A279259 - OEIS

A279259

Smallest positive integer m such that m, m+1, m+2, m+3 are divisible by 2n+1, 2n+3, 2n+5, 2n+7 respectively.

53, 159, 1735, 4508, 3222, 18238, 31499, 16965, 78013, 114722, 54348, 225124, 303425, 133515, 519187, 662408, 277794, 1035370, 1272023, 515697, 1864393, 2228174, 880920, 3112528, 3642317, 1412343, 4901599, 5641460, 2154030, 7368982, 8368163, 3155229, 10667605, 11980538

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,1

LINKS

Joseph Myers, Table of n, a(n) for n = 0..1000

United Kingdom Mathematics Trust, 2016/17 British Mathematical Olympiad Round 1, Problem 6.

Index entries for linear recurrences with constant coefficients, signature (0,0,5,0,0,-10,0,0,10,0,0,-5,0,0,1).

Index to sequences related to Olympiads and other Mathematical competitions.

FORMULA

a(n) = (2*n+1 + lcm(2*n+1, 2*n+3, 2*n+5, 2*n+7))/2.

G.f.: (8*x^14 +10*x^12 -89*x^11 -153*x^10 -1777*x^9 -4173*x^8 -2445*x^7 -9489*x^6 -9563*x^5 -2427*x^4 -4243*x^3 -1735*x^2 -159*x-53) / ((x-1)^5*(x^2+x+1)^5). - Alois P. Heinz, Dec 08 2016

From Bernard Schott, Dec 08 2020: (Start)

If n == 1 (mod 3), a(n) = (2*n+1)* ((2*n+3)*(2*n+5)*(2*n+7)/3 + 1)/2.

If n == 0, 2 (mod 3), a(n) = (2*n+1)* ((2*n+3)*(2*n+5)*(2*n+7) + 1)/2. (End)

EXAMPLE

53 is the smallest positive integer such that 53, 54, 55, 56 are divisible by 1, 3, 5, 7 respectively, hence a(0) = 53. - Bernard Schott, Dec 08 2020

MATHEMATICA

LinearRecurrence[{0, 0, 5, 0, 0, -10, 0, 0, 10, 0, 0, -5, 0, 0, 1}, {53, 159, 1735, 4508, 3222, 18238, 31499, 16965, 78013, 114722, 54348, 225124, 303425, 133515, 519187}, 40] (* Harvey P. Dale, Dec 29 2017 *)

CROSSREFS

Sequence in context: A053070 A140655 A142508 * A142639 A141880 A142698

Adjacent sequences: A279256 A279257 A279258 * A279260 A279261 A279262

KEYWORD

nonn,easy

AUTHOR

Joseph Myers, Dec 08 2016

STATUS

approved