624D4BDA86 = A331549. x^5 = ...2AA66B44A40E43797853AD13 = x.
PROG
(PARI) \\ after Paul D. Hanna's program in A120817{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]}
(PARI) ( A331548_vec(n)=Vecrev(digits(lift(Mod(3, 15^n)^5^(n-1)), 15)))(99) \\ M. F. Hasler, Jan 26 2020
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