A165351 -id:A165351 - cp's OEIS frontend

A165520 Antidiagonal writing from three rows trio A165351,A165355,A165367 (first,second and third trisections of A026741).

Original entry on oeis.org

0, 1, 3, 1, 2, 3, 5, 7, 9, 4, 5, 6, 11, 13, 15, 7, 8, 9, 17, 19, 21, 10, 11, 12, 23, 25, 27, 13, 14, 15, 29, 31, 33

Offset: 1

Paul Curtz, Sep 21 2009

(6n+3)-th term is 6n+3=A016945. 6n-th term is 3n=3*n=A008585.

Mix (A004273(3n),A004273(3n+1),A004273(3n+2)), (A000027(3n),A000027(3n+1),A000027(3n+2)).

A165355 a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.

Original entry on oeis.org

1, 2, 7, 5, 13, 8, 19, 11, 25, 14, 31, 17, 37, 20, 43, 23, 49, 26, 55, 29, 61, 32, 67, 35, 73, 38, 79, 41, 85, 44, 91, 47, 97, 50, 103, 53, 109, 56, 115, 59, 121, 62, 127, 65, 133, 68, 139, 71, 145, 74, 151, 77, 157, 80, 163, 83, 169, 86, 175, 89, 181, 92, 187, 95, 193, 98

Offset: 0

Paul Curtz, Sep 16 2009

Second trisection of A026741.
A111329(n+1) = A000041(a(n)). - Reinhard Zumkeller, Nov 19 2009
We observe that this sequence is a particular case of sequence for which there exists q: a(n+3) = (a(n+2)*a(n+1)+q)/a(n) for every n >= n0. Here q=-9 and n0=0. - Richard Choulet, Mar 01 2010
The entries are also encountered via the bilinear transform approximation to the natural log (unit circle). Specifically, evaluating 2(z-1)/(z+1) at z = 2, 3, 4, ..., A165355 entries stem from the pair (sums) seen 2 ahead of each new successive prime. For clarity, the evaluation output is 2, 3, 1, 1, 6, 5, 4, 3, 10, 7, 3, 2, 14, 9, 8, 5, 18, 11, ..., where (1+1), (4+3), (3+2), (8+5), ... generate the A165355 entries (after the first). As an aside, the same mechanism links A165355 to A140777. - Bill McEachen, Jan 08 2015
As a follow-up to the previous comment, it appears that the numerators and denominators of 2(z-1)/(z+1) are respectively given by A145979 and A060819, but with different offsets. - Michel Marcus, Jan 14 2015
Odd parts of the terms give A067745. E.g.: 1, 2/2, 7, 5, 13, 8/8 .... - Joe Slater, Nov 30 2016

G. C. Greubel, Table of n, a(n) for n = 0..1000
D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A165351, A165367, A067745, A007814.

f[n_] := If[ OddQ@ n, (3n +1)/2, (3n +1)]; Array[f, 66, 0] (* Robert G. Wilson v, Jan 26 2015 *)
f[n_] := (3 (-1)^(2n) + (-1)^(1 + n)) (-2 + 3n)/4; Array[f, 66] (* or *)
CoefficientList[ Series[(x^3 + 5x^2 + 2x + 1)/(x^2 - 1)^2, {x, 0, 65}], x] (* or *)
LinearRecurrence[{0, 2, 0, -1}, {1, 2, 7, 5}, 66] (* Robert G. Wilson v, Apr 13 2017 *)

a(n)=n+=2*n+1; if(n%2,n,n/2) \\ Charles R Greathouse IV, Jan 13 2015

a(n) = A026741(3*n+1).
a(n)*A026741(n) = A005449(n).
a(n)*A022998(n+1) = A000567(n+1).
a(n) = A026741(n+1) + A022998(n).
a(2n) = A016921(n). a(2n+1) = A016789(n).
a(2n+1)*A026741(2n) = A045944(n).
G.f.: (1+2*x+5*x^2+x^3)/((x-1)^2 * (1+x)^2). - R. J. Mathar, Sep 26 2009
a(n) = (3+9*n)/4 + (-1)^n*(1+3*n)/4. - R. J. Mathar, Sep 26 2009
a(n) = 2*(3n+1)/(4-((2n+2) mod 4)). - Bill McEachen, Jan 09 2015
If a(2n-1) = x then a(2n) = 2x+3. - Robert G. Wilson v, Jan 26 2015
Let the reduced Collatz procedure be defined as Cr(n) = (3*n+1)/2. For odd n, a(n) = Cr(n). For even n, a(n) = Cr(4*n+1)/2. - Joe Slater, Nov 29 2016
a(n) = A067745(n+1) * 2^A007814((3n+1)/2). - Joe Slater, Nov 30 2016
a(n) = 2*a(n-2) - a(n-4). - G. C. Greubel, Apr 13 2017

All comments changed to formulas by R. J. Mathar, Sep 26 2009
New name from Charles R Greathouse IV, Jan 13 2015
Name corrected by Joe Slater, Nov 29 2016

A098832 Square array read by antidiagonals: even-numbered rows of the table are of the form n(n+m) and odd-numbered rows are of the form n(n+m)/2.

Original entry on oeis.org

1, 3, 3, 6, 8, 2, 10, 15, 5, 5, 15, 24, 9, 12, 3, 21, 35, 14, 21, 7, 7, 28, 48, 20, 32, 12, 16, 4, 36, 63, 27, 45, 18, 27, 9, 9, 45, 80, 35, 60, 25, 40, 15, 20, 5, 55, 99, 44, 77, 33, 55, 22, 33, 11, 11, 66, 120, 54, 96, 42, 72, 30, 48, 18, 24, 6, 78, 143, 65, 117, 52, 91, 39, 65, 26, 39, 13, 13

Offset: 1

Eugene McDonnell (eemcd(AT)mac.com), Nov 02 2004

The rows of this table and that in A098737 are related. Given a function f = n/( 1 + (1+n) mod(2) ), row n of A098737 can be derived from row n of T by multiplying the latter by f(n); row n of T can be derived from row n of A098737 by dividing the latter by f(n).

			Array begins as:
  1,  3,  6, 10, 15, 21,  28,  36,  45 ... A000217;
  3,  8, 15, 24, 35, 48,  63,  80,  99 ... A005563;
  2,  5,  9, 14, 20, 27,  35,  44,  54 ... A000096;
  5, 12, 21, 32, 45, 60,  77,  96, 117 ... A028347;
  3,  7, 12, 18, 25, 33,  42,  52,  63 ... A027379;
  7, 16, 27, 40, 55, 72,  91, 112, 135 ... A028560;
  4,  9, 15, 22, 30, 39,  49,  60,  72 ... A055999;
  9, 20, 33, 48, 65, 84, 105, 128, 153 ... A028566;
  5, 11, 18, 26, 35, 45,  56,  68,  81 ... A056000;
Antidiagonals begin as:
   1;
   3,  3;
   6,  8,  2;
  10, 15,  5,  5;
  15, 24,  9, 12,  3;
  21, 35, 14, 21,  7,  7;
  28, 48, 20, 32, 12, 16,  4;
  36, 63, 27, 45, 18, 27,  9,  9;
  45, 80, 35, 60, 25, 40, 15, 20,  5;
  55, 99, 44, 77, 33, 55, 22, 33, 11, 11;

G. C. Greubel, Antidiagonals n = 1..50, flattened

Row m of array: A000217 (m=1), A005563 (m=2), A000096 (m=3), A028347 (m=4), A027379 (m=5), A028560 (m=6), A055999 (m=7), A028566 (m=8), A056000 (m=9), A098603 (m=10), A056115 (m=11), A098847 (m=12), A056119 (m=13), A098848 (m=14), A056121 (m=15), A098849 (m=16), A056126 (m=17), A098850 (m=18), A051942 (m=19).
Column m of array: A026741 (m=1), A022998 (m=2), A165351 (m=3).
Cf. A098737, A168509, A181900.

A098832:= func< n,k | (1/4)*(3+(-1)^k)*(n+1)*(n-k+1) >;
[A098832(n,k): k in [1..n], n in [1..15]]; // G. C. Greubel, Jul 31 2022

A098832[n_, k_]:= (1/4)*(3+(-1)^k)*(n+1)*(n-k+1);
Table[A098832[n,k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Jul 31 2022 *)

def A098832(n,k): return (1/4)*(3+(-1)^k)*(n+1)*(n-k+1)
flatten([[A098832(n,k) for k in (1..n)] for n in (1..15)]) # G. C. Greubel, Jul 31 2022

Item m of row n of T is given (in infix form) by: n T m = n * (n + m) / (1 + m (mod 2)). E.g. Item 4 of row 3 of T: 3 T 4 = 14.
From G. C. Greubel, Jul 31 2022: (Start)
A(n, k) = (1/4)*(3 + (-1)^n)*k*(k+n) (array).
T(n, k) = (1/4)*(3 + (-1)^k)*(n+1)*(n-k+1) (antidiagonal triangle).
Sum_{k=1..n} T(n, k) = (1/8)*(n+1)*( (3*n-1)*(n+1) + (1+(-1)^n)/2 ).
T(2*n-1, n) = A181900(n).
T(2*n+1, n) = 2*A168509(n+1). (End)

Missing terms added by G. C. Greubel, Jul 31 2022

A165367 Trisection a(n) = A026741(3n + 2).

Original entry on oeis.org

1, 5, 4, 11, 7, 17, 10, 23, 13, 29, 16, 35, 19, 41, 22, 47, 25, 53, 28, 59, 31, 65, 34, 71, 37, 77, 40, 83, 43, 89, 46, 95, 49, 101, 52, 107, 55, 113, 58, 119, 61, 125, 64, 131, 67, 137, 70, 143, 73, 149, 76, 155, 79, 161, 82, 167, 85, 173, 88, 179, 91, 185, 94, 191, 97, 197

Offset: 0

Paul Curtz, Sep 17 2009

The other trisections are A165351 and A165355.

John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1)

Cf. A000326, A016777, A016969, A022998, A026741, A045944, A165351, A165355.

A026741 := proc(n) if type(n,'odd') then n; else n/2 ; fi; end:
A165367 := proc(n) A026741(3*n+2) ; end: seq(A165367(n),n=0..100) ; # R. J. Mathar, Nov 22 2009

LinearRecurrence[{0, 2, 0, -1}, {1, 5, 4, 11}, 66] (* Jean-François Alcover, Nov 15 2017 *)

a(n) = (3*n+2)>>!(n%2); \\ Ruud H.G. van Tol, Oct 09 2023

a(n)*A022998(n) = A045944(n).
a(n)*A026741(n+1) = A000326(n+1).
a(2n) = A016777(n); a(2n+1) = A016969(n).
From R. J. Mathar Nov 22 2009: (Start)
a(n) = 2*a(n-2) - a(n-4).
G.f.: (1 + 5*x + 2*x^2 + x^3)/((1-x)^2*(1+x)^2). (End)

All comments rewritten as formulas by R. J. Mathar, Nov 22 2009

A165988 First trisection of A022998.

Original entry on oeis.org

0, 3, 12, 9, 24, 15, 36, 21, 48, 27, 60, 33, 72, 39, 84, 45, 96, 51, 108, 57, 120, 63, 132, 69, 144, 75, 156, 81, 168, 87, 180, 93, 192, 99, 204, 105, 216, 111, 228, 117, 240, 123, 252, 129, 264, 135, 276, 141, 288, 147, 300, 153, 312, 159, 324, 165, 336, 171, 348, 177

Offset: 0

Paul Curtz, Oct 03 2009

Read modulo 10, this yields a sequence with a period of length 10 containing all 10 digits: 0, 3, 2, 9, 4, 5, 6, 1, 8, 7.
The other two trisections start 1, 8, 7, 20, 13, 32, 19, 44.... and 4, 5, 16, 11, 28, 17, 40, 23....
The Pisano period lengths for reading the sequence modulo m>=1 are 1,  2,  1,  4, 10,  2, 14,  8,  6, 10, 22,  4, 26, 14, 10, 16, 34,  6, 38, 20, 14, 22, 46,  8, 50, 26, 18, 28, 58... - R. J. Mathar, Oct 08 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A165351, A165355, A165367.

LinearRecurrence[{0, 2, 0, -1}, {0, 3, 12, 9}, 50] (* G. C. Greubel, Apr 20 2016 *)

a(n) = my(n=3*n); if (n % 2, n, 2*n); \\ Michel Marcus, Apr 21 2016

a(n) = A022998(3n) = 3*A022998(n) = 3*n*(3 +(-1)^n)/2 .
a(n) = 2*a(n-2) - a(n-4).
G.f.: 3*x*(1+4*x+x^2)/((x-1)^2 *(1+x)^2).
E.g.f.: 3*x*(-1 + 3*exp(2*x))*exp(-x)/2. - Ilya Gutkovskiy, Apr 21 2016

A166720 Trisection A165342(3n).

Original entry on oeis.org

0, 3, 3, 9, 12, 15, 9, 21, 6, 27, 15, 33, 18, 39, 21, 45, 12, 51, 27, 57, 240, 63, 33, 69, 18, 75, 39, 81, 42, 87, 45, 93, 24, 99, 51, 105, 108, 111, 57, 117, 30, 123, 63, 129, 66, 135, 69, 141, 36, 147, 75, 153, 312, 159

Offset: 0

Paul Curtz, Oct 20 2009

nonn

See A165351 from A026741.

Edited by N. J. A. Sloane, Oct 22 2009

cp's OEIS Frontend

A165520 Antidiagonal writing from three rows trio A165351,A165355,A165367 (first,second and third trisections of A026741).

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A165355 a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.

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A098832 Square array read by antidiagonals: even-numbered rows of the table are of the form n*(n+m) and odd-numbered rows are of the form n*(n+m)/2.

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A165367 Trisection a(n) = A026741(3n + 2).

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A165988 First trisection of A022998.

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A166720 Trisection A165342(3n).

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A098832 Square array read by antidiagonals: even-numbered rows of the table are of the form n(n+m) and odd-numbered rows are of the form n(n+m)/2.