Displaying 1-10 of 28 results found.
10-adic integer x=...07839804103263499879186432 satisfying x^5 = x; also x^3 = -x = A120818; (x^2)^3 = x^2 = A091664; (x^4)^2 = x^4 = A018248.
+20
15
2, 3, 4, 6, 8, 1, 9, 7, 8, 9, 9, 4, 3, 6, 2, 3, 0, 1, 4, 0, 8, 9, 3, 8, 7, 0, 4, 0, 3, 5, 5, 6, 1, 4, 2, 2, 1, 4, 4, 1, 5, 4, 2, 3, 0, 3, 5, 5, 4, 0, 3, 3, 2, 2, 3, 2, 5, 9, 4, 6, 9, 3, 8, 3, 9, 5, 2, 6, 8, 6, 0, 9, 5, 7, 2, 0, 9, 1, 4, 6, 4, 3, 6, 4, 9, 6, 3, 3, 3, 0, 8, 2, 0, 3, 3, 5, 8, 8, 3, 4, 0, 4, 3, 5, 5
FORMULA
x = 10-adic limit_{n->infinity} 2^(5^n).
EXAMPLE
x equals the limit of the (n+1) trailing digits of 2^(5^n):
2^(5^0)=(2), 2^(5^1)=(32), 2^(5^2)=33554(432),
2^(5^3)=4253529586511730793292182592897102(6432), ...
x=...93839649523223304553032451441224165530407839804103263499879186432.
x^2=...0557423423230896109004106619977392256259918212890624 ( A091664). x^3=...6695446967548558775834469592160195896736500120813568 ( A120818). x^4=...9442576576769103890995893380022607743740081787109376 ( A018248). x^5=...3304553032451441224165530407839804103263499879186432 = x.
PROG
(PARI) {a(n)=local(b=2, v=[]); for(k=1, n+1, b=b^5%10^k; v=concat(v, (10*b\10^k))); v[n+1]}
(PARI) {a(n)=if(n<0, 0, lift(chinese(Mod(truncate( teichmuller(2+O(5^(n+1)))), 5^(n+1)), Mod(0, 2^(n+1))))\10^n)} /* Michael Somos, Oct 03 2006 */
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
12
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]}
1, 5, -3, 9, -9, -7, 5, 7, 5, -7, 7, -9, -9, -1, 5, -3, -1, 5, 5, -9, 3, -5, -5, -9, -9, 7, -3, -3, 9, 7, 1, 9, 9, -9, 9, 7, -3, -9, -7, 9, 3, 5, 3, 5, 1, 3, 5, 1, -5, -1, -1, 9, 9, 9, 7, 7, -7, 3, -3, -7, 9, -7, -1, -9, 9, -1, -3, -3, 7, 5, -3, 9, 9, -9, -7, -9, 9, -1, -7, 3, -9, 5, 9, -7
COMMENTS
Numbers in A018247 and A018248 are known as automorphic numbers in base 10, meaning that the infinite integers a=(...256259918212890625) or b=(...743740081787109376) provides a nontrivial solution to x*x == x (mod any power of 10). All entries must be odd.
Is the accumulative sum equally positive and negative, i.e. does the sum equal 0 infinitely often?
MATHEMATICA
(* execute the programming in both A018247 & A018248 *) Reverse[b - a] aa[n_] := For[t = 5; k = 1, True, k++, t = Mod[t^2, 10^k]; If[k == n, Return[ Quotient[t, 10^(n-1)]]]]; bb[n_] := Reap[ For[t = 6; k = 1, k <= n , k++, t = Mod[t^5, 10^k]; Sow[ Quotient[10*t, 10^k]]]][[2, 1, n]]; a[n_] := bb[n] - aa[n]; Table[a[n], {n, 1, 84}](* Jean-François Alcover, May 25 2012, after Paul D. Hanna *)
3, 5, 4, 5, 3, 8, 5, 2, 5, 8, 5, 1, 5, 9, 4, 9, 8, 9, 4, 8, 7, 3, 8, 2, 9, 2, 2, 2, 2, 7, 0, 6, 4, 4, 7, 2, 3, 5, 0, 9, 7, 2, 0, 1, 7, 7, 7, 5, 4, 7, 8, 6, 9, 0, 1, 8, 9, 7, 3, 3, 8, 5, 1, 4, 1, 2, 4, 7, 0, 5, 8, 0, 6, 3, 6, 2, 4, 1, 6, 4, 6, 8, 3, 6, 8, 5, 1, 1, 1, 9, 3, 6, 3, 7, 0, 4, 0, 3, 9, 0, 0, 2, 5, 4, 0
MATHEMATICA
(* execute the programming in both A018247 & A018248 *) IntegerDigits[ FromDigits[ Reverse[a]]*FromDigits[ Reverse[b]]]
Automorphic numbers: m^2 ends with m.
(Formerly M3752)
+10
48
0, 1, 5, 6, 25, 76, 376, 625, 9376, 90625, 109376, 890625, 2890625, 7109376, 12890625, 87109376, 212890625, 787109376, 1787109376, 8212890625, 18212890625, 81787109376, 918212890625, 9918212890625, 40081787109376, 59918212890625, 259918212890625, 740081787109376
COMMENTS
Also called curious numbers.
For entries after the second, two successive terms sum up to a total having the form 10^n + 1. - Lekraj Beedassy, Apr 29 2005 [This comment is clearly wrong as stated. The sums of two consecutive terms are 1, 6, 11, 31, 101, 452, 1001, 10001, 100001, 200001, 1000001, 3781250, .... - T. D. Noe, Nov 14 2010] If a d-digit number n is in the sequence, then so is 10^d+1-n. However, the same number can be 10^d+1-n for different n in the sequence (e.g., 10^3+1-376 = 10^4+1-9376 = 625), which spoils Beedassy's comment. - Robert Israel, Jun 19 2015 This is also the sequence of numbers such that the n-th m-gonal number ends in n for any m == 0,4,8,16 (mod 20). - Robert Dawson, Jul 09 2018
REFERENCES
J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 76, p. 26, Ellipses, Paris 2008.
V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.
R. A. Fairbairn, More on automorphic numbers, J. Rec. Math., 2 (No. 3, 1969), 170-174.
Jan Gullberg, Mathematics, From the Birth of Numbers, W. W. Norton & Co., NY, page 253-254.
B. A. Naik, 'Automorphic numbers' in 'Science Today'(subsequently renamed '2001') May 1982 pp. 59, Times of India, Mumbai.
Ya. I. Perelman, Algebra can be fun, pp. 97-98.
Clifford A. Pickover, A Passion for Mathematics, John Wiley & Sons, Hoboken, 2005, p. 64.
C. P. Schut, Idempotents. Report AM-R9101, Centrum voor Wiskunde en Informatica, Amsterdam, 1991.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Beeler, M., Gosper, R. W. and Schroeppel, R., HAKMEM. MIT AI Memo 239, Feb 29 1972 (Item 59 provides a 40-digit member of this sequence). C. P. Schut, Idempotents, Report AM-R9101, Centre for Mathematics and Computer Science, Amsterdam, 1991. (Annotated scanned copy) Xiaolong Ron Yu, Curious Numbers, Pi Mu Epsilon Journal, Spring 1999, pp. 819-823.
MAPLE
V:= proc(m) option remember;
select(t -> t^2 - t mod 10^m = 0, map(s -> seq(10^(m-1)*j+s, j=0..9), V(m-1)))
end proc:
V(0):= {0, 1}:
V(1):= {5, 6}:
sort(map(op, [V(0), seq(V(i) minus V(i-1), i=1..50)])); # Robert Israel, Jun 19 2015
MATHEMATICA
f[k_] := (r = Reduce[0 < 10^k < n < 10^(k + 1) && n^2 == m*10^(k + 1) + n, {n, m}, Integers]; If[Head[r] === And, n /. ToRules[r], n /. {ToRules[r]}]); Flatten[ Join[{0, 1}, Table[f[k], {k, 0, 13}]]] (* Jean-François Alcover, Dec 01 2011 *) Union@ Join[{1}, Array[PowerMod[5, 2^#, 10^#] &, 16, 0], Array[PowerMod[16, 5^#, 10^#] &, 16, 0]] (* Robert G. Wilson v, Jul 23 2018 *)
PROG
(Haskell)
import Data.List (isSuffixOf)
a003226 n = a003226_list !! (n-1)
a003226_list = filter (\x -> show x `isSuffixOf` show (x^2)) a008851_list
(PARI) A003226(n)={ n<3 & return(n-1); my(i=10, j=10, b=5, c=6, a=b); for( k=4, n, while(b<=a, b=b^2%i*=10); while(c<=a, c=(2-c)*c%j*=10); a=min(b, c)); a } \\ M. F. Hasler, Dec 06 2012 (Sage)
def automorphic(maxdigits, pow, base=10) :
morphs = [[0]]
for i in range(maxdigits):
T=[d*base^i+x for x in morphs[-1] for d in range(base)]
morphs.append([x for x in T if x^pow % base^(i+1) == x])
res = list(set(sum(morphs, []))); res.sort()
return res
(Magma) [n: n in [0..10^7] | Intseq(n^2)[1..#Intseq(n)] eq Intseq(n)]; // Vincenzo Librandi, Jul 03 2015 (Python)
from itertools import count, islice
from sympy.ntheory.modular import crt
def A003226_gen(): # generator of terms a = 0
yield from (0, 1)
for n in count(0):
b = sorted((int(crt(m:=(1<<n, 5**n), (0, 1))[0]), int(crt(m, (1, 0))[0])))
if b[0] > a:
yield from b
a = b[1]
elif b[1] > a:
yield b[1]
a = b[1]
KEYWORD
nonn,base,nice,easy,changed
EXTENSIONS
Incorrect statement removed from title by Robert Dawson, Jul 09 2018
The 10-adic integer x = ...8212890625 satisfying x^2 = x.
+10
38
5, 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, 0, 9, 3, 6, 2, 4, 6, 7
COMMENTS
The 10-adic numbers a and b defined in this sequence and A018248 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.
V. deGuerre and R. A. Fairbairn, Jnl. Rec. Math., No. 3, (1968), 173-179.
M. Kraitchik, Sphinx, 1935, p. 1.
LINKS
V. deGuerre and R. A. Fairbairn, Automorphic numbers, Jnl. Rec. Math., 1 (No. 3, 1968), 173-179.
FORMULA
x = 10-adic lim_{n->oo} 5^(2^n) mod 10^(n+1). - Paul D. Hanna, Jul 08 2006
EXAMPLE
x = ...0863811000557423423230896109004106619977392256259918212890625.
MATHEMATICA
a = {5}; f[n_] := Block[{k = 0, c}, While[c = FromDigits[Prepend[a, k]]; Mod[c^2, 10^n] != c, k++ ]; a = Prepend[a, k]]; Do[ f[n], {n, 2, 105}]; Reverse[a]
With[{n = 150}, Reverse[IntegerDigits[PowerMod[5, 2^n, 10^n]]]] (* IWABUCHI Yu(u)ki, Feb 16 2024 *)
PROG
(PARI) a(n)=local(t=5); for(k=1, n+1, t=t^2%10^k); t\10^n \\ Paul D. Hanna, Jul 08 2006 (PARI) Vecrev(digits(lift(chinese(Mod(1, 2^100), Mod(0, 5^100))))) \\ Seiichi Manyama, Aug 07 2019
CROSSREFS
A007185 gives associated automorphic numbers.
AUTHOR
Yoshihide Tamori (yo(AT)salk.edu)
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
36
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.
a(n) = 16^(5^n) mod 10^n: Automorphic numbers ending in digit 6, with repetitions.
+10
32
6, 76, 376, 9376, 9376, 109376, 7109376, 87109376, 787109376, 1787109376, 81787109376, 81787109376, 81787109376, 40081787109376, 740081787109376, 3740081787109376, 43740081787109376, 743740081787109376, 7743740081787109376, 7743740081787109376
COMMENTS
Also called congruent numbers.
a(n)^2 == a(n) (mod 10^n), that is, a(n) is idempotent of Z[10^n].
Conjecture: For any m coprime to 10 and for any k, the density of n such that a(n) == k (mod m) is 1/m. - Eric M. Schmidt, Aug 01 2012 a(n) is the unique positive integer less than 10^n such that a(n) is divisible by 2^n and a(n) - 1 is divisible by 5^n. - Eric M. Schmidt, Aug 18 2012
REFERENCES
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.
R. A. Fairbairn, More on automorphic numbers, J. Rec. Math., 2 (No. 3, 1969), 170-174.
Jan Gullberg, Mathematics, From the Birth of Numbers, W. W. Norton & Co., NY, page 253-4.
Ya. I. Perelman, Algebra can be fun, pp. 97-98.
A. M. Robert, A Course in p-adic Analysis, Springer, 2000; see pp. 63, 419.
C. P. Schut, Idempotents. Report AM-R9101, Centrum voor Wiskunde en Informatica, Amsterdam, 1991.
LINKS
C. P. Schut, Idempotents, Report AM-R9101, Centre for Mathematics and Computer Science, Amsterdam, 1991. (Annotated scanned copy) Xiaolong Ron Yu, Curious Numbers, Pi Mu Epsilon Journal, Spring 1999, pp. 819-823.
FORMULA
a(n) = 16^(5^n) mod 10^n.
a(2*n) = (3*a(n)^2 - 2*a(n)^3) mod 10^(2*n). - Sylvie Gaudel, Mar 12 2018 a(n) = 2^(10^n) mod 10^n for n >= 2. - Peter Bala, Nov 10 2022
EXAMPLE
a(5) = 09376 because 09376^2 == 87909376 ends in 09376.
PROG
(Sage) [crt(0, 1, 2^n, 5^n) for n in range(1, 1001)] # Eric M. Schmidt, Aug 18 2012 (Magma) [Modexp(16, 5^n, 10^n): n in [1..30]]; // Bruno Berselli, Mar 13 2018 (GAP) List([1..22], n->PowerModInt(16, 5^n, 10^n)); # Muniru A Asiru, Mar 20 2018
CROSSREFS
A018248 gives the associated 10-adic number.
10-adic integer x=.....06619977392256259918212890624 satisfying x^3 = x.
+10
19
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=.....93380022607743740081787109375 satisfying x^3 = x.
+10
17
5, 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
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.
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.
AUTHOR
Edoardo Gueglio (egueglio(AT)yahoo.it), Jan 28 2004
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