Displaying 1-10 of 47 results found.
Subsequence of the automorphic numbers ( A003226) with initial term 5 and such that a(n+1) ends with the digits of a(n).
+20
2
5, 25, 625, 12890625, 6259918212890625, 4106619977392256259918212890625, 80863811000557423423230896109004106619977392256259918212890625
COMMENTS
An automorphic number is a number whose square ends in the same digits as the number itself.
The next term has 126 digits.
PROG
(PARI)
seq(m, maxn) = L=List(); for(j=1, maxn, listput(L, m); m=(3*m^2-2*m^3)%10 ^ (2*sizedigit(m))); Vec(L)
seq(5, 8)
Subsequence of the automorphic numbers ( A003226) with initial term 6 and such that a(n+1) ends with the digits of a(n).
+20
2
6, 76, 139376, 114087109376, 792415373740081787109376, 88398678125844615295893380022607743740081787109376, 3724919229963099270422168663257939520419136188999442576576769103890995893380022607743740081787109376
COMMENTS
An automorphic number is a number whose square ends in the same digits as the number itself.
The next term has 200 digits.
PROG
(PARI)
seq(m, maxn) = L=List(); for(j=1, maxn, listput(L, m); m=(3*m^2-2*m^3)%10 ^ (2*sizedigit(m))); Vec(L)
seq(6, 8)
a(n) is the number which, when concatenated with A003226(n), the n-th automorphic number, gives ( A003226(n))^2.
+20
0
0, 0, 2, 3, 6, 57, 141, 390, 8790, 82128, 11963, 793212, 835571, 5054322, 1661682, 75880433, 45322418, 619541169, 319375992, 6745157241, 3317093849, 66891312600, 843114912509, 9837094694375, 16065496578813, 35901922360062, 67557477392256, 547721051611007
COMMENTS
Let na and nb represent the indices of the preceding and next A003226(n)'s beginning with a 9, and where (na - nb) >= 3 (note that the first such 'zone' begins with an exception for which the index A003226(na) = 1). Then for na < n < nb and such that n == (na + 1) mod 2, it appears that A003226(n) - a(n) = A003226(n+1) - a(n+1) = k. In such cases, it also appears that a(n)*a(n+1) = k^2 - k.
EXAMPLE
For n=13, A003226(13)=2890625, ( A003226(13))^2=8355712890625. So a(13)=835571.
PROG
(PARI) auto(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; } \\ A003226a(n) = {my(m = auto(n), dm = digits(m), dm2 = digits(m^2)); fromdigits(vector(#dm2 - #dm, k, dm2[k])); } \\ Michel Marcus, May 18 2019
Congruent to 0 or 1 mod 5.
+10
38
0, 1, 5, 6, 10, 11, 15, 16, 20, 21, 25, 26, 30, 31, 35, 36, 40, 41, 45, 46, 50, 51, 55, 56, 60, 61, 65, 66, 70, 71, 75, 76, 80, 81, 85, 86, 90, 91, 95, 96, 100, 101, 105, 106, 110, 111, 115, 116, 120, 121, 125, 126, 130, 131, 135, 136, 140, 141, 145, 146, 150, 151
COMMENTS
Numbers k that have the same last digit as k^2.
REFERENCES
L. E. Dickson, History of the Theory of Numbers, I, p. 459.
FORMULA
G.f.: x^2*(1+4*x) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Oct 07 2011 E.g.f.: 4 + ((10*x - 13)*exp(x) - 3*exp(-x))/4. - David Lovler, Sep 11 2022 Sum_{n>=2} (-1)^n/a(n) = sqrt(1+2/sqrt(5))*Pi/10 + log(phi)/(2*sqrt(5)) + log(5)/4, where phi is the golden ratio ( A001622). - Amiram Eldar, Oct 12 2022
MAPLE
a[0]:=0:a[1]:=1:for n from 2 to 100 do a[n]:=a[n-2]+5 od: seq(a[n], n=0..61); # Zerinvary Lajos, Mar 16 2008
MATHEMATICA
Select[Range[0, 151], MemberQ[{0, 1}, Mod[#, 5]] &] (* T. D. Noe, Mar 31 2013 *)
PROG
(Haskell)
a008851 n = a008851_list !! (n-1)
a008851_list = [10*n + m | n <- [0..], m <- [0, 1, 5, 6]]
(PARI) a(n) = 5*(n\2)+bitand(n, 1); /* Joerg Arndt, Mar 31 2013 */ (PARI) a(n) = floor((5/3)*floor(3*(n-1)/2)); /* Joerg Arndt, Mar 31 2013 */
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)
Automorphic numbers ending in digit 5: a(n) = 5^(2^n) mod 10^n.
(Formerly M3940)
+10
35
5, 25, 625, 625, 90625, 890625, 2890625, 12890625, 212890625, 8212890625, 18212890625, 918212890625, 9918212890625, 59918212890625, 259918212890625, 6259918212890625, 56259918212890625, 256259918212890625, 2256259918212890625, 92256259918212890625
COMMENTS
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 5^n and a(n) - 1 is divisible by 2^n. - Eric M. Schmidt, Aug 18 2012
REFERENCES
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.
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
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) = 5^(2^n) mod 10^n.
a(n)^2 == a(n) (mod 10^n), that is, a(n) is an idempotent in Z[10^n].
a(2n) = (3*a(n)^2 - 2*a(n)^3) mod 10^(2n). - Sylvie Gaudel, Mar 10 2018
EXAMPLE
625 is in the sequence because 625^2 = 390625, which ends in 625.
90625 is in the sequence because 90625^2 = 8212890625, which ends in 90625.
90635 is not in the sequence because 90635^2 = 8214703225, which does not end in 90635.
MAPLE
a:= n-> 5&^(2^n) mod 10^n: seq(a(n), n=1..25); # Alois P. Heinz, Mar 11 2018
PROG
(Sage) [crt(1, 0, 2^n, 5^n) for n in range(1, 1001)] # Eric M. Schmidt, Aug 18 2012
CROSSREFS
A018247 gives the associated 10-adic number.
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.
The 10-adic integer x = ...1787109376 satisfies x^2 = x.
+10
29
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)
Numbers k such that the decimal expansion of k^2 contains k as a substring.
+10
27
0, 1, 5, 6, 10, 25, 50, 60, 76, 100, 250, 376, 500, 600, 625, 760, 1000, 2500, 3760, 3792, 5000, 6000, 6250, 7600, 9376, 10000, 14651, 25000, 37600, 50000, 60000, 62500, 76000, 90625, 93760, 100000, 109376, 250000, 376000, 495475, 500000, 505025
EXAMPLE
25^2 = 625 which contains 25.
3792^2 = 14_3792_64, 14651^2 = 2_14651_801.
MATHEMATICA
Select[Range[510000], MemberQ[FromDigits /@ Partition[IntegerDigits[#^2], IntegerLength[#], 1], #] &] (* Jayanta Basu, Jun 29 2013 *) Select[Range[0, 510000], StringPosition[ToString[#^2], ToString[#]]!={}&] (* Ivan N. Ianakiev, Oct 02 2016 *)
PROG
(Haskell)
import Data.List (isInfixOf)
a018834 n = a018834_list !! (n-1)
a018834_list = filter (\x -> show x `isInfixOf` show (x^2)) [0..]
(Python)
from itertools import count, islice
def A018834_gen(startvalue=0): # generator of terms >= startvalue return filter(lambda n:str(n) in str(n**2), count(max(startvalue, 0)))
Automorphic numbers: n^2 ends with n in base 6.
+10
12
0, 1, 3, 4, 9, 28, 81, 136, 1216, 6561, 16768, 29889, 76545, 203392, 636417, 1043200, 3995649, 6082048, 24151041, 36315136, 326481921, 689278977, 1487503360, 11573190657, 76876660737, 155240824833, 314944159744, 785129144320, 2035980763137, 4857090670593
EXAMPLE
a(3) = (3)_6 = 3 since 3^2 = 9 = (13)_6 ends with 3 in base 6.
a(4) = (4)_6 = 4 since 4^2 = 16 = (24)_6 ends with 4 in base 6.
a(5) = (13)_6 = 9 since 9^2 = 81 = (213)_6 ends with 13 in base 6.
PROG
(PARI) isok(n) = ((n^2-n) % 6^(#digits(n, 6))) == 0; \\ Michel Marcus, Mar 08 2014
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