A109613 -id:A109613 - cp's OEIS frontend

A244587 a(n) = n OR 5.

5, 5, 7, 7, 5, 5, 7, 7, 13, 13, 15, 15, 13, 13, 15, 15, 21, 21, 23, 23, 21, 21, 23, 23, 29, 29, 31, 31, 29, 29, 31, 31, 37, 37, 39, 39, 37, 37, 39, 39, 45, 45, 47, 47, 45, 45, 47, 47, 53, 53, 55, 55, 53, 53, 55, 55, 61, 61, 63, 63, 61, 61, 63, 63, 69, 69, 71, 71

Offset: 0

Gary Detlefs, Jun 30 2014

Cf. similar sequences of the type n OR k: A109613 (k=1), A174091 (k=2), A244584 (k=3), A244586 (k=4), this sequence (k=5), A244588 (k=6).

[BitwiseOr(n, 5): n in [0..80]]; // Bruno Berselli, Jul 01 2014

Maple
```
with(Bits): seq(Or(n,5), n = 0..60);
```

a(n) = bitor(n, 5); \\ Michel Marcus, Jan 19 2023

def A244587(n): return n|5 # Chai Wah Wu, Jan 18 2023

a(n) = (n+5) - (n AND 5).
a(n) = (n XOR 5) + (n AND 5).
a(n) = n +((n+1) mod 2)+4*floor(((n+4) mod 8)/4).
a(n) = (2*n + 4*(-1)^floor(n/4) + (-1)^n + 5)/2. - Bruno Berselli, Jul 01 2014
G.f.: (x^6+x^4-3*x^2+5)/((x+1)*(x^4+1)*(x-1)^2). - Alois P. Heinz, Jul 01 2014

Some terms corrected by Bruno Berselli, Jul 01 2014

A295609 a(n) = least prime number p such that p AND n = n (where AND denotes the binary AND operator).

Original entry on oeis.org

2, 3, 2, 3, 5, 5, 7, 7, 11, 11, 11, 11, 13, 13, 31, 31, 17, 17, 19, 19, 23, 23, 23, 23, 29, 29, 31, 31, 29, 29, 31, 31, 37, 37, 43, 43, 37, 37, 47, 47, 41, 41, 43, 43, 47, 47, 47, 47, 53, 53, 59, 59, 53, 53, 127, 127, 59, 59, 59, 59, 61, 61, 127, 127, 67, 67

Offset: 0

Rémy Sigrist, Nov 24 2017

For any n > 0: gcd(A109613(n), A062383(n)) = 1, hence, by Dirichlet's theorem on arithmetic progressions, we have a prime number, say p, of the form A109613(n) + k * A062383(n) with k > 0; this prime number satisfies p AND n = n; also a(0) = 2, hence the sequence is well defined for any n >= 0.
a(n) = n iff n is prime.
Each prime number appears 2*k times in this sequence for some k > 0.

			a(42) = 42 + A295335(42) = 42 + 1 = 43.

Rémy Sigrist, Table of n, a(n) for n = 0..8192
Rémy Sigrist, Scatterplot of the first 2^17 terms

Cf. A062383, A109613, A295335.

Table[Block[{p = 2}, While[BitAnd[p, n] != n, p = NextPrime@ p]; p], {n, 0, 65}] (* Michael De Vlieger, Nov 26 2017 *)

avoid(n,i) = if (i, if (n%2, 2*avoid(n\2,i), 2*avoid(n\2,i\2)+(i%2)), 0) \\ (i+1)-th number k such that k AND n = 0
a(n) = for (i=0, oo, my (k=avoid(n,i)); if (isprime(n+k), return (n+k)))

a(n) = n + A295335(n).

For any k > 1, a(2*k) = a(2*k+1).

A372601 The maximal exponent in the prime factorization of the largest exponentially odd divisor of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1

Offset: 1

Amiram Eldar, May 07 2024

First differs from A331273 at n = 64.

Differs from A363332 at n = 1, 216, 432, 648, 864, 1000, ... .

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A051903, A109613, A350390, A368711.
Cf. A331273, A363332.
Similar sequences: A007424, A368781, A372602, A372603, A372604.

f[n_] := n - If[EvenQ[n], 1, 0]; a[n_] := f[Max[FactorInteger[n][[;; , 2]]]]; a[1] = 0; Array[a, 100]

s(n) = (n+1) \ 2 * 2 - 1;
a(n) = if(n>1, s(vecmax(factor(n)[,2])), 0);

a(n) = A051903(A350390(n)).
a(n) = A109613(A051903(n)-1) for n >= 2.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + 2 * Sum_{i>=1} (1 - (1/zeta(2*i+1))) = 1.42929441950714075659... .

A168276 a(n) = 2*n - (-1)^n - 1.

Original entry on oeis.org

2, 2, 6, 6, 10, 10, 14, 14, 18, 18, 22, 22, 26, 26, 30, 30, 34, 34, 38, 38, 42, 42, 46, 46, 50, 50, 54, 54, 58, 58, 62, 62, 66, 66, 70, 70, 74, 74, 78, 78, 82, 82, 86, 86, 90, 90, 94, 94, 98, 98, 102, 102, 106, 106, 110, 110, 114, 114, 118, 118, 122, 122, 126, 126, 130, 130

Offset: 1

Vincenzo Librandi, Nov 22 2009

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A039722, A168277.

Cf. A063210. - R. J. Mathar, Nov 25 2009

[2*n-1-(-1)^n: n in [1..70]]; // Vincenzo Librandi, Sep 16 2013

CoefficientList[Series[2 (1 + x^2) / ((1 + x) (1 - x)^2), {x, 0, 80}], x] (* Vincenzo Librandi, Sep 16 2013 *)
Table[2 n - 1 - (-1)^n, {n, 70}] (* Bruno Berselli, Sep 17 2013 *)
LinearRecurrence[{1,1,-1},{2,2,6},70] (* Harvey P. Dale, Oct 22 2014 *)

a(n) = 4*n - a(n-1) - 4, with n>1, a(1)=2.
from R. J. Mathar, Nov 25 2009: (Start)
a(n) = 2*n - (-1)^n - 1.
a(n) = 2*A109613(n-1).
G.f.: 2*x*(1 + x^2)/((1+x)*(1-x)^2). (End)
a(n) = a(n-1) + a(n-2) - a(n-3). - Vincenzo Librandi, Sep 16 2013
a(n) = A168277(n) + 1. - Vincenzo Librandi, Sep 17 2013
E.g.f.: (-1 + 2*exp(x) + (2*x -1)*exp(2*x))*exp(-x). - G. C. Greubel, Jul 16 2016
Sum_{n>=1} 1/a(n)^2 = Pi^2/16. - Amiram Eldar, Aug 21 2022

Previous definition replaced with closed-form expression by Bruno Berselli, Sep 17 2013

A168329 a(n) = (3/2)(2n - (-1)^n - 1).

Original entry on oeis.org

3, 3, 9, 9, 15, 15, 21, 21, 27, 27, 33, 33, 39, 39, 45, 45, 51, 51, 57, 57, 63, 63, 69, 69, 75, 75, 81, 81, 87, 87, 93, 93, 99, 99, 105, 105, 111, 111, 117, 117, 123, 123, 129, 129, 135, 135, 141, 141, 147, 147, 153, 153, 159, 159, 165, 165, 171, 171, 177, 177, 183, 183

Offset: 1

Vincenzo Librandi, Nov 23 2009

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A016945, A109613, A198392.

[(3/2)*(2*n-(-1)^n-1): n in [1..70]]; // Vincenzo Librandi, Nov 15 2011

LinearRecurrence[{1,1,-1},{3,3,9},80 ] (* Vincenzo Librandi, Nov 15 2011 *)
Table[(3/2) (2 n - (-1)^n - 1), {n, 70}] (* Bruno Berselli, Sep 17 2013 *)

a(n) = 6*n - a(n-1) - 6 for n>1, a(1)=3.
G.f.: 3*x*(1 + x^2)/((1+x)*(1-x)^2). - Bruno Berselli, Nov 06 2011
a(n) = -a(-n+1) = 3*A109613(n-1) = A198392(n-1) - A198392(-n). - Bruno Berselli, Nov 06 2011 - Sep 17 2013
E.g.f.: (3/2)*(-1 + 2*exp(x) + (2*x - 1)*exp(2*x))*exp(-x). - G. C. Greubel, Jul 18 2016

New definition by Bruno Berselli, Sep 17 2013

A181878 Coefficient array for square of Chebyshev S-polynomials.

Original entry on oeis.org

1, 1, 1, -2, 1, 4, -4, 1, 1, -6, 11, -6, 1, 9, -24, 22, -8, 1, 1, -12, 46, -62, 37, -10, 1, 16, -80, 148, -128, 56, -12, 1, 1, -20, 130, -314, 367, -230, 79, -14, 1, 25, -200, 610, -920, 771, -376, 106, -16, 1, 1, -30, 295, -1106, 2083, -2232, 1444, -574, 137, -18, 1, 36, -420, 1897, -4352, 5776, -4744, 2486, -832, 172, -20, 1, 1, -42, 581, -3108, 8518, -13672, 13820, -9142, 4013, -1158, 211, -22, 1

Offset: 0

Wolfdieter Lang, Dec 22 2010

For the coefficients of Chebyshev polynomials S(n,x) see A049310.
The row length sequence for this array is A109613 = {1,1,3,3,5,5,...}.
The row polynomials (in x^2) for even row numbers are
  S(2*k,x)^2 = Sum_{m=0..2*k} a(2*k,m)*x^(2*m), k >= 0.
For odd row numbers the row polynomials (in x^2) are
  (S(2*k+1,x)^2)/x^2 = Sum_{m=0..2*k} a(2*k+1,m)*x^(2*m), k >= 0.
The o.g.f. for the polynomials S(n,x)^2 is
  S(x,z):=((1+z)/(1-z))/(1 + (2-x^2)z +z^2). See the link for a proof. Therefore the coefficients constitute the Riordan array (1/(1-x^2),x/(1+x)^2) found as A158454.
The o.g.f. for S(2*k,sqrt(x))^2 is
  (1-2*(1-x)*z+z^2)/((1-z)*(1 - (2-4*x+x^2)*z + z^2)).
The o.g.f. for (S(2*k+1,sqrt(x))^2)/x is
  ((1+z)/(1-z))/(1 - (2-4*x+x^2)*z + z^2).
The row sums A011655(n+1) are the same as those for the triangle A158454.
The alternating row sums for even numbered rows (-1)^n*A007598(n+1) coincide with those of triangle A158454. For odd row numbers n=2k+1 these sums are A049684(k+1), k >= 0 (squares of even-indexed Fibonacci numbers).

			The irregular triangle a(n,m) begins:
  n\m  0    1    2      3     4      5     6    7   8   9  10 ...
  0:   1
  1:   1
  2:   1   -2    1
  3:   4   -4    1
  4:   1   -6   11     -6     1
  5:   9  -24   22     -8     1
  6:   1  -12   46    -62    37    -10     1
  7:  16  -80  148   -128    56    -12     1
  8:   1  -20  130   -314   367   -230    79  -14   1
  9:  25 -200  610   -920   771   -376   106  -16   1
  10:  1  -30  295  -1106  2083  -2232  1444 -574 137 -18   1
  ... Reformatted and extended by _Wolfdieter Lang_, Nov 24 2012

Wolfdieter Lang, First ten rows with more details and proofs.

Cf. A158454, A129818.

Join[{{1}, {1}}, CoefficientList[Table[ChebyshevU[n, Sqrt[x]/2]^2, {n, 2, 10}], x]] // Flatten (* Eric W. Weisstein, Apr 04 2018 *)
Join[{{1}, {1}}, CoefficientList[ChebyshevU[Range[2, 10], Sqrt[x]/2]^2, x]]  // Flatten (* Eric W. Weisstein, Apr 04 2018 *)

a(2*k,m) = (-1)^m*Sum_{j=0..k} binomial(2*k+m-1-2*j, 2*m-1), k >= 0.
a(2*k+1,m) = (-1)^m*Sum_{j=0..k} binomial(2*k+1+m-2*j, 2*m+1), k >= 0.
This derives from the formula for the entries of the Riordan array A158454.
For the o.g.f.s see the comment.

Corrected by Wolfdieter Lang, Jan 21 2011

A215495 a(4n) = a(4n+2) = a(2n+1) = 2n + 1.

Original entry on oeis.org

1, 1, 1, 3, 3, 5, 3, 7, 5, 9, 5, 11, 7, 13, 7, 15, 9, 17, 9, 19, 11, 21, 11, 23, 13, 25, 13, 27, 15, 29, 15, 31, 17, 33, 17, 35, 19, 37, 19, 39, 21, 41, 21, 43, 23, 45, 23, 47, 25, 49, 25, 51, 27, 53, 27, 55, 29, 57, 29, 59, 31, 61, 31, 63, 33, 65, 33, 67, 35, 69, 35, 71, 37, 73, 37, 75, 39, 77, 39, 79, 41, 81, 41, 83, 43, 85, 43

Offset: 0

Paul Curtz, Aug 13 2012

A214282(n) and -A214283(n) are companions. Separately or together, they have many links with the Catalan's numbers A000108(n). Examples:
A214282(n+1) - 2*A214282(n) = -1, -1,  1,  0, -2, -5,  5,   0, -14, -42,  42,    0, -132, ....
2*A214283(n) - A214283(n+1) =  1,  0, -1, -2,  2,  0, -5, -14,  14,   0, -42, -132, 132, ....
A214282(n) + A214283(n) = 1, 0, -1, 0, 2, 0, -5, 0, 14, 0, -42,... (A126120).
The companion to a(n) is b(n) = -A214283(n)/(1,1,1,1,2,2,5,5,...) = 0, 1, 2, 3, 2, 5, 4, 7, 4, 9, 6, ....
a(n) - b(n) = A056594(n).
Discovered as a(n) = A214282(n+1)/A000108([n/2]). See abs(A129996(n-2)).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1,0,-1).

Cf. A130823, A090192.

I:=[1,1,1,3,3,5]; [n le 6 select I[n] else Self(n-2) +Self(n-4) -Self(n-6): n in [1..30]]; // G. C. Greubel, Apr 23 2018

a[n_?EvenQ] := n/2 + Boole[Mod[n, 4] == 0]; a[n_?OddQ] := n; Table[a[n], {n, 0, 86}] (* Jean-François Alcover, Aug 14 2012 *)
LinearRecurrence[{0,1,0,1,0,-1}, {1,1,1,3,3,5}, 50] (* G. C. Greubel, Apr 23 2018 *)

x='x+O('x^30); Vec(( 1+x+2*x^3+x^4+x^5 )/( (x^2+1)*(x-1)^2*(1+x)^2 )) \\ G. C. Greubel, Apr 23 2018

a(n+3) = (A185048(n+3)=2,2,4,2,... ) + 1.
a(n+2) - a(n) = 0, 2, 2, 2. (Period 4).
a(n) = 2*a(n-4) - a(n-8).
a(2*n) = A109613(n).
a(n+1) - a(n) = 2* (-1)^n * A059169(n).
G.f. : ( 1+x+2*x^3+x^4+x^5 ) / ( (x^2+1)*(x-1)^2*(1+x)^2 ). - Jean-François Alcover, Aug 14 2012

A230584 Either two less than a square or two more than a square.

Original entry on oeis.org

2, 3, 6, 7, 11, 14, 18, 23, 27, 34, 38, 47, 51, 62, 66, 79, 83, 98, 102, 119, 123, 142, 146, 167, 171, 194, 198, 223, 227, 254, 258, 287, 291, 322, 326, 359, 363, 398, 402, 439, 443, 482, 486, 527, 531, 574, 578, 623, 627, 674, 678, 727, 731, 782, 786, 839, 843, 898, 902, 959, 963

Offset: 1

Ralf Stephan, Oct 24 2013

Numbers n such that the polynomial x^4 - n*x^2 + 1 is reducible.
The corresponding factorizations are (x^2 + k*x - 1)*(x^2 - k*x - 1) == x^4 - (k^2 + 2)*x^2 + 1 and (x^2 + k*x + 1)*(x^2 - k*x + 1) == x^4 - (k^2 - 2)*x^2 + 1. - Joerg Arndt, Feb 07 2015
Union of A008865 and A059100.
For k > 1: a(2*k+1) - a(2*k) = 4 and a(2*k) - a(2*k-1) = k - 1; for n > 4: a(n) - a(n-2) = 2*floor(n/2) + 1 = A109613(n). - Reinhard Zumkeller, Feb 10 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A008865, A059100, A000290, A109613.

import Data.List (transpose)
a230584 n = a230584_list !! (n-1)
a230584_list = 2 : 3 : concat
               (transpose [drop 2 a059100_list, drop 2 a008865_list])
-- Reinhard Zumkeller, Feb 10 2015

PARI
```
is(n)=issquare(n-2)||issquare(n+2)
```

A230584_vec(N)=Vec((2+x-x^2-x^3+2*x^5-x^6)/((1-x)^3*(1+x)^2)+O(x^N)) \\ M. F. Hasler, Oct 26 2013

From Colin Barker, Oct 24 2013: (Start)
  a(n) = (5-13*(-1)^n+2*(3+(-1)^n)*n+2*n^2)/8 for n>2.
  a(n) = (n^2+4*n-4)/4 for n>2 and even.
  a(n) = (n^2+2*n+9)/4 for n>2 and odd.
  a(n) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5) for n>7.
  G.f.: x*(x^6-2*x^5+x^3+x^2-x-2) / ((x-1)^3*(x+1)^2). (End)
After the first two terms 0^2+2 = 2^2-2, 1^2+2, the squares are sufficiently spaced to ensure that the sequence continues 2^2+2, 3^2-2, 3^2+2, 4^2-2, 4^2+2,..., i.e., a(2n-1) = n^2+2, a(2n)=(n+1)^2-2. - M. F. Hasler, Oct 26 2013

A238303 Triangle T(n,k), 0<=k<=n, read by rows given by T(n,0) = 1, T(n,k) = 2 if k>0.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 0

Philippe Deléham, Feb 24 2014

Row sums are A005408(n).
Diagonals sums are A109613(n).
Sum_{k=0..n} T(n,k)*x^k = A033999(n), A000012(n), A005408(n), A036563(n+2), A058481(n+1), A083584(n), A137410(n), A233325(n), A233326(n), A233328(n), A211866(n+1), A165402(n+1) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*x^(n-k) = A151575(n), A000012(n), A040000(n), A005408(n), A033484(n), A048473(n), A020989(n), A057651(n), A061801(n), A238275(n), A238276(n), A138894(n), A090843(n), A199023(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively.
Sum_{k=0..n} T(n,k)^x = A000027(n+1), A005408(n), A016813(n), A017077(n) for x = 0, 1, 2, 3 respectively.
Sum_{k=0..n} k*T(n,k) = A002378(n).
Sum_{k=0..n} A000045(k)*T(n,k) = A019274(n+2).
Sum_{k=0..n} A000142(k)*T(n,k) = A066237(n+1).

			Triangle begins:
1;
1, 2;
1, 2, 2;
1, 2, 2, 2;
1, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2;
1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2;
...

Antti Karttunen, Table of n, a(n) for n = 0..22154; the first 210 rows of triangle

Cf. Diagonals: A040000.
Cf. Columns: A000012, A007395.
First differences of A001614.

A238303(n) = (2-ispolygonal(n,3)); \\ Antti Karttunen, Jan 19 2025

T(n,0) = A000012(n) = 1, T(n+k,k) = A007395(n) = 2 for k>0.

Data section extended to a(104) by Antti Karttunen, Jan 19 2025

A289296 a(n) = (n - 1)(2floor(n/2) + 1).

Original entry on oeis.org

-1, 0, 3, 6, 15, 20, 35, 42, 63, 72, 99, 110, 143, 156, 195, 210, 255, 272, 323, 342, 399, 420, 483, 506, 575, 600, 675, 702, 783, 812, 899, 930, 1023, 1056, 1155, 1190, 1295, 1332, 1443, 1482, 1599, 1640, 1763, 1806, 1935, 1980, 2115, 2162, 2303, 2352, 2499, 2550, 2703, 2756, 2915

Offset: 0

Jean-François Alcover and Paul Curtz, Jul 02 2017

Summing a(n) by pairs, one gets -1, 9, 35, 77, 135, ... = A033566.

A198442(k) is a member of this sequence if k == 0 or 1 (mod 4). - Bruno Berselli, Jul 04 2017

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Subsequence of A214297.

Cf. A023443, A033566, A093178, A109613.

Table[(n - 1) (2 Floor[n/2] + 1), {n, 0, 60}] (* or *) LinearRecurrence[{1, 2, -2, -1, 1}, {-1, 0, 3, 6, 15}, 61]

a(n)=(n\2*2+1)*(n-1) \\ Charles R Greathouse IV, Jul 02 2017

Vec(-(1 - x - 5*x^2 - x^3 - 2*x^4) / ((1 - x)^3*(1 + x)^2) + O(x^60)) \\ Colin Barker, Jul 02 2017

def A289296(n): return (n-1)*(n|1) # Chai Wah Wu, Jan 18 2023

a(n) = A023443(n) * A109613(n).
a(n) = n^2-1 if n is even and n^2-n if n is odd.
n^2 - a(n) = A093178(n).
From Colin Barker, Jul 02 2017: (Start)
G.f.: -(1 - x - 5*x^2 - x^3 - 2*x^4) / ((1 - x)^3*(1 + x)^2).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>4. (End)

cp's OEIS Frontend

A244587 a(n) = n OR 5.

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A295609 a(n) = least prime number p such that p AND n = n (where AND denotes the binary AND operator).

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A372601 The maximal exponent in the prime factorization of the largest exponentially odd divisor of n.

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A168276 a(n) = 2*n - (-1)^n - 1.

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Magma

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A168329 a(n) = (3/2)*(2*n - (-1)^n - 1).

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Magma

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A181878 Coefficient array for square of Chebyshev S-polynomials.

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A215495 a(4*n) = a(4*n+2) = a(2*n+1) = 2*n + 1.

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Magma

Mathematica

A168329 a(n) = (3/2)(2n - (-1)^n - 1).

A215495 a(4n) = a(4n+2) = a(2n+1) = 2n + 1.

A289296 a(n) = (n - 1)(2floor(n/2) + 1).