A081350 -id:A081350 - cp's OEIS frontend

A081352 Main diagonal of square maze arrangement of natural numbers A081349.

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

Offset: 0

Paul Barry, Mar 19 2003

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 co-domain 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 co-domain is 2*C(n-1,2) - (-1)^(n-1). So, a(n-3) = D for n >= 4. Example: given n = 4, we have D = 2*(4-1,2) - (-1)^(4-1) = 2*3 + 1 = 7 = a(4-3) = a(1). Partial order of domain. 1/4 < 1/3 < 1/2 = 2/4 < 2/3 < 3/4. Total order of co-domain. 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 co-domain 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

B. Berselli, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A081350, A081351, A054569.

I:=[1,7,11,21]; [n le 4 select I[n] else 2*Self(n-1)-2*Self(n-3)+Self(n-4): n in [1..50]]; // Vincenzo Librandi, Aug 08 2013

A081352:=n->(n + 1)*(n + 2) - (-1)^n; seq(A081352(n), n=0..50); # Wesley Ivan Hurt, Feb 26 2014

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

x='x+O('x^99); Vec((1+5*x-3*x^2+x^3)/((1+x)*(1-x)^3)) \\ Altug Alkan, Mar 26 2016

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)*(1-x)^3). [Bruno Berselli, Aug 01 2010]
a(n) -2*a(n-1) +2*a(n-3) -a(n-4) = 0 with n>3. [Bruno Berselli, Aug 01 2010]
a(n) = 3*A000982(n + 2) - A000982(n + 3). - Miko Labalan, Mar 26 2016
a(n) = A116940(n) + A236283(n + 1). - Miko Labalan, Dec 04 2016
a(n) = (2*n^2 + 6*n - 2*(-1)^n + (-1)^(2*n) + 3)/2. - Kritsada Moomuang, Oct 24 2019

A081349 Square maze arrangement of the natural numbers, read by antidiagonals.

Original entry on oeis.org

1, 2, 8, 3, 7, 9, 4, 6, 10, 24, 15, 5, 11, 23, 25, 16, 14, 12, 22, 26, 48, 35, 17, 13, 21, 27, 47, 49, 36, 34, 18, 20, 28, 46, 50, 80, 63, 37, 33, 19, 29, 45, 51, 79, 81, 64, 62, 38, 32, 30, 44, 52, 78, 82, 120, 99, 65, 61, 39, 31, 43, 53, 77, 83, 119, 121, 100, 98, 66, 60, 40

Offset: 0

Paul Barry, Mar 19 2003

			Starting with 1,2,3,4, turn (LL) and then repeat (RRR)(LLL) to get:
.
   1   8---9  24 ...
   |   |   |   |
   2   7  10  23 ...
   |   |   |   |
   3   6  11  22 ...
   |   |   |   |
   4---5  12  21 ...
           |   |
  15--14--13  20 ...
   |           |
  16--17--18--19 ...
  ...

Index entries for sequences that are permutations of the natural numbers

Cf. A081344, A093650.
First row is A081351. First column is A081350.
Main diagonal is A081352. Main subdiagonal is 2*A000217(n+1).

A081351 First row in square maze array of natural numbers A081349.

Original entry on oeis.org

1, 8, 9, 24, 25, 48, 49, 80, 81, 120, 121, 168, 169, 224, 225, 288, 289, 360, 361, 440, 441, 528, 529, 624, 625, 728, 729, 840, 841, 960, 961, 1088, 1089, 1224, 1225, 1368, 1369, 1520, 1521, 1680, 1681, 1848, 1849, 2024, 2025, 2208, 2209, 2400, 2401, 2600

Offset: 0

Paul Barry, Mar 19 2003

Interleaves the odd squares A016754 with 8 times the triangular numbers A000217.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A081350, A081352.

[(n+1)*(n+2)-(n+1)*(-1)^n: n in [0..50]]; // Vincenzo Librandi, Sep 06 2011

a(n) = (n+1)*(n+2)-(n+1)*(-1)^n = 2*C(n+2,2)-C(n+1,1)*(-1)^n.

G.f.: (x^3-x^2+7*x+1)/((1-x)^3*(1+x)^2). [Colin Barker, Sep 03 2012]

A081353 Diagonal of square maze arrangement of natural numbers A081349.

Original entry on oeis.org

3, 5, 13, 19, 31, 41, 57, 71, 91, 109, 133, 155, 183, 209, 241, 271, 307, 341, 381, 419, 463, 505, 553, 599, 651, 701, 757, 811, 871, 929, 993, 1055, 1123, 1189, 1261, 1331, 1407, 1481, 1561, 1639, 1723, 1805, 1893, 1979, 2071, 2161, 2257, 2351, 2451, 2549

Offset: 0

Paul Barry, Mar 19 2003

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A081349, A081350, A081351, A081352.
Cf. A109613, A137932.
Bisections are in A054554, A125202.

[(n + 1)*(n + 2) + (-1)^n: n in [0..50]]; // Vincenzo Librandi, Sep 06 2011

A081353:=n->(n+1)*(n+2)+(-1)^n: seq(A081353(n), n=0..100); # Wesley Ivan Hurt, Aug 09 2015

Table[(n + 1) (n + 2) + (-1)^n, {n, 0, 70}] (* Wesley Ivan Hurt, Aug 09 2015 *)
LinearRecurrence[{2,0,-2,1},{3,5,13,19},50] (* Harvey P. Dale, Aug 02 2021 *)

a(n) = (n+1)*(n+2)+(-1)^n = 2*binomial(n+2,2)+(-1)^n.
G.f.: (3-x)*(1+x^2)/((1-x)^3*(1+x)). [Colin Barker, Sep 03 2012]
From Wesley Ivan Hurt, Aug 09 2015: (Start)
a(n) = 2*a(n-1)-2*a(n-3)+a(n-4), n>4.
a(n) = n^2+3n+3 if n is even, otherwise n^2+3n+1.
a(n) = A137932(n+3) - A109613(n+1). (End)

A341779 Numbers k such that k and k+1 are both anti-tau numbers (A046642).

Original entry on oeis.org

3, 4, 15, 16, 64, 100, 195, 196, 255, 256, 483, 484, 676, 783, 784, 1023, 1024, 1155, 1156, 1295, 1296, 1443, 1444, 1599, 1600, 1936, 2116, 2703, 2704, 3363, 3364, 3844, 4096, 4623, 4624, 4899, 4900, 5183, 5184, 5476, 5776, 6399, 6400, 6723, 6724, 7395, 7396

Offset: 1

Amiram Eldar, Feb 19 2021

nonn

Since the even anti-tau numbers (A268066) are square numbers, all the terms are either of the form 4*k^2 - 1 or 4*k^2.

			3 is a term since 3 and 4 are both anti-tau numbers: gcd(3, tau(3)) = gcd(3, 2) = 1 and gcd(4, tau(4)) = gcd(4, 3) = 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Subsequence of A046642 and A081350.

Cf. A000010, A009191, A114617, A268066, A341780.

antiTauQ[n_] := CoprimeQ[n, DivisorSigma[0, n]]; s = {}; Do[k = 4*n^2; If[antiTauQ[k], If[antiTauQ[k - 1], AppendTo[s, k - 1]]; If[antiTauQ[k + 1], AppendTo[s, k]]], {n, 1, 50}]; s

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A081352 Main diagonal of square maze arrangement of natural numbers A081349.

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A081349 Square maze arrangement of the natural numbers, read by antidiagonals.

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A081351 First row in square maze array of natural numbers A081349.

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A081353 Diagonal of square maze arrangement of natural numbers A081349.

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A341779 Numbers k such that k and k+1 are both anti-tau numbers (A046642).

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