A004767 -id:A004767 - cp's OEIS frontend

A142463 a(n) = 2n^2 + 2n - 1.

-1, 3, 11, 23, 39, 59, 83, 111, 143, 179, 219, 263, 311, 363, 419, 479, 543, 611, 683, 759, 839, 923, 1011, 1103, 1199, 1299, 1403, 1511, 1623, 1739, 1859, 1983, 2111, 2243, 2379, 2519, 2663, 2811, 2963, 3119, 3279, 3443, 3611, 3783, 3959, 4139, 4323, 4511, 4703, 4899, 5099

Offset: 0

Roger L. Bagula, Sep 19 2008

Essentially the same as A132209.
From Vincenzo Librandi, Nov 25 2010: (Start)
Numbers k such that 2*k + 3 is a square.
First diagonal of A144562. (End)
The terms a(n) give the values for c of indefinite binary quadratic forms [a, b, c] = [2, 4n+2, a(n)] of discriminant D = 12, where a and c can be switched. The positive numbers represented by these forms are given in A084917. - Klaus Purath, Aug 31 2023

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Leo Tavares, Illustration: Hexagonic Diamonds.
Leo Tavares, Illustration: Hexagonic Rectangles.
Leo Tavares, Illustration: Hexagonic Crosses.
Leo Tavares, Illustration: Hexagonic Columns.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A005408, A132209, A144562, A188653.

Cf. A005563, A016945, A001105, A000290, A004767, A046092, A002378, A028387.

Magma
```
[2*n^2+2*n-1: n in [0..100]];
```

A142463:= n-> 2*n^2 +2*n -1; seq(A142463(n), n=0..50); # G. C. Greubel, Mar 01 2021

Array[ -#*(2-#*2)-1&,5!,1] (* Vladimir Joseph Stephan Orlovsky, Dec 21 2008 *)
Table[2n^2+2n-1,{n,0,50}] (* Harvey P. Dale, Feb 29 2024 *)

a(n)=2*n^2+2*n-1 \\ Charles R Greathouse IV, Sep 24 2015

[2*n^2 +2*n -1 for n in (0..50)] # G. C. Greubel, Mar 01 2021

a(n) = a(n-1) + 4*n.
From Paul Barry, Nov 03 2009: (Start)
G.f.: (1 - 6*x + x^2)/(1-x)^3.
a(n) = 4*C(n+1,2) - 1. (End)
a(n) = -A188653(2*n+1). - Reinhard Zumkeller, Apr 13 2011
a(n) = 3*( Sum_{k=1..n} k^5 )/( Sum_{k=1..n} k^3 ), n > 0. - Gary Detlefs, Oct 18 2011
a(n) = (A005408(n)^2 - 3)/2. - Zhandos Mambetaliyev, Feb 11 2017
E.g.f.: (-1 + 4*x + 2*x^2)*exp(x). - G. C. Greubel, Mar 01 2021
From Leo Tavares, Nov 22 2021: (Start)
a(n) = 2*A005563(n) - A005408(n). See Hexagonic Diamonds illustration.
a(n) = A016945(n-1) + A001105(n-1). See Hexagonic Rectangles illustration.
a(n) = A004767(n-1) + A046092(n-1). See Hexagonic Crosses illustration.
a(n) = A002378(n) + A028387(n-1). See Hexagonic Columns illustration. (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Wesley Ivan Hurt, Dec 03 2021
Sum_{n>=0} 1/a(n) = tan(sqrt(3)*Pi/2)*Pi/(2*sqrt(3)). - Amiram Eldar, Sep 16 2022

Edited by the Associate Editors of the OEIS, Sep 02 2009

A017629 a(n) = 12*n + 9.

Original entry on oeis.org

9, 21, 33, 45, 57, 69, 81, 93, 105, 117, 129, 141, 153, 165, 177, 189, 201, 213, 225, 237, 249, 261, 273, 285, 297, 309, 321, 333, 345, 357, 369, 381, 393, 405, 417, 429, 441, 453, 465, 477, 489, 501, 513, 525, 537, 549, 561, 573, 585, 597, 609, 621, 633

Offset: 0

N. J. A. Sloane

Numbers k such that k mod 2 = (k+1) mod 3 = 1 and (k+2) mod 4 != 1. - Klaus Brockhaus, Jun 15 2004
For n > 3, the number of squares on the infinite 3-column chessboard at <= n knight moves from any fixed point. - Ralf Stephan, Sep 15 2004
A016946 is the subsequence of squares (for n = 3*k*(k+1) = A028896(k), then a(n) = (6k+3)^2 = A016946(k)). - Bernard Schott, Apr 05 2021

Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Subsequences include A072065, A016945, A016813, A008585, and A005408.

Cf. A008594, A017533, A017545, A017137, A004767, A001019, A028896, A016946, A089911.

a017629 = (+ 9) . (* 12)  -- Reinhard Zumkeller, Jul 05 2013

12*Range[0,200]+9 (* Vladimir Joseph Stephan Orlovsky, Feb 19 2011 *)
LinearRecurrence[{2,-1},{9,21},60] (* Harvey P. Dale, Apr 14 2019 *)

a(n)=12*n+9 \\ Charles R Greathouse IV, Jul 10 2016

[i+9 for i in range(525) if gcd(i,12) == 12] # Zerinvary Lajos, May 21 2009

a(n) = 6*(4*n+1) - a(n-1) (with a(0)=9). - Vincenzo Librandi, Dec 17 2010
A089911(2*a(n)) = 4. - Reinhard Zumkeller, Jul 05 2013
G.f.: (9 + 3*x)/(1 - x)^2. - Alejandro J. Becerra Jr., Jul 08 2020
Sum_{n>=0} (-1)^n/a(n) = (Pi + log(3-2*sqrt(2)))/(12*sqrt(2)). - Amiram Eldar, Dec 12 2021
E.g.f.: 3*exp(x)*(3 + 4*x). - Stefano Spezia, Feb 25 2023

A089911 a(n) = Fibonacci(n) mod 12.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0, 1, 1

Offset: 0

Casey Mongoven, Nov 14 2003

From Reinhard Zumkeller, Jul 05 2013: (Start)
Sequence has been applied by several composers to 12-tone equal temperament pitch structure. The complete Fibonacci mod 12 system (a set of 10 periodic sequences) exhausts all possible ordered dyads; that is, every possible combination of two pitches is found in these sets.
a(A008594(n)) = 0;
a(A227144(n)) = 1;
a(3*A047522(n)) = 2;
a(A017569(n)) = a(2*A016933(n)) = a(4*A016777(n)) = 3;
a(2*A017629(n)) = a(3*A017137(n)) = a(6*A004767(n)) = 4;
a(A227146(n)) = 5;
a(nonexistent) = 6;
a(2*A017581(n)) = 7;
a(2*A017557(n)) = a(4*A016813(n)) = 8;
a(A017617(n)) = a(2*A016957(n)) = a(4*A016789(n)) = 9;
a(3*A047621(n)) = 10;
a(2*A017653(n)) = 11. (End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..1199
Ron Knott, Fibonacci Numbers and the Golden Section
M. Renault, The Fibonacci Sequence Modulo M
A. P. Shah, Fibonacci Sequence Modulo m, Fibonacci Quarterly, Vol.6, No.2 (1968), 139-141.
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
Index entries for linear recurrences with constant coefficients, signature (1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1, 0, -1, 1).

Cf. A000045, A001175, A003893, A010881.

a089911 n = a089911_list !! n
a089911_list = 0 : 1 : zipWith (\u v -> (u + v) `mod` 12)
                       (tail a089911_list) a089911_list
-- Reinhard Zumkeller, Jul 01 2013

[Fibonacci(n) mod 12: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

with(combinat,fibonacci); A089911 := proc(n) fibonacci(n) mod 12; end;

Table[Mod[Fibonacci[n], 12], {n, 0, 100}] (* Vincenzo Librandi, Feb 04 2014 *)

a(n)=fibonacci(n)%12 \\ Charles R Greathouse IV, Feb 03 2014

Has period of 24, restricted period 12 and multiplier 5.

a(n) = (a(n-1) + a(n-2)) mod 12, a(0) = 0, a(1) = 1.

More terms from Ray Chandler, Nov 15 2003

A002516 Earliest sequence with a(a(n)) = 2n.

Original entry on oeis.org

0, 3, 6, 2, 12, 7, 4, 10, 24, 11, 14, 18, 8, 15, 20, 26, 48, 19, 22, 34, 28, 23, 36, 42, 16, 27, 30, 50, 40, 31, 52, 58, 96, 35, 38, 66, 44, 39, 68, 74, 56, 43, 46, 82, 72, 47, 84, 90, 32, 51, 54, 98, 60, 55, 100, 106, 80, 59, 62, 114, 104, 63, 116, 122, 192, 67, 70, 130

Offset: 0

Colin Mallows

T. D. Noe, Table of n, a(n) for n = 0..1000
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Index entries for sequences of the a(a(n)) = 2n family

Cf. A002517, A004767, A007379, A017089, A091067.

import Data.List (transpose)
a002516 n = a002516_list !! n
a002516_list = 0 : concat (transpose
[a004767_list, f a002516_list, a017089_list, g $ drop 2 a002516_list])
where f [z] = []; f (_:z:zs) = 2 * z : f zs
g [z] = [z]; g (z:_:zs) = 2 * z : g zs
-- Reinhard Zumkeller, Jun 08 2015

a[0] = 0; a[n_ /; Mod[n, 2] == 0] := a[n] = 2*a[n/2]; a[n_ /; Mod[n, 4] == 1] := n+2; a[n_ /; Mod[n, 4] == 3] := 2(n-2); Table[a[n], {n, 0, 67}] (* Jean-François Alcover, Feb 06 2012, after Henry Bottomley *)

v2(n)=valuation(n,2)
a(n)=2^v2(n)*(-1+3/2*n/2^v2(n)-(-3+1/2*n/2^v2(n))*(-1)^((n/2^v2(n)-1)/2))

a(n)=local(t); if(n<1,0,if(n%2==0,2*a(n/2),t=(n-1)/2; 3*t+1/2-(t-5/2)*(-1)^t)) \\ Ralf Stephan, Feb 22 2004

a(4n) = 2*(a(2n)), a(4n+1) = 4n+3, a(4n+2) = 2*(a(2n+1)), a(4n+3) = 8n+2. - Henry Bottomley, Apr 27 2000
From Ralf Stephan, Feb 22 2004: (Start)
a(n) = n + 2*A006519(n) if odd part of n is of form 4k+1, or 2n - 4*A006519(n) otherwise.
a(2n) = 2*a(n), a(2n+1) = 2n + 3 + (2n - 5)*[n mod 2].
G.f.: Sum_{k>=0} 2^k*t(6t^6 + t^4 + 2t^2 + 3)/(1 - t^4)^2, t = x^2^k. (End)

A257852 Array A read by upward antidiagonals in which the entry A(n,k) in row n and column k is defined by A(n,k) = (2^n(6k - 3 - 2*(-1)^n) - 1)/3, n,k >= 1.

Original entry on oeis.org

3, 1, 7, 13, 9, 11, 5, 29, 17, 15, 53, 37, 45, 25, 19, 21, 117, 69, 61, 33, 23, 213, 149, 181, 101, 77, 41, 27, 85, 469, 277, 245, 133, 93, 49, 31, 853, 597, 725, 405, 309, 165, 109, 57, 35, 341, 1877, 1109, 981, 533, 373, 197, 125, 65, 39

Offset: 1

L. Edson Jeffery, Jul 12 2015

Sequence is a permutation of the odd natural numbers.

Let N_1 denote the set of odd natural numbers, and let |y|_2 denote 2-adic valuation of y. Define the map F : N_1 -> N_1 by F(x) = (3*x + 1)/2^|3*x+1|_2 (cf. A075677). Then row n of A is the set of all x in N_1 for which |3*x + 1|_2 = A371093(x) = n. Hence F(A(n,k)) = 6*k - 3 - 2*(-1)^n.

			From _Ruud H.G. van Tol_, Oct 17 2023, corrected and extended by _Antti Karttunen_, Apr 18 2024: (Start)
Array A begins:
n\k|   1|   2|   3|   4|   5|   6|   7|   8| ...
---+---------------------------------------------
1  |   3,   7,  11,  15,  19,  23,  27,  31, ...
2  |   1,   9,  17,  25,  33,  41,  49,  57, ...
3  |  13,  29,  45,  61,  77,  93, 109, 125, ...
4  |   5,  37,  69, 101, 133, 165, 197, 229, ...
5  |  53, 117, 181, 245, 309, 373, 437, 501, ...
6  |  21, 149, 277, 405, 533, 661, 789, 917, ...
... (End)

Cf. A006370, A075677, A096773 (after its initial 0, column 1 of this array).
Cf. A004767, A017077, A082285, A238477 (rows 1-4).
Cf. A371092, A371093 (column and row indices for odd numbers).
Cf. also arrays A371095, A371096, A371097, A371100, A371101, A371102, A371103.

(* Array: *)
Grid[Table[(2^n*(6*k - 3 - 2*(-1)^n) - 1)/3, {n, 10}, {k, 10}]]
(* Array antidiagonals flattened: *)
Flatten[Table[(2^(n - k + 1)*(6*k - 3 - 2*(-1)^(n - k + 1)) - 1)/ 3, {n, 10}, {k, n}]]

up_to = 105;
A257852sq(n,k) = ((2^n * (6*k - 3 - 2*(-1)^n) - 1)/3);
A257852list(up_to) = { my(v = vector(up_to), i=0); for(a=1,oo, for(col=1,a, i++; if(i > up_to, return(v)); v[i] = A257852sq((a-(col-1)),col))); (v); };
v257852 = A257852list(up_to);
A257852(n) = v257852[n]; \\ Antti Karttunen, Apr 18 2024

From Ruud H.G. van Tol, Oct 17 2023: (Start)
A(n,k+1) = A(n,k) + 2^(n+1).
A(n+2,k) = A(n,k)*4 + 1.
A(1,k) = A004767(k-1).
A(2,k) = A017077(k-1).
A(3,k) = A082285(k-1).
A(4,k) = A238477(k). (End)
For all odd positive numbers n, A(A371093(n), A371092(n)) = n. - Antti Karttunen, Apr 24 2024

A078703 Number of ways of subtracting twice a triangular number from a perfect square to obtain the integer n.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 3, 1, 1, 3, 1, 1, 2, 2, 2, 3, 1, 1, 2, 2, 2, 2, 1, 1, 4, 1, 2, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 1, 4, 1, 2, 3, 1, 2, 2, 1, 1, 4, 2, 1, 3, 2, 1, 4, 2, 1, 2, 1, 3, 3, 1, 2, 2, 2, 2, 2, 1, 1, 6, 2, 2, 2, 1, 2, 2, 2, 1, 4, 2, 1, 3, 1, 2, 4, 1, 1, 3, 2, 2, 4, 2, 2, 2, 2, 1

Offset: 1

R. L. Coffman, K. W. McLaughlin and R. J. Dawson (robert.l.coffman(AT)uwrf.edu), Dec 19 2002

Also number of symmetric unimodal consecutive integer sequences that sum to the integer n (e.g., 4+5+6+5+4 = 24 = n). Also number of double trapezoidal arrangements of n objects, denoted SDT(n); i.e., the number of ways to arrange n objects into symmetrically-placed, congruent isosceles trapezoids adjoined at overlapping largest bases.

Also number of divisors of 4*n-1 of form 4*k+1 (or 4*k+3). - Vladeta Jovovic, Jan 05 2004. Therefore a(n) is one half of the number of divisors of A004767(n-1) (numbers 3 (mod 4)). - Wolfdieter Lang, Jul 29 2016

			SDT(34) = 4 since we have 34 or 11+12+11 or 6+7+8+7+6 or 2+3+4+5+6+5+4+3+2, Also 4*34 - 1 = 135 = (3^3)*(5^1) so that r1=3 and r2=1 (p1=3 and p2=5), resulting in SDT(34) = (3+1)*(1+1)/2 = 4.
a(4) = 2 since 4 = 2^2 - 2*0 = 4^2 - 2*6. Also A034178(4*4 - 1) = 2 since 15 = 4^2 - 1^2 = 8^2 - 7^2. - _Michael Somos_, May 11 2011
G.f. = x + x^2 + x^3 + 2*x^4 + x^5 + x^6 + 2*x^7 + x^8 + 2*x^9 + 2*x^10 + x^11 + ...
Number of divisors of numbers 3 (mod 4) (see the Jovovic Jan 05 2004 comment): a(16) = 3 from the 2*3 = 6 divisors [1, 3, 7, 9, 21, 63] of 63 = A004767(15), being 1, -1, -1, 1, 1, -1 (mod 4). - _Wolfdieter Lang_, Jul 29 2016

Seiichi Manyama, Table of n, a(n) for n = 1..10000
T. Verhoeff, Rectangular and Trapezoidal Arrangements, J. Integer Sequences, Vol. 2 (1999), Article 99.1.6.

Cf. A000005, A001620, A004767, A001227, A034178, A359227, A359240.

(* This defines SDT(n): *)
SDT[n_] := Length[Cases[Range[1, n], j_ /; Cases[Range[1, j], k_ /; Plus @@ Join[Range[k, j], Range[j - 1, k, -1]] == n] != {}]] The restricted factorization technique for obtaining SDT(n) is encoded as follows: SDT[n_] := (Times @@ Cases[FactorInteger[4 n - 1], {p_, r_} -> r + 1])/2
Rest[ CoefficientList[ Series[ Sum[x^k/(1 - x^(4k - 1)), {k, 111}], {x, 0, 110}], x]] (* Robert G. Wilson v, Sep 20 2005 *)
a[ n_] := If[ n < 1, 0, With[{m = 4 n - 1}, Sum[1 - Sign@Mod[m - k^2, 2 k], {k, Sqrt@m}]]]; (* Michael Somos, Aug 01 2016 *)
a[n_] := DivisorSigma[0, 4*n - 1]/2; Array[a, 100] (* Amiram Eldar, Dec 26 2022 *)

{a(n) = if( n<1, 0, n = 4*n-1; sum(k=1, sqrtint(n), 0 == (n - k^2) % (2*k)))}; /* Michael Somos, Aug 01 2016 */

a(n) = ((r1 + 1)*(r2 + 1)*...*(rk + 1))/2, where ((p1^r1)*(p2^r2)*...*(pk^rk)) is the factorization of 4*n - 1 into (odd) primes.
G.f.: Sum_{n>0} x^n/(1-x^(4*n-1)). - Vladeta Jovovic, Jan 05 2004
a(n) = A034178(4*n - 1). - Michael Somos, May 11 2011
G.f.: Sum_{n >= 1} x^(3*n-2)/(1 - x^(4*n-3)). - Peter Bala, Jan 08 2021
From Amiram Eldar, Dec 26 2022: (Start)
a(n) = A000005(A004767(n-1))/2.
Sum_{k=1..n} a(k) = (log(n) + 2*gamma - 1 + 4*log(2))*n/4 + O(n^(1/3)*log(n)), where gamma is Euler's constant (A001620). (End)
G.f.: Sum_{n >= 1} x^(n^2)/(1-x^(2*n-1)) (conjecture). - Joerg Arndt, Jan 04 2024

A191670 Dispersion of A042968 (>1 and congruent to 1 or 2 or 3 mod 4), by antidiagonals.

Original entry on oeis.org

1, 2, 4, 3, 6, 8, 5, 9, 11, 12, 7, 13, 15, 17, 16, 10, 18, 21, 23, 22, 20, 14, 25, 29, 31, 30, 27, 24, 19, 34, 39, 42, 41, 37, 33, 28, 26, 46, 53, 57, 55, 50, 45, 38, 32, 35, 62, 71, 77, 74, 67, 61, 51, 43, 36, 47, 83, 95, 103, 99, 90, 82, 69, 58, 49, 40, 63

Offset: 1

Clark Kimberling, Jun 11 2011

For a background discussion of dispersions, see A191426.
...
Each of the sequences (4n, n>2), (4n+1, n>0), (3n+2, n>=0), generates a dispersion. Each complement (beginning with its first term >1) also generates a dispersion. The six sequences and dispersions are listed here:
...
A191452=dispersion of A008586 (4k, k>=1)
A191667=dispersion of A016813 (4k+1, k>=1)
A191668=dispersion of A016825 (4k+2, k>=0)
A191669=dispersion of A004767 (4k+3, k>=0)
A191670=dispersion of A042968 (1 or 2 or 3 mod 4 and >=2)
A191671=dispersion of A004772 (0 or 1 or 3 mod 4 and >=2)
A191672=dispersion of A004773 (0 or 1 or 2 mod 4 and >=2)
A191673=dispersion of A004773 (0 or 2 or 3 mod 4 and >=2)
...
EXCEPT for at most 2 initial terms (so that column 1 always starts with 1):
A191452 has 1st col A042968, all else A008486
A191667 has 1st col A004772, all else A016813
A191668 has 1st col A042965, all else A016825
A191669 has 1st col A004773, all else A004767
A191670 has 1st col A008486, all else A042968
A191671 has 1st col A016813, all else A004772
A191672 has 1st col A016825, all else A042965
A191673 has 1st col A004767, all else A004773
...
Regarding the dispersions A191670-A191673, there is a formula for sequences of the type "(a or b or c mod m)", (as in the Mathematica program below):
   If f(n)=(n mod 3), then (a,b,c,a,b,c,a,b,c,...) is given by
   a*f(n+2)+b*f(n+1)+c*f(n), so that "(a or b or c mod m)" is given by
   a*f(n+2)+b*f(n+1)+c*f(n)+m*floor((n-1)/3)), for n>=1.

			Northwest corner:
1....2....3....5....7
4....6....9....13...18
8....11...15...21...29
12...17...23...31...42
16...22...30...41...55

Ivan Neretin, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals, flattened)

Row 1: A155167, Row 2: A171861.

Cf. A008486, A042968, A191673, A191452.

(* Program generates the dispersion array T of the increasing sequence f[n] *)
r = 40; r1 = 12; c = 40; c1 = 12;
a = 2; b = 3; c2 = 5; m[n_] := If[Mod[n, 3] == 0, 1, 0];
f[n_] := a*m[n + 2] + b*m[n + 1] + c2*m[n] + 4*Floor[(n - 1)/3]
Table[f[n], {n, 1, 30}] (* A042968 *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]] (* A191670 *)
Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191670 *)

A191673 Dispersion of A004773 (>1 and congruent to 0 or 1 or 2 mod 4), by antidiagonals.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 6, 8, 10, 11, 9, 12, 14, 16, 15, 13, 17, 20, 22, 21, 19, 18, 24, 28, 30, 29, 26, 23, 25, 33, 38, 41, 40, 36, 32, 27, 34, 45, 52, 56, 54, 49, 44, 37, 31, 46, 61, 70, 76, 73, 66, 60, 50, 42, 35, 62, 82, 94, 102, 98, 89, 81, 68, 57, 48, 39, 84

Offset: 1

Clark Kimberling, Jun 11 2011

For a background discussion of dispersions, see A191426.
...
Each of the sequences (4n, n>2), (4n+1, n>0), (3n+2, n>=0), generates a dispersion. Each complement (beginning with its first term >1) also generates a dispersion. The six sequences and dispersions are listed here:
...
A191452=dispersion of A008586 (4k, k>=1)
A191667=dispersion of A016813 (4k+1, k>=1)
A191668=dispersion of A016825 (4k+2, k>=0)
A191669=dispersion of A004767 (4k+3, k>=0)
A191670=dispersion of A042968 (1 or 2 or 3 mod 4 and >=2)
A191671=dispersion of A004772 (0 or 1 or 3 mod 4 and >=2)
A191672=dispersion of A004773 (0 or 1 or 2 mod 4 and >=2)
A191673=dispersion of A004773 (0 or 2 or 3 mod 4 and >=2)
...
EXCEPT for at most 2 initial terms (so that column 1 always starts with 1):
A191452 has 1st col A042968, all else A008486
A191667 has 1st col A004772, all else A016813
A191668 has 1st col A042965, all else A016825
A191669 has 1st col A004773, all else A004767
A191670 has 1st col A008486, all else A042968
A191671 has 1st col A016813, all else A004772
A191672 has 1st col A016825, all else A042965
A191673 has 1st col A004767, all else A004773
...
Regarding the dispersions A191670-A191673, there is a formula for sequences of the type "(a or b or c mod m)", (as in the Mathematica program below):
   If f(n)=(n mod 3), then (a,b,c,a,b,c,a,b,c,...) is given by
   a*f(n+2)+b*f(n+1)+c*f(n), so that "(a or b or c mod m)" is given by
   a*f(n+2)+b*f(n+1)+c*f(n)+m*floor((n-1)/3)), for n>=1.

			Northwest corner:
1....2....4....6....9
3....5....8....12...17
7....10...14...20...28
11...16...22...30...41
15...21...29...40...54

Ivan Neretin, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals, flattened)

Cf. A004767, A004773, A191669, A191452.

(* Program generates the dispersion array T of the increasing sequence f[n] *)
r = 40; r1 = 12; c = 40; c1 = 12;
a = 2; b = 4; c2 = 5; m[n_] := If[Mod[n, 3] == 0, 1, 0];
f[n_] := a*m[n + 2] + b*m[n + 1] + c2*m[n] + 4*Floor[(n - 1)/3]
Table[f[n], {n, 1, 30}] (* A004773 *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]] (* A191673 *)
Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191673 *)

A001539 a(n) = (4n+1)(4*n+3).

Original entry on oeis.org

3, 35, 99, 195, 323, 483, 675, 899, 1155, 1443, 1763, 2115, 2499, 2915, 3363, 3843, 4355, 4899, 5475, 6083, 6723, 7395, 8099, 8835, 9603, 10403, 11235, 12099, 12995, 13923, 14883, 15875, 16899, 17955, 19043, 20163, 21315, 22499, 23715, 24963, 26243, 27555, 28899

Offset: 0

N. J. A. Sloane

Sequence arises from reading the line from 3, in the direction 3, 35, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
log(sqrt(2)+1)/sqrt(2) = 0.62322524... = 2/3 - 2/35 + 2/99 - 2/195 + 2/323, ... = (1 - 1/3) + (1/7 - 1/5) + (1/9 - 1/11) + (1/15 - 1/13) + (1/17 - 1/19) + (1/23 - 1/21) + ... - Gary W. Adamson, Mar 01 2009
Numbers k such that k+1 is a square and k+5 is divisible by 8. - Bruno Berselli, Sep 27 2017
The concatenation of 8*A000217(n) and 99 is a term of the sequence. Example: for A000217(5) = 15, 8*15 = 120 and 12099 = a(27). In general, a(5*n+2) = 800*A000217(n) + 99. - Bruno Berselli, Sep 29 2017

T. D. Noe, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A000466.

Cf. A000217, A000290, A001533, A001538, A004767, A016286, A016813, A016826, A157142, A133766, A154633.

CoefficientList[Series[(3 + 26 x + 3 x^2)/(1 - x)^3, {x, 0, 41}], x] (* or *) Table[(4 n + 1) (4 n + 3), {n, 0, 41}] (* Michael De Vlieger, Sep 29 2017 *)

makelist((4*n+1)*(4*n+3), n, 0, 30); /* Martin Ettl, Nov 12 2012 */

a(n)=(4*n+1)*(4*n+3) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = A016826(n) - 1 = (A001533(n)+5)/4 = (A001538(n)+16)/9.
Sum_{k>=0} 1/a(k) = Pi/8. - Benoit Cloitre, Aug 20 2002
G.f.: (3 + 26*x + 3*x^2)/(1 - x)^3. - Jaume Oliver Lafont, Mar 07 2009
a(n) = 32*n + a(n-1) for n > 0, a(0)=3. - Vincenzo Librandi, Nov 12 2010
a(n) = a(m) + 16*(n-m)*(n+m+1). The previous formula is obtained for m = n-1. - Bruno Berselli, Sep 29 2017
From Amiram Eldar, Feb 19 2023: (Start)
a(n) = A016813(n)*A004767(n).
Product_{n>=0} (1 - 1/a(n)) = sqrt(2)*cos(Pi/(2*sqrt(2))).
Product_{n>=0} (1 + 1/a(n)) = sqrt(2). (End)
From Elmo R. Oliveira, Oct 23 2024: (Start)
E.g.f.: exp(x)*(3 + 16*x*(2 + x)).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A106535 Numbers k such that the smallest x > 1 for which Fibonacci(x) == 0 mod k is x = k - 1.

Original entry on oeis.org

11, 19, 31, 59, 71, 79, 131, 179, 191, 239, 251, 271, 311, 359, 379, 419, 431, 439, 479, 491, 499, 571, 599, 631, 659, 719, 739, 751, 839, 971, 1019, 1039, 1051, 1091, 1171, 1259, 1319, 1399, 1439, 1451, 1459, 1499, 1531, 1559, 1571, 1619, 1759, 1811, 1831

Offset: 1

Peter K. Pearson (ppearson+att(AT)spamcop.net), May 06 2005

nonn

This is a sister sequence to A000057, because this sequence, since {k : A001177(k) = k-1}, might be called a subdiagonal sequence of A001177, and {k : A001177(k) = k+1}, which might be called a superdiagonal sequence of A001177. Sequences A000057 and A106535 are disjoint. Is this sequence the set of all divisors of some family of sequences, like A000057 is? - Art DuPre, Jul 11 2012
Are all members of this sequence prime? Using A069106, any composite members must exceed 89151931. - Robert Israel, Oct 13 2015
From Jianing Song, Jul 02 2019: (Start)
Yes, all terms are primes. See a brief proof below.
Also, if p == 1 (mod 4) then b(p) divides (p-Legendre(p,5))/2. So terms in this sequence are congruent to 11 or 19 modulo 20.
Primes p such that ord(-(3+sqrt(5))/2,p) = p-1, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer. (End)
Comments from Amiram Eldar, Jan 30 2022 (Start)
Sequence A003147, "Primes p with a Fibonacci primitive root", is defined in the paper: Daniel Shanks, Fibonacci primitive roots, Fibonacci Quarterly, Vol.  10, No. 2 (1972), pp. 163-168, and 181.
A second paper on this subject Daniel Shanks and Larry Taylor, An Observation of Fibonacci Primitive Roots, Fibonacci Quarterly, Vol. 11, No. 2 (1973), pp. 159-160,
deals with terms p == 3 (mod 4) of A003147, i.e., the intersection of A003147 and A002145 (or A004767).
It states that if g is a Fibonacci primitive root of a prime p such that p == 3 (mod 4) then g-1 and g-2 are also primitive roots of p.
The first 2000 terms of (A003147 intersect A002145) agree with the present sequence, although the definitions are quite different. Are these two sequences the same? (End)

Robert Israel, Table of n, a(n) for n = 1..2000
Alfred Brousseau, Primes which are factors of all Fibonacci sequences, Fib. Quart., 2 (1964), 33-38.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972.
Jianing Song, Proof that all terms are prime

Cf. A000057, A001175, A069106.

Similar sequences that give primes p such that A001177(p) = (p-1)/s: this sequence (s=1), A308795 (s=2), A308796 (s=3), A308797 (s=4), A308798 (s=5), A308799 (s=6), A308800 (s=7),A308801 (s=8), A308802 (s=9).

Filtered([2..2000], n -> Fibonacci(n-1) mod n = 0 and Filtered( [2..n-2], x -> Fibonacci(x) mod n = 0 ) = [] );

A106535 := proc(n)
        option remember;
        if n = 1 then
                11;
        else
                for a from procname(n-1)+1 do
                        if A001177(a) = a-1 then
                                return a;
                        end if;
                end do:
        end if;
end proc: # R. J. Mathar, Jul 09 2012
# Alternative:
fmod:= proc(a,b) local A;
  uses LinearAlgebra[Modular];
  A:= Mod(b, <<1,1>|<1,0>>,integer[8]);
  MatrixPower(b,A,a)[1,2];
end proc:
filter:= proc(n)
  local cands;
  if fmod(n-1,n) <> 0 then return false fi;
  cands:= map(t -> (n-1)/t, numtheory:-factorset(n-1));
  andmap(c -> (fmod(c,n) > 0), cands);
end proc:
select(filter, [$2..10^4]); # Robert Israel, Oct 13 2015

f[n_] := Block[{x = 2}, While[Mod[Fibonacci@ x, n] != 0, x++]; x];Select[Range@ 1860, f@ # == # - 1 &] (* Michael De Vlieger, Oct 13 2015 *)

isok(n) = {x = 2; while(fibonacci(x) % n, x++); x == n-1;} \\ Michel Marcus, Oct 20 2015

{n: A001177(n) = n-1}. - R. J. Mathar, Jul 09 2012

Corrected by T. D. Noe, Oct 25 2006

cp's OEIS Frontend

A142463 a(n) = 2*n^2 + 2*n - 1.

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A089911 a(n) = Fibonacci(n) mod 12.

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A002516 Earliest sequence with a(a(n)) = 2n.

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A257852 Array A read by upward antidiagonals in which the entry A(n,k) in row n and column k is defined by A(n,k) = (2^n*(6*k - 3 - 2*(-1)^n) - 1)/3, n,k >= 1.

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A078703 Number of ways of subtracting twice a triangular number from a perfect square to obtain the integer n.

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A142463 a(n) = 2n^2 + 2n - 1.

A257852 Array A read by upward antidiagonals in which the entry A(n,k) in row n and column k is defined by A(n,k) = (2^n(6k - 3 - 2*(-1)^n) - 1)/3, n,k >= 1.

A001539 a(n) = (4n+1)(4*n+3).