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

1, 5, 23, 53, 1523, 29243, 299513, 4383593, 188677703, 5765999453, 5765999453, 2211931390883, 165468170356703, 8075975022064163, 361310530977154973, 20037783573808880093, 1779852341342071295513, 40235059344426324076913

Offset

1,2

Comments

Suggested by Chinese Remainder Theorem.

References

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

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

Links

Nick Hobson and Robert G. Wilson v, Table of n, a(n) for n = 1..350 (first 100 terms from Nick Hobson)

Project Euler, Problem 552: Chinese leftovers II

Example

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).
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.

Mathematica

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

Prog

(PARI) for(n=1, 20, m=1; while(sum(i=1, n, abs(m%prime(i)-i))>0, m++); print1(m, ", "))
(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 */
(Python)
from sympy.ntheory.modular import crt
from sympy import prime
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

Crossrefs

Cf. A192363.

Sequence in context: A289154 A155851 A019267 * A186030 A092544 A319087

Adjacent sequences: A053661 A053662 A053663 * A053665 A053666 A053667

Keyword

nonn,easy,nice

Author

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

Extensions

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

Edited by N. J. A. Sloane and Robert G. Wilson v, May 03 2002

Status

approved