%I #24 Jul 01 2023 14:25:25

%S 301,721,1141,1561,1981,2401,2821,3241,3661,4081,4501,4921,5341,5761,

%T 6181,6601,7021,7441,7861,8281,8701,9121,9541,9961,10381,10801,11221,

%U 11641,12061,12481,12901,13321,13741,14161,14581,15001,15421,15841

%N The broken eggs problem.

%C This is a problem of byzantine mathematics appearing in the Codex Vinobonensis Phil. Gr. 65. See the Hunger-Vogel reference, p. 73, problem 86.

%C It also appears in the Tropfke reference on p. 640, 4.3.5.2, Die Eierfrau, as a special case of the Chinese Ta-yen rule (method of the great extension) treated on p. 636.

%C This is also a problem posed in the Alten et al. reference, p. 203, Aufgabe 3.1.6 (taken from Tropfke).

%C For the statement of the problem (in another setting) see the Cherowitzo link, where it is considered as an application of the Chinese remainder theorem.

%C The problem is to find all solutions of the common congruences: N congruent to 1 modulo 2, 3, 4, 5 and 6, and 0 modulo 7. For the application of the Chinese remainder theorem one first disposes of the moduli 2 and 6 (these congruences follow from the others).

%C The egg-selling woman had 301 eggs before they were broken according to problem 86 with this special solution in the Hunger-Vogel reference.

%D H.-W. Alten et al., 4000 Jahre Algebra, 2. Auflage, Springer, 2014, p. 203.

%D H. Hunger and K. Vogel, Ein byzantinisches Rechenbuch des 15.Jahrhunderts. 100 Aufgaben aus dem Codex Vindobonensis Phil. Gr. 65. (in Greek and German translation), Hermann Böhlaus Nachf., Wien, 1963 (Österr. Akad. d. Wiss., phil.-hist. Kl., Denkschriften, 78. Band, 2. Abhandlung), p. 73.

%D J. Tropfke, Geschichte der Elementarmathematik, Band 1, Arithmetik und Algebra, 4. Auflage, Walter de Gruyter, Berlin, New York , 1980, p. 640.

%H Adam Hugill, <a href="/A256101/b256101.txt">Table of n, a(n) for n = 1..10000</a>

%H Bill Cherowitzo, <a href="http://www-math.ucdenver.edu/~wcherowi/courses/m5410/crt.pdf">Chinese remainder Theorem</a>.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2, -1).

%F a(n) = 420*n-119, n >= 1, (note that 420 = 3*4*5*7, with pairwise coprime factors needed for the Chinese remainder theorem).

%F a(n) = 60*A017041(n-1) + 1, n >= 1.

%F G.f.: x*7*(43+17*x)/(1-x)^2. (Corrected by _Vincenzo Librandi_, Apr 11 2015

%p A256101:=n->420*n-119: seq(A256101(n), n=1..50); # _Wesley Ivan Hurt_, Apr 11 2015

%t CoefficientList[Series[(301 + 119 x) / (1 - x)^2, {x, 0, 40}], x] (* _Vincenzo Librandi_, Apr 11 2015 *)

%o (Magma) [420*n-119: n in [1..40]]; // _Vincenzo Librandi_, Apr 11 2015

%o (Python)

%o terms=[]

%o n=50 #terms here

%o for i in range(1, n+1):

%o ans=420*i-119

%o terms.append(ans)

%o print(terms)

%o # _Adam Hugill_, Feb 22 2022

%Y Cf. A017041.

%K nonn,easy

%O 1,1

%A _Wolfdieter Lang_, Apr 10 2015

%E Corrected G.f. rewritten. - _Wolfdieter Lang_, Apr 15 2015