A136289 -id:A136289 - cp's OEIS frontend

A047266 Numbers that are congruent to {0, 1, 5} mod 6.

0, 1, 5, 6, 7, 11, 12, 13, 17, 18, 19, 23, 24, 25, 29, 30, 31, 35, 36, 37, 41, 42, 43, 47, 48, 49, 53, 54, 55, 59, 60, 61, 65, 66, 67, 71, 72, 73, 77, 78, 79, 83, 84, 85, 89, 90, 91, 95, 96, 97, 101, 102, 103, 107, 108, 109, 113, 114, 115, 119, 120, 121, 125

Offset: 1

N. J. A. Sloane

a(n+3) is the Hankel transform of A005773(n+3). - Paul Barry, Nov 04 2008
The numbers m == 0, 2 or 10 mod 12 (the doubles of this sequence, that is, 10, 12, 14, 22, 24, 26, 34, ...) have the property that exactly 1/4 of the 3-part partitions of m form the sides of a triangle. See Mathematics Stack Exchange, 2013, link. - Ed Pegg Jr, Dec 19 2013
Row sum of a triangle where two rules build the triangle. #1 Start with the value "1" at the top of the triangle. #2 Require every "triple" to contain the values 1,2,3 (see link below). Compare with A136289 that has "3" at the apex. - Craig Knecht, Oct 17 2015
Nonnegative m such that floor(k*m^2/6) = k*floor(m^2/6), where k = 2, 3, 4 or 5. - Bruno Berselli, Dec 03 2015

Craig Knecht, Triangular row sum of A204259.
Mathematics Stack Exchange, Prove: Exactly a quarter of 3-part partitions of numbers >2 equal to 0, 2, 10 mod 12 will make a triangle, 2013.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A005773, A047240, A047242, A057078, A136289, A204259.

[n : n in [0..150] | n mod 6 in [0, 1, 5]]; // Wesley Ivan Hurt, Jun 13 2016

seq(seq(6*s+j, j=[0,1,5]), s=0..100); # Robert Israel, Dec 01 2014

Select[Range[0, 200], Mod[#, 6] == 0 || Mod[#, 6] == 1 || Mod[#, 6] == 5 &] (* Vladimir Joseph Stephan Orlovsky, Jul 07 2011 *)

concat(0, Vec(x^2*(1+4*x+x^2)/((1+x+x^2)*(x-1)^2) + O(x^100))) \\ Altug Alkan, Oct 17 2015

G.f.: x^2*(1+4*x+x^2) / ((1+x+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
a(n) = 2*(n-1) + A057078(n). - Robert Israel, Dec 01 2014
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4. - Wesley Ivan Hurt, Nov 09 2015
From Wesley Ivan Hurt, Jun 13 2016: (Start)
a(n) = 2*n-2+cos(2*n*Pi/3)+sin(2*n*Pi/3)/sqrt(3).
a(3k) = 6k-1, a(3k-1) = 6k-5, a(3k-2) = 6k-6. (End)
Sum_{n>=2} (-1)^n/a(n) = log(2)/6 + log(2 + sqrt(3))/sqrt(3). - Amiram Eldar, Dec 14 2021

A264788 a(n) is the number of circles added at n-th iteration of the pattern starting with 2 circles. (See comment.)

Original entry on oeis.org

2, 2, 4, 4, 6, 10, 10, 12, 16, 16, 18, 22, 22, 24, 28, 28, 30, 34, 34, 36, 40, 40, 42, 46, 46, 48, 52, 52, 54, 58, 58, 60, 64, 64, 66, 70, 70, 72, 76, 76, 78, 82, 82, 84, 88, 88, 90, 94, 94, 96, 100, 100, 102, 106, 106, 108, 112, 112, 114, 118, 118, 120, 124

Offset: 0

Kival Ngaokrajang, Nov 25 2015

Pattern construction rules: (i) At n = 0, there are two circles of radius s with centers at the ends of a straight line of length s. (ii) At n > 0, draw circles by placing center at the intersection points of the circumferences of circles in the previous iteration, with overlaps forbidden. The pattern seems to be the flower of life. See illustration.

Colin Barker, Table of n, a(n) for n = 0..1000
Kival Ngaokrajang, Illustration of initial terms
Sacred Geometry, Flower of Life
Eric Weisstein's World of Mathematics, Flower of Life
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A052380, A136289.

LinearRecurrence[{1,0,1,-1},{2,2,4,4,6,10},100] (* Paolo Xausa, Nov 17 2023 *)

{a = 4; print1("2, 2, ", a, ", "); for(n = 2, 100, if (Mod(n,3)==0, d1 = 2); if (Mod(n,3)==1, d1 = 4);  if (Mod(n,3)==2, d1 = 0); a = a + d1; print1(a, ", "))}

Vec(2*(1+x^2-x^3+x^4+x^5)/((1-x)^2*(1+x+x^2)) + O(x^100)) \\ Colin Barker, Dec 10 2015

From Colin Barker, Dec 10 2015: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n>5.
G.f.: 2*(1+x^2-x^3+x^4+x^5) / ((1-x)^2*(1+x+x^2)).
(End)

A136290 a(0)=1, a(1)=3, a(2)=9, a(3)=12, a(4)=15; thereafter a(n) = a(n-1) + a(n-3) - a(n-4).

Original entry on oeis.org

1, 3, 9, 12, 15, 21, 24, 27, 33, 36, 39, 45, 48, 51, 57, 60, 63, 69, 72, 75, 81, 84, 87, 93, 96, 99, 105, 108, 111, 117, 120, 123, 129, 132, 135, 141, 144, 147, 153, 156, 159, 165, 168, 171, 177, 180, 183, 189, 192, 195, 201, 204, 207, 213, 216, 219, 225, 228, 231, 237

Offset: 0

Colin Mallows, Apr 13 2008

nonn

This is the coordination sequence for Marjorie Rice's tiling of the plane shown in Fig. 15 of Schattschneider (1981), with respect to the central vertex. The Schattschneider illustration below shows that the first differences of the coordination sequence are 2, 6, 3, 3, 6, 3, 3, 6, 3, 3, ..., and so the sequence itself satisfies the recurrence in the definition. The tiling has symmetry group D_6 (the dihedral group of order 6).

Continuing from the arrangement of pennies described in A136289, we also wish to place dimes over the holes in the array, where the n-th generation of dimes can be placed only when all three of its supporting pennies are in place already; then a(n-1) is the number of dimes in generation n for >= 1. - Colin Mallows, Apr 13 2008

Doris Schattschneider, In Praise of Amateurs, pp. 140-166 in D. A. Klarner, editor, The Mathematical Gardner. Prindle, Weber and Schmidt, Boston, 1981.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Colin Mallows, Analysis of coordination sequence for Marjorie Rice tiling
Doris Schattschneider, Illustration of terms a(0) to a(12) of the coordination sequence for the Rice tiling
N. J. A. Sloane, Illustration of terms a(0) to a(9) of the coordination sequence for the Rice tiling [Annotated copy of Fig. 15 of Schattschneider (1981)]
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A042965, A136289.

a:=[1,3,9,12,15]; [n le 5 select a[n] else Self(n-1)+Self(n-3)-Self(n-4):n in [1..60]]; // Marius A. Burtea, Oct 23 2019

1, seq(4*n -simplify(ChebyshevU(n-1, -1/2)), n = 1..20); # G. C. Greubel, Apr 13 2021

{1}~Join~LinearRecurrence[{1, 0, 1, -1}, {3, 9, 12, 15}, 59] (* Jean-François Alcover, Oct 23 2019 *)

[1]+[4*n-chebyshev_U(n-1,-1/2) for n in (1..60)] # G. C. Greubel, Apr 13 2021

G.f.: (1 + 2*x + 6*x^2 + 2*x^3 + x^4)/((1-x)^2*(1+x+x^2)). - Colin Barker, Jul 12 2014
a(n) = [n=0] + 4*n - ChebyshevU(n-1, -1/2). - G. C. Greubel, Apr 13 2021
For n > 0, a(n) = 3*A042965(n+1). - Jon E. Schoenfield, Jun 03 2022

Entry revised by N. J. A. Sloane, Apr 06 2019, replacing the old definition with Colin Barker's recurrence.

A263536 Row sum of an equilateral triangle tiled with the 3,4,5 Pythagorean triple.

Original entry on oeis.org

5, 7, 12, 17, 19, 24, 29, 31, 36, 41, 43, 48, 53, 55, 60, 65, 67, 72, 77, 79, 84, 89, 91, 96, 101, 103, 108, 113, 115, 120, 125, 127, 132, 137, 139, 144, 149, 151, 156, 161, 163, 168, 173, 175, 180, 185, 187, 192, 197, 199, 204, 209, 211, 216, 221, 223, 228

Offset: 1

Craig Knecht, Oct 20 2015

Maximum number of Pythagorean triples in an equilateral triangle.
Two rules are used to construct this equilateral triangle: #1. Start with the number 5 at the top. #2. Require every "triple" to contain the Pythagorean triple 3, 4, 5 (see link below).
Up and down Pythagorean triples consist of two terms below and one above when k is odd (an up triple), and two terms above and one below when k is even (a down triple). Three adjacent terms in a straight line within the triangle form a linear triple.

			Triangle T(n,k):           Row sum
  5;                          5
  3, 4;                       7
  4, 5, 3;                   12
  5, 3, 4, 5;                17
  3, 4, 5, 3, 4;             19
  4, 5, 3, 4, 5, 3;          24

Colin Barker, Table of n, a(n) for n = 1..1000
Craig Knecht, Equilateral triangle tiled with 3,4,5 Pythagorean triples.
Craig Knecht, Interlocked up/down Pythagorean pairs.
Craig Knecht, Linear and triangular triples.
Craig Knecht, Incarcerated numbers.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A136289 (every triple contains 1,2,3), A008854 (every triple contains 1,2,2), A259052 (sum of Pascal triples).

Vec(x*(5*x^2+2*x+5)/((x-1)^2*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Oct 26 2015

From Colin Barker, Oct 26 2015: (Start)
a(n) = a(n-1)+a(n-3)-a(n-4) for n>4.
G.f.: x*(5*x^2+2*x+5) / ((x-1)^2*(x^2+x+1)).
(End)

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A047266 Numbers that are congruent to {0, 1, 5} mod 6.

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A264788 a(n) is the number of circles added at n-th iteration of the pattern starting with 2 circles. (See comment.)

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A136290 a(0)=1, a(1)=3, a(2)=9, a(3)=12, a(4)=15; thereafter a(n) = a(n-1) + a(n-3) - a(n-4).

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A263536 Row sum of an equilateral triangle tiled with the 3,4,5 Pythagorean triple.

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