

A081352


Main diagonal of square maze arrangement of natural numbers A081349.


10



1, 7, 11, 21, 29, 43, 55, 73, 89, 111, 131, 157, 181, 211, 239, 273, 305, 343, 379, 421, 461, 507, 551, 601, 649, 703, 755, 813, 869, 931, 991, 1057, 1121, 1191, 1259, 1333, 1405, 1483, 1559, 1641, 1721, 1807, 1891, 1981, 2069, 2163, 2255, 2353, 2449, 2551
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OFFSET

0,2


COMMENTS

Conjecture: let a and b be integers such that 0 < a < b so that 0 < a/b is a proper fraction. Define the map f(a,b,D) = a/b + gcd(a,b)/D. Of course, all such a/b can be partially ordered by value, i.e., 1/2 = 0.5 < 2/3 = 4/6 = 6/9 = 0.6666... < 3/4 = 6/8 = 0.75 < 4/5 = 0.8 etc. The map f appears to specify a total strict order on the codomain for all a/b that is consistent with the given partial order of the domain, i.e., the partial order remains intact, while equivalent fractions are given a total strict order themselves. Moreover, equivalent fractions are strictly ordered by numerator (or denominator), e.g., 1/2 < 2/4 < 3/6 etc. The conditions are that for n >= 4 all of the fractions with denominator b <= n are listed and the minimum integer value of D to achieve the total strict order of the codomain is 2*C(n1,2)  (1)^(n1). So, a(n3) = D for n >= 4. Example: given n = 4, we have D = 2*(41,2)  (1)^(41) = 2*3 + 1 = 7 = a(43) = a(1). Partial order of domain. 1/4 < 1/3 < 1/2 = 2/4 < 2/3 < 3/4. Total order of codomain. f(1,4,7) = 1/4 + 1/7 = 33/84 < f(1,3,7) = 1/3 + 1/7 = 40/84 < f(1,2,7) = 1/2 + 1/7 = 54/84 < f(2,4,7) = 2/4 + 2/7 = 66/84 < f(2,3,7) = 2/3 + 2/7 = 68/84 < f(3,4,7) = 3/4 + 1/7 = 75/84. Observe that if D = 6, then f(2,4,6) = 2/4 + 2/6 = 10/12 = f(2,3,6) = 2/3 + 1/6. Computation shows the same failure to achieve total strict order of the codomain for D = 2..5. (As a >= 1, then b >=2, from the above). Computation also shows that the conjecture holds for n = 4..17.  Ross La Haye, Oct 02 2016


LINKS



FORMULA

a(n) = (n + 1)*(n + 2)  (1)^n = 2*C(n+2, 2)  (1)^n.
G.f.: (1 +5*x 3*x^2 +x^3) / ((1+x)*(1x)^3). [Bruno Berselli, Aug 01 2010]
a(n) 2*a(n1) +2*a(n3) a(n4) = 0 with n>3. [Bruno Berselli, Aug 01 2010]


MAPLE



MATHEMATICA

CoefficientList[Series[(1 + 5 x  3 x^2 + x^3) / ((1 + x) (1  x)^3), {x, 0, 60}], x] (* Vincenzo Librandi, Aug 08 2013 *)


PROG

(Magma) I:=[1, 7, 11, 21]; [n le 4 select I[n] else 2*Self(n1)2*Self(n3)+Self(n4): n in [1..50]]; // Vincenzo Librandi, Aug 08 2013
(PARI) x='x+O('x^99); Vec((1+5*x3*x^2+x^3)/((1+x)*(1x)^3)) \\ Altug Alkan, Mar 26 2016


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



