A007583 -id:A007583 - cp's OEIS frontend

A083066 5th row of number array A083064.

1, 5, 29, 173, 1037, 6221, 37325, 223949, 1343693, 8062157, 48372941, 290237645, 1741425869, 10448555213, 62691331277, 376147987661, 2256887925965, 13541327555789, 81247965334733, 487487792008397, 2924926752050381

Offset: 0

Paul Barry, Apr 21 2003

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=8, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)*charpoly(A,2). - Milan Janjic, Feb 21 2010

An Engel expansion of 3/2 to the base b := 6/5 as defined in A181565, with the associated series expansion 3/2 = b + b^2/5 + b^3/(5*29) + b^4/(5*29*173) + .... Cf. A007051. - Peter Bala, Oct 29 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Mudit Aggarwal and Samrith Ram, Generating Functions for Straight Polyomino Tilings of Narrow Rectangles, J. Int. Seq., Vol. 26 (2023), Article 23.1.4.
R. J. Mathar, Tilings of rectangular regions by rectangular tiles: Counts derived from transfer matrices, arXiv:1406.7788 [math.CO], 2014, eq (43).
Index entries for linear recurrences with constant coefficients, signature (7,-6).

Cf. A083064, A083065, A083067, A007051, A007583.

[(4*6^n+1)/5: n in [0..30]]; // Vincenzo Librandi, Nov 06 2011

f[n_]:=6^n; lst={}; Do[a=f[n]; Do[a-=f[m],{m,n-1,1,-1}]; AppendTo[lst,a/6],{n,1,30}]; lst (* Vladimir Joseph Stephan Orlovsky, Feb 10 2010 *)

a(n) = (4*6^n+1)/5.
G.f.: (1-2*x)/((1-6*x)*(1-x)).
E.g.f.: (4*exp(6*x)+exp(x))/5.
a(n) = 6*a(n-1)-1 with n>0, a(0)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = 7*a(n-1)-6*a(n-2). - Vincenzo Librandi, Nov 04 2011
a(n) = 6^n - Sum_{i=0..n-1} 6^i for n>0. - Bruno Berselli, Jun 20 2013

A127936 Numbers k such that 1 + Sum_{i=1..k} 2^(2*i-1) is prime.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 9, 11, 15, 21, 30, 39, 50, 63, 83, 95, 99, 156, 173, 350, 854, 1308, 1769, 2903, 5250, 5345, 5639, 6195, 7239, 21368, 41669, 47684, 58619, 63515, 69468, 70539, 133508, 134993, 187160, 493095

Offset: 1

Artur Jasinski, Feb 08 2007, Feb 09 2007

If this sequence is infinite then so is A124401.
Equals A127965(n)/2.
The sum has the simple closed form 1 + 2/3*(4^n-1). - Stefan Steinerberger, Nov 24 2007
Terms beyond a(30) correspond to probable primes, cf. A000978. - M. F. Hasler, Aug 29 2008

			a(1)=1 because 1 + 2 = 3 is prime;
a(2)=2 because 1 + 2 + 2^3 = 11 is prime;
a(3)=3 because 1 + 2 + 2^3 + 2^5 = 43 is prime;
a(4)=5 because 1 + 2 + 2^3 + 2^5 + 2^7 + 2^9 = 683 is prime;
...

Cf. A127962, A127963, A127964, A127965, A127961, A000979, A000978, A124400, A126614, A127955, A127956, A127957, A127958, A127936, A127936, A124401, A010051, A007583.

import Data.List (findIndices)
a127936 n = a127936_list !! (n-1)
a127936_list = findIndices ((== 1) . a010051'') a007583_list
-- Reinhard Zumkeller, Mar 24 2013

a = {}; Do[If[PrimeQ[1 + Sum[2^(2n - 1), {n, 1, x}]], AppendTo[a, x]], {x, 1, 1000}]; a
b = {}; Do[c = 1 + Sum[2^(2n - 1), {n, 1, x}]; If[PrimeQ[c], AppendTo[b, c]], {x, 0, 1000}]; a = {}; Do[AppendTo[a, FromDigits[IntegerDigits[b[[x]], 2]]], {x, 1, Length[b]}]; d = {}; Do[AppendTo[d, (1/2)(DigitCount[a[[x]], 10, 0]+DigitCount[a[[x]], 10, 1])], {x, 1, Length[a]}]; d
Position[Accumulate[2^(2*Range[1000]-1)],?(PrimeQ[#+1]&)]//Flatten (* The program generates the first 21 terms of the sequence. To generate more, increase the Range constant. *) (* _Harvey P. Dale, Mar 23 2022 *)

for(n=1,999, ispseudoprime(2^(2*n+1)\3+1) & print1(n",")) \\ M. F. Hasler, Aug 29 2008

from sympy import isprime
A127936 = [i for i in range(1,10**3) if isprime(int('01'*i+'1',2))]
# Chai Wah Wu, Sep 05 2014

a(n) = floor(A000978(n)/2) = ceiling(log(4)(A000979(n))); A000978(n) = 2 a(n) + 1; A000979(n) = (2*4^a(n)+1)/3. - M. F. Hasler, Aug 29 2008

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 11 2007
2 more terms from Stefan Steinerberger, Nov 24 2007
6 more terms from Dmitry Kamenetsky, Jul 12 2008
a(30)-a(40) calculated from A000978 by M. F. Hasler, Aug 29 2008

A136412 a(n) = (5*4^n + 1)/3.

Original entry on oeis.org

2, 7, 27, 107, 427, 1707, 6827, 27307, 109227, 436907, 1747627, 6990507, 27962027, 111848107, 447392427, 1789569707, 7158278827, 28633115307, 114532461227, 458129844907, 1832519379627, 7330077518507, 29320310074027

Offset: 0

Paul Curtz, Mar 31 2008

An Engel expansion of 4/5 to the base b := 4/3 as defined in A181565, with the associated series expansion 4/5 = b/2 + b^2/(2*7) + b^3/(2*7*27) + b^4/(2*7*27*107) + .... Cf. A199115 and A140660. - Peter Bala, Oct 29 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Sequences of the form (m*4^n + 1)/3: A007583 (m=2), this sequence (m=5), A199210 (m=11), A199210 (m=11), A206373 (m=14).

Cf. A007302, A140660, A181565, A199115.

a136412 = (`div` 3) . (+ 1) . (* 5) . (4 ^)
-- Reinhard Zumkeller, Jun 17 2012

[(5*4^n+1)/3: n in [0..30]]; // Vincenzo Librandi, Nov 04 2011

LinearRecurrence[{5,-4}, {2,7}, 31] (* G. C. Greubel, Jan 19 2023 *)

a(n)=(5*4^n+1)/3 \\ Charles R Greathouse IV, Oct 07 2015

[(5*4^n+1)/3 for n in range(31)] # G. C. Greubel, Jan 19 2023

a(n) = 4*a(n-1) - 1.
a(n) = A199115(n)/3.
O.g.f.: (2-3*x)/((1-x)*(1-4*x)). - R. J. Mathar, Apr 04 2008
a(n) = 5*a(n-1) - 4*a(n-2). - Vincenzo Librandi, Nov 04 2011
E.g.f.: (1/3)*(5*exp(4*x) + exp(x)). - G. C. Greubel, Jan 19 2023

Formula in definition and more terms from R. J. Mathar, Apr 04 2008

A083064 Square number array T(n,k) = (k*(k+2)^n+1)/(k+1) read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 11, 14, 1, 1, 5, 19, 43, 41, 1, 1, 6, 29, 94, 171, 122, 1, 1, 7, 41, 173, 469, 683, 365, 1, 1, 8, 55, 286, 1037, 2344, 2731, 1094, 1, 1, 9, 71, 439, 2001, 6221, 11719, 10923, 3281, 1, 1, 10, 89, 638, 3511, 14006, 37325, 58594, 43691, 9842, 1

Offset: 0

Paul Barry, Apr 21 2003

			Rows begin:
1  1   1    1     1      1       1        1         1 ...
1  2   5   14    41    122     365     1094      3281 ...  A007051
1  3  11   43   171    683    2731    10923     43691 ...  A007583
1  4  19   94   469   2344   11719    58594    292969 ...  A083065
1  5  29  173  1037   6221   37325   223949   1343693 ...  A083066
1  6  41  286  2001  14006   98041   686286   4804001 ...  A083067
1  7  55  439  3511  28087  224695  1797559  14380471 ...  A083068
1  8  71  638  5741  51668  465011  4185098  37665881 ...  A187709
1  9  89  889  8889  88889  888889  8888889  88888889 ...  A059482
1 10 109 1198 13177 144946 1594405 17538454 192922993 ...  A199760, etc.
Column 2: A000027;
column 3: A028387;
column 4: A083074;
column 5: A125082;
column 6: A125083.
Diagonals:
1,  2,  11,   94,  1037,  14006, ... A083069;
1,  3,  19,  173,  2001,  28087, ... A083071;
1,  4,  29,  286,  3511,  51668, ... A083072;
1,  5,  41,  439,  5741,  88889, ... A083073;
1,  5,  43,  469,  6221,  98041, ... A083070;
1, 14, 171, 2344, 37325, 686286, ... A191690.
Triangle begins:
1;
1, 1;
1, 2, 1;
1, 3, 5, 1;
1, 4, 11, 14, 1;
1, 5, 19, 43, 41, 1;
1, 6, 29, 94, 171, 122, 1; etc.

Cf. rows: A007051, A007583, A059482, A083065 - A083068, A187709, A199760; columns: A000027, A028387, A083074, A125082, A125083; diagonals: A083069 - A083073, A191690.

Edited by Bruno Berselli, Jun 21 2013

A039301 Number of distinct quadratic residues mod 4^n.

Original entry on oeis.org

1, 2, 4, 12, 44, 172, 684, 2732, 10924, 43692, 174764, 699052, 2796204, 11184812, 44739244, 178956972, 715827884, 2863311532, 11453246124, 45812984492, 183251937964, 733007751852, 2932031007404, 11728124029612, 46912496118444, 187649984473772, 750599937895084

Offset: 0

David W. Wilson

Number of distinct n-digit suffixes of base 4 squares.

Vincenzo Librandi, Table of n, a(n) for n = 0..170
Index entries for linear recurrences with constant coefficients, signature (5,-4).

[Floor((4^n+10)/6): n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011

Floor[(4^Range[0, 30] + 10)/6] (* Paolo Xausa, Jan 29 2025 *)

a(n) = floor((4^n+10)/6).
a(n) = A007583(n-1)+1 = A020988(n-2)+2 = A083584(n-2)+3. - Ralf Stephan, Jun 14 2003
Also, a(0)=1 and, for n>0, a(n) = (4^n+8)/6. - Bruno Berselli, Jul 27 2010
G.f.: (1-3*x-2*x^2)/((1-x)*(1-4*x)). - Bruno Berselli, Jul 27 2010
a(n)-5*a(n-1)+4*a(n-2) = 0 for n>1. - Bruno Berselli, Jul 27 2010

A083065 4th row of number array A083064.

Original entry on oeis.org

1, 4, 19, 94, 469, 2344, 11719, 58594, 292969, 1464844, 7324219, 36621094, 183105469, 915527344, 4577636719, 22888183594, 114440917969, 572204589844, 2861022949219, 14305114746094, 71525573730469, 357627868652344

Offset: 0

Paul Barry, Apr 21 2003

Inverse binomial transform of A090040. [Paul Curtz, Jan 11 2009]
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=7, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)*charpoly(A,2). [Milan Janjic, Feb 21 2010]
For an integer x, consider the sequence P(x) of polynomials p_{1}, p_{2}, p_{3}, . . . defined by p_{1} = x-1, p_{n+1} = x*p_{1} - 1. P(5) = This sequence. P(1), P(2), P(3), P(4) are A000004, A123412, A007051, A007583 resp. [K.V.Iyer, Jun 22 2010]
It appears that if s(n) is a first order rational sequence of the form s(0)=2, s(n)= (3*s(n-1)+2)/(2*s(n-1)+3), n>0, then s(n)=2*a(n)/(2*a(n)-1), n>0.
An Engel expansion of 5/3 to the base b := 5/4 as defined in A181565, with the associated series expansion 5/3 = b + b^2/4 + b^3/(4*19) + b^4/(4*19*94) + .... Cf. A007051. - Peter Bala, Oct 29 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-5).

Cf. A007583, A083066, A007051.

[(3*5^n+1)/4: n in [0..30]]; // Vincenzo Librandi, Nov 04 2011

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]*5-1 od: seq(a[n], n=1..22); # Zerinvary Lajos, Feb 22 2008

CoefficientList[Series[(1-2x)/((1-5x)(1-x)),{x,0,30}],x] (* or *) LinearRecurrence[{6,-5},{1,4},30] (* Harvey P. Dale, Jul 27 2022 *)

a(n) = (3*5^n+1)/4.
G.f.: (1-2*x)/((1-5*x)(1-x)).
E.g.f.: (3*exp(5*x) + exp(x))/4.
a(n) = 5*a(n-1)-1 with n>0, a(0)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = 6*a(n-1)-5*a(n-2). - Vincenzo Librandi, Nov 04 2011
a(n) = 5^n - Sum_{i=0..n-1} 5^i. - Bruno Berselli, Jun 20 2013

A060920 Bisection of Fibonacci triangle A037027: even-indexed members of column sequences of A037027 (not counting leading zeros).

Original entry on oeis.org

1, 2, 1, 5, 5, 1, 13, 20, 9, 1, 34, 71, 51, 14, 1, 89, 235, 233, 105, 20, 1, 233, 744, 942, 594, 190, 27, 1, 610, 2285, 3522, 2860, 1295, 315, 35, 1, 1597, 6865, 12473, 12402, 7285, 2534, 490, 44, 1, 4181, 20284, 42447, 49963, 36122, 16407, 4578, 726, 54, 1

Offset: 0

Wolfdieter Lang, Apr 20 2001

Companion triangle (odd-indexed members) A060921.

			Triangle begins as:
      1;
      2,     1;
      5,     5,      1;
     13,    20,      9,      1;
     34,    71,     51,     14,      1;
     89,   235,    233,    105,     20,     1;
    233,   744,    942,    594,    190,    27,     1;
    610,  2285,   3522,   2860,   1295,   315,    35,    1;
   1597,  6865,  12473,  12402,   7285,  2534,   490,   44,    1;
   4181, 20284,  42447,  49963,  36122, 16407,  4578,  726,   54,  1;
  10946, 59155, 140109, 190570, 163730, 91959, 33705, 7776, 1035, 65, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Yidong Sun, Numerical Triangles and Several Classical Sequences, Fib. Quart. 43, no. 4, Nov. 2005, pp. 359-370.

Column sequences: A001519 (k=0), A054444 (k=1), A061178 (k=2), A061179 (k=3), A061180 (k=4), A061181 (k=5).

Cf. A000045, A001519, A007583, A037027, A060921, A061176.

A060920:= func< n,k | (&+[Binomial(2*n-k-j, j)*Binomial(2*n-k-2*j, k): j in [0..2*n-k]]) >;
[A060920(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 06 2021

A060920[n_, k_]:= Sum[Binomial[2*n-k-j, j]*Binomial[2*n-k-2*j, k], {j,0,2*n-k}];
Table[A060920[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 06 2021 *)

def A060920(n,k): return sum(binomial(2*n-k-j, j)*binomial(2*n-k-2*j, k) for j in (0..2*n-k))
flatten([[A060920(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 06 2021

T(n, k) = A037027(2*n-k, k).
T(n, k) = ((2*(n-k) + 1)*A060921(n-1, k-1) + 4*n*T(n-1, k-1))/(5*k), n >= k >= 1.
T(n, 0) = F(n)^2 + F(n+1)^2 = A001519(n), with the Fibonacci numbers F(n) = A000045(n).
Sum_{k=0..n} T(n, k) = (2^(2*n + 1) + 1)/3 = A007583(n).
G.f. for column m >= 0: x^m*pFe(m+1, x)/(1-3*x+x^2)^(m+1), where pFe(n, x) := Sum_{m=0..n} A061176(n, m)*x^m (row polynomials of signed triangle A061176).
G.f.: (1-x*(1+y))/(1 - (3+2*y)*x + (1+y)^2*x^2). - Vladeta Jovovic, Oct 11 2003

A083584 a(n) = (8*4^n - 5)/3.

Original entry on oeis.org

1, 9, 41, 169, 681, 2729, 10921, 43689, 174761, 699049, 2796201, 11184809, 44739241, 178956969, 715827881, 2863311529, 11453246121, 45812984489, 183251937961, 733007751849, 2932031007401, 11728124029609, 46912496118441

Offset: 0

Paul Barry, May 01 2003

a(n) = A007583(n+1) - 2 = A020988(n) - 1 = A039301(n+2) - 3. - Ralf Stephan, Jun 14 2003

Sum of n-th row of triangle of powers of 4: 1; 4 1 4; 16 4 1 4 16; 64 16 4 1 4 16 64; .... - Philippe Deléham, Feb 24 2014

			a(0) = 1;
a(1) = 4 + 1 + 4 = 9;
a(2) = 16 + 4 + 1 + 4 + 16 = 41;
a(3) = 64 + 16 + 4 + 1 + 4 + 16 + 64 = 169; etc. - _Philippe Deléham_, Feb 24 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..170
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A083855.

[(8*4^n-5)/3: n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011

(8 4^Range[0,30]-5)/3 (* or *) LinearRecurrence[{5,-4},{1,9},30] (* Harvey P. Dale, Oct 23 2011 *)

a(n)=(8*4^n-5)/3 \\ Charles R Greathouse IV, Oct 07 2015

print([8*4**n//3 - 1 for n in range(50)]) # Karl V. Keller, Jr., May 21 2022

a(n) = (8*4^n - 5)/3.
G.f.: (1+4*x)/((1-x)*(1-4*x)).
E.g.f.: (8*exp(4*x) - exp(x))/3.
a(0)=1, a(1)=9, a(n) = 5*a(n-1) - 4*a(n-2). - Harvey P. Dale, Oct 23 2011
a(n) = 4*a(n-1) + 5, a(0) = 1. - Philippe Deléham, Feb 24 2014
a(n+1) = 2^(2^n+1) + a(n), a(1)=1. - Ben Paul Thurston, Dec 27 2015

A299960 a(n) = (4^(2*n+1) + 1) / 5.

Original entry on oeis.org

1, 13, 205, 3277, 52429, 838861, 13421773, 214748365, 3435973837, 54975581389, 879609302221, 14073748835533, 225179981368525, 3602879701896397, 57646075230342349, 922337203685477581, 14757395258967641293, 236118324143482260685, 3777893186295716170957

Offset: 0

M. F. Hasler, Feb 22 2018

nonn

It is easily seen that 4^(2n+1)+1 is divisible by 5 for all n, since 4 = -1 (mod 5). For even powers this does not hold.
The aerated sequence 1, 0, 13, 0, 205, 0, 3277, ... is a linear divisibility sequence of order 4. It is the case P1 = 0, P2 = -5^2, Q = 4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. Cf. A007583, A095372 and A100706. - Peter Bala, Aug 28 2019
Let G be a sequence satisfying G(i) = 2*G(i-1) + G(i-2) - 2*G(i-3) for arbitrary integers i and without regard to the initial values of G. Then a(n) = (G(i)*2^(4*n+2) + G(i+8*n+4))/(5*G(i+4*n+2)) as long as G(i+4*n+2) != 0. - Klaus Purath, Feb 02 2021
Ch. Gerbr asks (personal comm.) whether we can prove that 13 is the only prime in this sequence. We can prove divisibility conditions for many residue classes of the index n (cf. formulas), but have not yet found a complete covering set. - M. F. Hasler, Jan 07 2025

			For n = 0, a(0) = (4^1+1)/5 = 5/5 = 1.
For n = 1, a(1) = (4^3+1)/5 = 65/5 = 13.

H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A299959 for the smallest prime factor.

Cf. A052539, A007583, A095372, A100706.

A299960 := n -> (4^(2*n+1)+1)/5: seq(A299960(n), n=0..20);

LinearRecurrence[{17, -16}, {1, 13}, 20] (* Jean-François Alcover, Feb 22 2018 *)

PARI
```
A299960(n)=4^(2*n+1)\5+1
```

def A299960(n): return ((1<<(n<<2)+2)+1)//5 # Chai Wah Wu, Jul 29 2022

a(n) = A052539(2*n+1)/5 = A015521(2*n+1) = A014985(2*n+1) = A007910(4*n+1) = A007909(4*n+1) = A207262(n+1)/5.
O.g.f.: (1 - 4*x)/(1 - 17*x + 16*x^2). - Peter Bala, Aug 28 2019
a(n) = 17*a(n-1) - 16*a(n-2). - Wesley Ivan Hurt, Oct 02 2020
From Klaus Purath, Feb 02 2021: (Start)
a(n) = (2^(4*n+2)+1)/5.
a(n) = (A061654(n) + A001025(n))/2.
a(n) = A091881(n+1) + 7*A131865(n-1) for n > 0.
(End)
E.g.f.: (exp(x) + 4*exp(16*x))/5. - Stefano Spezia, Feb 02 2021
We have d | a(n) for all n in R, for the following pairs (d, R) of divisors d and residue classes R: (13, 1 + 3Z), (5, 2 + 5Z), (29, 3 + 7Z), (397, 5 + 11Z),
  (53, 6 + 13Z), (137, 8 + 17Z), (229, 9 + 19Z), (277, 11 + 23Z),
  (107367629, 14 + 29Z), (5581, 15 + 31Z), (149, 18 + 27Z), (10169, 20 + 41Z),
  (173, 21 + 43Z), (3761, 23 + 47Z), (15358129, 26 + 53Z), (1181, 29 + 59Z),
  (733, 30 + 61Z), (269, 33 + 67Z), (569, 35 + 71Z),(293, 36 + 73Z), (317, 39 + 79Z),
  (997, 41 + 83Z), (1069, 44 + 89Z), (389, 48 + 97Z), (809, 50 + 101Z),
  (41201, 51 + 103Z), (857, 53 + 107Z), (5669, 54 + 109Z), (58309, 56 + 113Z),
  (509, 63 + 127Z), (269665073, 65 + 131Z), (189061, 68 + 137Z), (557, 69 + 139Z),
  (1789, 74 + 149Z), (653, 81 + 163Z), (9413, 90 + 181Z), (3821, 95 + 191Z),
  (773, 96 + 193Z), (4729, 98 + 197Z), (797, 99 + 199Z), ... - M. F. Hasler, Jan 07 2025

A263132 Positive values of m such that binomial(4*m - 1, m) is odd.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 11, 12, 16, 22, 24, 32, 43, 44, 48, 64, 86, 88, 96, 128, 171, 172, 176, 192, 256, 342, 344, 352, 384, 512, 683, 684, 688, 704, 768, 1024, 1366, 1368, 1376, 1408, 1536, 2048, 2731, 2732, 2736, 2752, 2816, 3072, 4096, 5462, 5464, 5472, 5504

Offset: 1

Peter Bala, Oct 10 2015

This sequence, when viewed as a set, equals the set of numbers of the form 4^n * ceiling(2^k/3) for n >= 0, k >= 1, i.e., the product subset in Z of A000302 and A005578 regarded as sets. See the example below.
Equivalently, this sequence, when viewed as a set, equals the set of numbers of the form 2^n * (2^(2*k + 1) + 1)/3 for n,k >= 0, i.e., the product subset in Z of A000079 and A007583 regarded as sets. See the example below.
2*a(n) gives the values of m such that binomial(4*m - 2,m) is odd. 4*a(n) gives the values of m such that binomial(4*m - 3,m) is odd (other than m = 1) and also the values of m such that binomial(4*m - 4,m) is odd.

			1) Notice how this sequence can be read from Table 1 below by moving through the table in a sequence of 'knight moves' (1 down and 2 to the left) starting from the first row. For example, starting at 11 on the top row we move in a series of knights moves 11 -> 12 -> 16, then return to the top row at 22 and move 22 -> 24 -> 32, return to the top row at 43 and move 43 -> 44 -> 48 -> 64, then return to top row at 86 and so on.
........................................................
.   Table 1: 4^n * ceiling(2^k/3) for n >= 0, k >= 1   .
........................................................
n\k|   1    2    3    4     5     6    7    8     9
---+----------------------------------------------------
0  |   1    2    3    6    11    22   43   86   171 ...
1  |   4    8   12   24    44    88  172  ...
2  |  16   32   48   96   176    ...
3  |  64  128  192  ...
4  | 256  ...
...
2) Notice how this sequence can be read from Table 2 below in a sequence of 'knight moves' (2 down and 1 to the left) starting from the first two rows. For example, starting at 43 in the first row we jump 43 -> 44 -> 48 -> 64, then return to the second row at 86 and jump 86 -> 88 -> 96 -> 128, followed by 171 -> 172 -> 176 -> 192 -> 256, and so on.
....................................................
.   Table 2: 2^n * (2^(2*k + 1) + 1)/3, n,k >= 0   .
....................................................
n\k|   0    1     2     3      4      5
---+----------------------------------------------
0  |   1    3    11    43    171    683  ...
1  |   2    6    22    86    342   1366  ...
2  |   4   12    44   172    684   2732  ...
3  |   8   24    88   344   1368   5464  ...
4  |  16   48   176   688   2736  10928  ...
5  |  32   96   352  1376   5472  21856  ...
6  |  64  192   704  2752  10944  43712  ...
7  | 128  384  1408  5504  21888  87424  ...
8  | 256 ...

Cf. A000079, A000267, A000302, A005578, A007583, A048716, A384688.

Other odd binomials: A002450 (4*m+1,m), A020988 (4*m+2,m), A263133 (4*m+3,m), A080674 (4*m+4,m), A118113 (3*m-2,m), A003714 (3*m,m).

[n: n in [1..6000] | Binomial(4*n-1, n) mod 2 eq 1]; // Vincenzo Librandi, Oct 12 2015

for n from 1 to 5000 do if mod(binomial(4*n-1, n), 2) = 1 then print(n) end if end do;

Select[Range[6000],OddQ[Binomial[4#-1,#]]&] (* Harvey P. Dale, Dec 26 2015 *)

for(n=1, 1e4, if (binomial(4*n-1, n) % 2 == 1, print1(n", "))) \\ Altug Alkan, Oct 11 2015

a(n) = my(r,s=sqrtint(4*n-3,&r)); (1<Kevin Ryde, Jun 14 2025

Python

A263132_list = [m for m in range(1,10**6) if not ~(4*m-1) & m] # Chai Wah Wu, Feb 07 2016

Formula

a(n) = A263133(n) + 1.

m is a term if and only if m AND NOT (4*m-1) = 0 where AND and NOT are bitwise operators. - Chai Wah Wu, Feb 07 2016

a(n) = (2^A000267(n-1) + 2^A384688(n-1)) / 3. - Kevin Ryde, Jun 14 2025

Extensions

More terms from Vincenzo Librandi, Oct 12 2015

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