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A053664 Smallest number m such that m == i (mod prime(i)) for all 1<=i<=n. 14

%I #41 May 04 2023 10:54:00

%S 1,5,23,53,1523,29243,299513,4383593,188677703,5765999453,5765999453,

%T 2211931390883,165468170356703,8075975022064163,361310530977154973,

%U 20037783573808880093,1779852341342071295513,40235059344426324076913

%N Smallest number m such that m == i (mod prime(i)) for all 1<=i<=n.

%C Suggested by Chinese Remainder Theorem.

%D Niven and Zuckerman, An Introduction to the Theory of Numbers, John Wiley, 1966, p. 40

%D Paulo Ribenboim, The New Book of Prime Numbers Records, Springer 1996, p. 33

%H Nick Hobson and Robert G. Wilson v, <a href="/A053664/b053664.txt">Table of n, a(n) for n = 1..350</a> (first 100 terms from Nick Hobson)

%H Project Euler, <a href="https://projecteuler.net/problem=552">Problem 552: Chinese leftovers II</a>

%e a(3) = 23 because this is the smallest number m such that m == 1 (mod 2), m == 2 (mod 3) and m == 3 (mod 5).

%e a(4) = 53 because 53 - 1 is divisible by 2, 53 - 2 is divisible by 3, 53 - 3 is divisible by 5 and 53 - 4 is divisible by 7.

%t f[n_] := ChineseRemainder[ Range[n], Prime[Range[n]]]; Array[f, 20]

%o (PARI) for(n=1,20,m=1; while(sum(i=1,n,abs(m%prime(i)-i))>0,m++); print1(m,","))

%o (PARI) x=Mod(1, 1); for(i=1, 18, x=chinese(x, Mod(i, prime(i))); print1(component(x, 2), ", ")) /* Nick Hobson (nickh(AT)qbyte.org), Jan 08 2007 */

%o (Python)

%o from sympy.ntheory.modular import crt

%o from sympy import prime

%o def A053664(n): return int(crt([prime(i) for i in range(1,n+1)],list(range(1,n+1)))[0]) # _Chai Wah Wu_, May 01 2023

%Y Cf. A192363.

%K nonn,easy,nice

%O 1,2

%A Joe K. Crump (joecr(AT)carolina.rr.com), Feb 16 2000

%E Additional comments from Luis A. Rodriguez (luiroto(AT)yahoo.com), Apr 23 2002

%E Edited by _N. J. A. Sloane_ and _Robert G. Wilson v_, May 03 2002

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Last modified March 29 09:59 EDT 2024. Contains 371268 sequences. (Running on oeis4.)