A087444 -id:A087444 - cp's OEIS frontend

A047241 Numbers that are congruent to {1, 3} mod 6.

1, 3, 7, 9, 13, 15, 19, 21, 25, 27, 31, 33, 37, 39, 43, 45, 49, 51, 55, 57, 61, 63, 67, 69, 73, 75, 79, 81, 85, 87, 91, 93, 97, 99, 103, 105, 109, 111, 115, 117, 121, 123, 127, 129, 133, 135, 139, 141, 145, 147, 151, 153, 157, 159, 163, 165, 169, 171, 175, 177, 181, 183

Offset: 1

N. J. A. Sloane

Also the numbers k such that 10^p+k could possibly be prime. - Roderick MacPhee, Nov 20 2011 This statement can be written as follows. If 10^m + k = prime, for any m >= 1, then k is in this sequence. See the pink box comments by Roderick MacPhee from Dec 09 2014. - Wolfdieter Lang, Dec 09 2014
The odd-indexed terms are one more than the arithmetic mean of their neighbors; the even-indexed terms are one less than the arithmetic mean of their neighbors. - Amarnath Murthy, Jul 29 2003
Partial sums are A212959. - Philippe Deléham, Mar 16 2014
12*a(n) is conjectured to be the length of the boundary after n iterations of the hexagon and square expansion shown in the link. The squares and hexagons have side length 1 in some units. The pattern is supposed to become the planar Archimedean net 4.6.12 when n -> infinity. - Kival Ngaokrajang, Nov 30 2014
Positive numbers k for which 1/2 + k/3 + k^2/6 is an integer. - Bruno Berselli, Apr 12 2018

L. Lovasz, J. Pelikan, K. Vesztergombi, Discrete Mathematics, Springer (2003); 14.4, p. 225.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
L. Lovász, J. Pelikán and K. Vesztergombi, Discrete Mathematics, Elementary and Beyond, Springer (2003); 14.4, p. 225.
Kival Ngaokrajang, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A047233, A056970, A007310, A047228, A047261, A047273, A212959.
Subsequence of A186422.
Union of A016921 and A016945. - Wesley Ivan Hurt, Sep 28 2013

a047241 n = a047241_list !! (n-1)
a047241_list = 1 : 3 : map (+ 6) a047241_list
-- Reinhard Zumkeller, Feb 19 2013

seq(3*k-2-((k+1) mod 2), k=1..100); # Wesley Ivan Hurt, Sep 28 2013

Table[{2, 4}, {30}] // Flatten // Prepend[#, 1]& // Accumulate (* Jean-François Alcover, Jun 10 2013 *)
Select[Range[200], MemberQ[{1, 3}, Mod[#, 6]]&] (* or *) LinearRecurrence[{1, 1, -1}, {1, 3, 7}, 70] (* Harvey P. Dale, Oct 01 2013 *)

a(n)=bitor(3*n-3,1) \\ Charles R Greathouse IV, Sep 28 2013

for n in range(1,10**5):print(3*n-2-((n+1)%2)) # Soumil Mandal, Apr 14 2016

From Paul Barry, Sep 04 2003: (Start)
O.g.f.: (1 + 2*x + 3*x^2)/((1 + x)*(1 - x)^2) = (1 + 2*x + 3*x^2)/((1 - x)*(1 - x^2)).
E.g.f.: (6*x + 1)*exp(x)/2 + exp(-x)/2;
a(n) = 3*n - 5/2 - (-1)^n/2. (End)
a(n) = 2*floor((n-1)/2) + 2*n - 1. - Gary Detlefs, Mar 18 2010
a(n) = 6*n - a(n-1) - 8 with n > 1, a(1)=1. - Vincenzo Librandi, Aug 05 2010
a(n) = 3*n - 2 - ((n+1) mod 2). - Wesley Ivan Hurt, Jun 29 2013
a(1)=1, a(2)=3, a(3)=7; for n>3, a(n) = a(n-1) + a(n-2) - a(n-3). - Harvey P. Dale, Oct 01 2013
From Benedict W. J. Irwin, Apr 13 2016: (Start)
A005408(a(n)+1) = A016813(A001651(n)),
A007310(a(n)) = A005408(A087444(n)-1),
A007310(A005408(a(n)+1)) = A017533(A001651(n)). (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(4*sqrt(3)) + log(3)/4. - Amiram Eldar, Dec 11 2021

Formula corrected by Bruno Berselli, Jun 24 2010

A087445 Numbers that are congruent to 1 or 5 mod 12.

Original entry on oeis.org

1, 5, 13, 17, 25, 29, 37, 41, 49, 53, 61, 65, 73, 77, 85, 89, 97, 101, 109, 113, 121, 125, 133, 137, 145, 149, 157, 161, 169, 173, 181, 185, 193, 197, 205, 209, 217, 221, 229, 233, 241, 245, 253, 257, 265, 269, 277, 281, 289, 293, 301, 305, 313, 317, 325, 329

Offset: 1

Paul Barry, Sep 04 2003

From Bob Selcoe, Jun 03 2015: (Start)
For k >= 1: all numbers congruent to A002450(k) mod 2^(2k+1) and A072197(k) mod 4^(k+1) not congruent to 0 mod 3. Equivalently, for k >= 3: all numbers congruent to A096773(k) mod 2^k not congruent to 0 mod 3.
Conjecture: at least one number in this sequence must appear in all Collatz sequences.
(End)
The sequence is composed of all numbers in congruence classes T(n,1) mod 2^(n+k) in A259663 (i.e., T"(1) in array T259663(n,k)) not congruent to 0 mod 3. Therefore the conjecture above is true (see A259663 for additional explanation). - Bob Selcoe, Jul 15 2017
Closure of {1} under the map (x,y)->2x+3y [Klarner-Rado, see Lagarias (2016), p. 755]. - N. J. A. Sloane, Oct 06 2016
The above conjecture is true: this is because even numbers and odd numbers divisible by 3 will lead to the set of odd numbers not divisible by 3. Odd numbers of the form 4k - 1 can also be ignored, as this consists of odd numbers that grow between themselves and the next odd term through Collatz iteration. No infinite sequence of growth between consecutive odd terms is possible, so all numbers of the form 4k - 1 will lead to an odd number that shrinks between itself and the next odd number. All numbers 4k - 1 will lead to a number in 4k - 3, the odd numbers that shrink between themselves and the following odd term. What we are left after that elimination is this sequence. - Aidan Simmons, Feb 25 2019

Jeffrey C. Lagarias, Erdős, Klarner and the 3x+ 1 Problem, Amer. Math. Monthly, Vol. 123, No. 8 (2016), pp. 753-776.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001651, A002450, A047241, A087444, A087446, A096773, A259663.

[k:k in [1..330]| k mod 12 in [1,5]]; // Marius A. Burtea, Feb 08 2020

seq(6*(n-1)-(-1)^n,n=1..100); # Robert Israel, Jun 10 2015

LinearRecurrence[{1,1,-1},{1,5,13},70] (* or *) Rest[CoefficientList[ Series[x (1+4x+7x^2)/((1+x)(1-x)^2),{x,0,70}],x]]  (* Harvey P. Dale, Jun 13 2011 *)

a(n)=(n-1)\2*12 + [5,1][n%2+1] \\ Charles R Greathouse IV, Jun 03 2015

G.f.: x*(1+4*x+7*x^2)/((1+x)*(1-x)^2).
E.g.f.: 6*(x-1)*exp(x) + 7 - exp(-x). - corrected by Robert Israel, Jun 10 2015
a(n) = 6*(n-1) - (-1)^n. - Rolf Pleisch, Aug 04 2009
a(n) = 12*n - a(n-1) - 18 (with a(1)=1). - Vincenzo Librandi, Aug 08 2010
a(n) = a(n-1) + a(n-2) - a(n-3), with a(0)=1, a(1)=5, a(2)=13. - Harvey P. Dale, Jun 13 2011
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/12 + log(2 + sqrt(3))/(2*sqrt(3)). - Amiram Eldar, Dec 28 2021

A087446 Numbers that are congruent to {1, 6} mod 15.

Original entry on oeis.org

1, 6, 16, 21, 31, 36, 46, 51, 61, 66, 76, 81, 91, 96, 106, 111, 121, 126, 136, 141, 151, 156, 166, 171, 181, 186, 196, 201, 211, 216, 226, 231, 241, 246, 256, 261, 271, 276, 286, 291, 301, 306, 316, 321, 331, 336, 346, 351, 361, 366, 376, 381, 391, 396, 406

Offset: 1

Paul Barry, Sep 04 2003

3*a(n) is conjectured to be the number of edges (bonds) visited when walking around the boundary of a certain equilateral triangle construction at the n-th iteration. See the illustration in the link. Note that isthmus edges (bridges) are counted twice. The pattern is supposed to become the planar Archimedean net 3.12.12 when n -> infinity. - Kival Ngaokrajang, Nov 30 2014

Kival Ngaokrajang, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (1,1,-1)

Cf. A001651, A047241, A087444, A087445.

#+{1,6}&/@(15*Range[0,30])//Flatten (* or *) LinearRecurrence[{1,1,-1},{1,6,16},60] (* Harvey P. Dale, Dec 05 2018 *)

G.f.: x*(1 + 5*x + 9*x^2)/((1 + x)*(1 - x)^2).
E.g.f.: (30*x-1)*exp(x)/4 + 5*exp(-x)/4.
a(n) = (18*n-1)/4 + 5*(-1)^n/4.
a(n) = 15*n - a(n-1) - 23, with a(1)=1. - Vincenzo Librandi, Aug 08 2010

Editing: rewording of Kival Ngaokrajang's comment. - Wolfdieter Lang, Dec 06 2014

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A047241 Numbers that are congruent to {1, 3} mod 6.

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A087445 Numbers that are congruent to 1 or 5 mod 12.

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A087446 Numbers that are congruent to {1, 6} mod 15.

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