A016090 a(n) = 16^(5^n) mod 10^n: Automorphic numbers ending in digit 6, with repetitions.
6, 76, 376, 9376, 9376, 109376, 7109376, 87109376, 787109376, 1787109376, 81787109376, 81787109376, 81787109376, 40081787109376, 740081787109376, 3740081787109376, 43740081787109376, 743740081787109376, 7743740081787109376, 7743740081787109376
Offset: 1
Examples
a(5) = 09376 because 09376^2 == 87909376 ends in 09376.
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
- Eric M. Schmidt, Table of n, a(n) for n = 1..1000
- Robert Dawson, On Some Sequences Related to Sums of Powers, J. Int. Seq., Vol. 21 (2018), Article 18.7.6.
- C. P. Schut, Idempotents, Report AM-R9101, Centre for Mathematics and Computer Science, Amsterdam, 1991. (Annotated scanned copy)
- Eric Weisstein's World of Mathematics, Automorphic Number
- Xiaolong Ron Yu, Curious Numbers, Pi Mu Epsilon Journal, Spring 1999, pp. 819-823.
- Index entries for sequences related to automorphic numbers
Crossrefs
Programs
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GAP
List([1..22], n->PowerModInt(16,5^n,10^n)); # Muniru A Asiru, Mar 20 2018
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Magma
[Modexp(16, 5^n, 10^n): n in [1..30]]; // Bruno Berselli, Mar 13 2018
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Maple
[seq(16 &^ 5^n mod 10^n, n=1..22)]; # Muniru A Asiru, Mar 20 2018
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Mathematica
Array[PowerMod[16, 5^#, 10^#] &, 18] (* Michael De Vlieger, Mar 13 2018 *)
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PARI
A016090(n)=lift(Mod(6,10^n)^5^(n-1)) \\ M. F. Hasler, Dec 05 2012, edited Jan 26 2020
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Sage
[crt(0, 1, 2^n, 5^n) for n in range(1, 1001)] # Eric M. Schmidt, Aug 18 2012
Formula
a(n) = 16^(5^n) mod 10^n.
a(n+1) == 2*a(n) - a(n)^2 (mod 10^(n+1)). - Eric M. Schmidt, Jul 28 2012
a(n) = 6^(5^n) mod 10^n. - Sylvie Gaudel, Feb 17 2018
a(2*n) = (3*a(n)^2 - 2*a(n)^3) mod 10^(2*n). - Sylvie Gaudel, Mar 12 2018
a(n) = 6^5^(n-1) mod 10^n. - M. F. Hasler, Jan 26 2020
a(n) = 2^(10^n) mod 10^n for n >= 2. - Peter Bala, Nov 10 2022
Extensions
Edited by David W. Wilson, Sep 26 2002
Definition corrected by M. F. Hasler, Dec 05 2012
Comments