A209308 -id:A209308 - cp's OEIS frontend

A000265 Remove all factors of 2 from n; or largest odd divisor of n; or odd part of n.

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

Offset: 1

N. J. A. Sloane

When n > 0 is written as k*2^j with k odd then k = A000265(n) and j = A007814(n), so: when n is written as k*2^j - 1 with k odd then k = A000265(n+1) and j = A007814(n+1), when n > 1 is written as k*2^j + 1 with k odd then k = A000265(n-1) and j = A007814(n-1).
Also denominator of 2^n/n (numerator is A075101(n)). - Reinhard Zumkeller, Sep 01 2002
Slope of line connecting (o, a(o)) where o = (2^k)(n-1) + 1 is 2^k and (by design) starts at (1, 1). - Josh Locker (joshlocker(AT)macfora.com), Apr 17 2004
Numerator of n/2^(n-1). - Alexander Adamchuk, Feb 11 2005
From Marco Matosic, Jun 29 2005: (Start)
"The sequence can be arranged in a table:
                          1
                       1  3  1
                    1  5  3  7  1
              1  9  5 11  3 13  7 15  1
  1 17  9 19  5 21 11 23  3 25 13 27  7 29 15 31  1
Every new row is the previous row interspaced with the continuation of the odd numbers.
Except for the ones; the terms (t) in each column are t+t+/-s = t_+1. Starting from the center column of threes and working to the left the values of s are given by A000265 and working to the right by A000265." (End)
This is a fractal sequence. The odd-numbered elements give the odd natural numbers. If these elements are removed, the original sequence is recovered. - Kerry Mitchell, Dec 07 2005
2k + 1 is the k-th and largest of the subsequence of k terms separating two successive equal entries in a(n). - Lekraj Beedassy, Dec 30 2005
It's not difficult to show that the sum of the first 2^n terms is (4^n + 2)/3. - Nick Hobson, Jan 14 2005
In the table, for each row, (sum of terms between 3 and 1) - (sum of terms between 1 and 3) = A020988. - Eric Desbiaux, May 27 2009
This sequence appears in the analysis of A160469 and A156769, which resemble the  numerator and denominator of the Taylor series for tan(x). - Johannes W. Meijer, May 24 2009
Indices n such that a(n) divides 2^n - 1 are listed in A068563. - Max Alekseyev, Aug 25 2013
From Alexander R. Povolotsky, Dec 17 2014: (Start)
With regard to the tabular presentation described in the comment by Marco Matosic: in his drawing, starting with the 3rd row, the first term in the row, which is equal to 1 (or, alternatively the last term in the row, which is also equal to 1), is not in the actual sequence and is added to the drawing as a fictitious term (for the sake of symmetry); an actual A000265(n) could be considered to be a(j,k) (where j >= 1 is the row number and k>=1 is the column subscript), such that a(j,1) = 1:
  1
  1  3
  1  5  3  7
  1  9  5 11  3 13  7 15
  1 17  9 19  5 21 11 23  3 25 13 27  7 29 15 31
and so on ... .
The relationship between k and j for each row is 1 <= k <= 2^(j-1). In this corrected tabular representation, Marco's notion that "every new row is the previous row interspaced with the continuation of the odd numbers" remains true. (End)
Partitions natural numbers to the same equivalence classes as A064989. That is, for all i, j: a(i) = a(j) <=> A064989(i) = A064989(j). There are dozens of other such sequences (like A003602) for which this also holds: In general, all sequences for which a(2n) = a(n) and the odd bisection is injective. - Antti Karttunen, Apr 15 2017
From Paul Curtz, Feb 19 2019: (Start)
This sequence is the truncated triangle:
   1,  1;
   3,  1,  5;
   3,  7,  1,  9;
   5, 11,  3, 13,  7;
  15,  1, 17,  9, 19,  5;
  21, 11, 23,  3, 25, 13, 27;
   7, 29, 15, 31,  1, 33, 17, 35;
  ...
The first column is A069834. The second column is A213671. The main diagonal is A236999. The first upper diagonal is A125650 without 0.
c(n) = ((n*(n+1)/2))/A069834 = 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1, 8, 8, 1, 1, ... for n > 0. n*(n+1)/2 is the rank of A069834. (End)
As well as being multiplicative, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 27 2019
a(n) is also the map n -> A026741(n) applied at least A007814(n) times. - Federico Provvedi, Dec 14 2021

			G.f. = x + x^2 + 3*x^3 + x^4 + 5*x^5 + 3*x^6 + 7*x^7 + x^8 + 9*x^9 + 5*x^10 + 11*x^11 + ...

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 10000 terms from T. D. Noe)
Peter Bala, A note on the sequence of numerators of a rational function
V. Daiev and J. L. Brown, Problem H-81, Fib. Quart., 6 (1968), 52.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Eric Weisstein's World of Mathematics, Odd Part
Eric Weisstein's World of Mathematics, Trigonometry Angles
Eric Weisstein's World of Mathematics, Sphere Line Picking

Cf. A000004, A000010, A000123, A000217, A000225, A000265, A001227.
Cf. A003602, A003961, A006516, A006519, A007814, A008683, A014577.
Cf. A020988, A025480, A026741, A027750, A035263, A038502, A049606.
Cf. A064989, A065330, A068563, A069834, A075101, A111929, A111930.
Cf. A111918, A111919, A111920, A111921, A111922, A111923, A125650.
Cf. A127793, A132739, A132740, A156769, A160469, A182469, A209308.
Cf. A213671, A220466, A236999, A242603.
Cf. A049606 (partial products), A135013 (partial sums), A099545 (mod 4), A326937 (Dirichlet inverse).
Cf. A026741 (map), A001511 (converging steps), A038550 (prime index).
Cf. A195056 (Dgf at s=3).

a000265 = until odd (`div` 2)
-- Reinhard Zumkeller, Jan 08 2013, Apr 08 2011, Oct 14 2010

int A000265(n){
    while(n%2==0) n>>=1;
    return n;
}
/* Aidan Simmons, Feb 24 2019 */

using IntegerSequences
[OddPart(n) for n in 1:77] |> println  # Peter Luschny, Sep 25 2021

A000265:= func< n | n/2^Valuation(n,2) >;
[A000265(n): n in [1..120]]; // G. C. Greubel, Jul 31 2024

A000265:=proc(n) local t1,d; t1:=1; for d from 1 by 2 to n do if n mod d = 0 then t1:=d; fi; od; t1; end: seq(A000265(n), n=1..77);
A000265 := n -> n/2^padic[ordp](n,2): seq(A000265(n), n=1..77); # Peter Luschny, Nov 26 2010

a[n_Integer /; n > 0] := n/2^IntegerExponent[n, 2]; Array[a, 77] (* Josh Locker *)
a[ n_] := If[ n == 0, 0, n / 2^IntegerExponent[ n, 2]]; (* Michael Somos, Dec 17 2014 *)

{a(n) = n >> valuation(n, 2)}; /* Michael Somos, Aug 09 2006, edited by M. F. Hasler, Dec 18 2014 */

from _future_ import division
def A000265(n):
    while not n % 2:
        n //= 2
    return n # Chai Wah Wu, Mar 25 2018

def a(n):
    while not n&1: n >>= 1
    return n
print([a(n) for n in range(1, 78)]) # Michael S. Branicky, Jun 26 2025

def A000265(n): return n//2^valuation(n,2)
[A000265(n) for n in (1..121)] # G. C. Greubel, Jul 31 2024

(define (A000265 n) (let loop ((n n)) (if (odd? n) n (loop (/ n 2))))) ;; Antti Karttunen, Apr 15 2017

a(n) = if n is odd then n, otherwise a(n/2). - Reinhard Zumkeller, Sep 01 2002
a(n) = n/A006519(n) = 2*A025480(n-1) + 1.
Multiplicative with a(p^e) = 1 if p = 2, p^e if p > 2. - David W. Wilson, Aug 01 2001
a(n) = Sum_{d divides n and d is odd} phi(d). - Vladeta Jovovic, Dec 04 2002
G.f.: -x/(1 - x) + Sum_{k>=0} (2*x^(2^k)/(1 - 2*x^(2^(k+1)) + x^(2^(k+2)))). - Ralf Stephan, Sep 05 2003
(a(k), a(2k), a(3k), ...) = a(k)*(a(1), a(2), a(3), ...) In general, a(n*m) = a(n)*a(m). - Josh Locker (jlocker(AT)mail.rochester.edu), Oct 04 2005
a(n) = Sum_{k=0..n} A127793(n,k)*floor((k+2)/2) (conjecture). - Paul Barry, Jan 29 2007
Dirichlet g.f.: zeta(s-1)*(2^s - 2)/(2^s - 1). - Ralf Stephan, Jun 18 2007
a(A132739(n)) = A132739(a(n)) = A132740(n). - Reinhard Zumkeller, Aug 27 2007
a(n) = 2*A003602(n) - 1. - Franklin T. Adams-Watters, Jul 02 2009
a(n) = n/gcd(2^n,n). (This also shows that the true offset is 0 and a(0) = 0.) - Peter Luschny, Nov 14 2009
a(-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2011
From Reinhard Zumkeller, May 01 2012: (Start)
A182469(n, k) = A027750(a(n), k), k = 1..A001227(n).
a(n) = A182469(n, A001227(n)). (End)
a((2*n-1)*2^p) = 2*n - 1, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 05 2013
G.f.: G(0)/(1 - 2*x^2 + x^4) - 1/(1 - x), where G(k) = 1 + 1/(1 - x^(2^k)*(1 - 2*x^(2^(k+1)) + x^(2^(k+2)))/(x^(2^k)*(1 - 2*x^(2^(k+1)) + x^(2^(k+2))) + (1 - 2*x^(2^(k+2)) + x^(2^(k+3)))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 06 2013
a(n) = A003961(A064989(n)). - Antti Karttunen, Apr 15 2017
Completely multiplicative with a(2) = 1 and a(p) = p for prime p > 2, i.e., the sequence b(n) = a(n) * A008683(n) for n > 0 is the Dirichlet inverse of a(n). - Werner Schulte, Jul 08 2018
From Peter Bala, Feb 27 2019: (Start)
O.g.f.: F(x) - F(x^2) - F(x^4) - F(x^8) - ..., where F(x) = x/(1 - x)^2 is the generating function for the positive integers.
O.g.f. for reciprocals: Sum_{n >= 1} x^n/a(n) = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4) + (1/2)*L(x^8) + ..., where L(x) = log(1/(1 - x)).
Sum_{n >= 1} x^n/a(n) =  1/2*log(G(x)), where G(x) = 1 + 2*x + 4*x^2 + 6*x^3 + 10*x^4 + ... is the o.g.f. of A000123. (End)
O.g.f.: Sum_{n >= 1} phi(2*n-1)*x^(2*n-1)/(1 - x^(2*n-1)), where phi(n) is the Euler totient function A000010. - Peter Bala, Mar 22 2019
a(n) = A049606(n) / A049606(n-1). - Flávio V. Fernandes, Dec 08 2020
a(n) = numerator of n/2^(floor(n/2)). - Federico Provvedi, Dec 14 2021
a(n) = Sum_{d divides n} (-1)^(d+1)*phi(2*n/d). - Peter Bala, Jan 14 2024
a(n) = A030101(A030101(n)). - Darío Clavijo, Sep 19 2024

Additional comments from Henry Bottomley, Mar 02 2000
More terms from Larry Reeves (larryr(AT)acm.org), Mar 14 2000
Name clarified by David A. Corneth, Apr 15 2017

A060096 Numerator of coefficients of Euler polynomials (rising powers).

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 1, 0, -3, 1, 0, 1, 0, -2, 1, -1, 0, 5, 0, -5, 1, 0, -3, 0, 5, 0, -3, 1, 17, 0, -21, 0, 35, 0, -7, 1, 0, 17, 0, -28, 0, 14, 0, -4, 1, -31, 0, 153, 0, -63, 0, 21, 0, -9, 1, 0, -155, 0, 255, 0, -126, 0, 30, 0, -5, 1, 691, 0, -1705, 0, 2805, 0, -231, 0, 165, 0, -11, 1, 0, 2073, 0, -3410, 0, 1683, 0, -396

Offset: 0

Wolfdieter Lang, Mar 29 2001

From S. Roman, The Umbral Calculus (see the reference in A048854), p. 101, (4.2.10) (corrected): E(n,x)= sum(sum(binomial(n,m)*((-1/2)^j)*j!*S2(n-m,j),j=0..k)*x^m,m=0..n), with S2(n,m)=A008277(n,m) and S2(n,0)=1 if n=0 else 0 (Stirling2).
From Wolfdieter Lang, Oct 31 2011: (Start)
This is the Sheffer triangle (2/(exp(x)+1),x) (which would be called in the above mentioned S. Roman reference Appell for (exp(t)+1)/2) (see p. 27).
  The e.g.f. for the row sums is 2/(1+exp(-x)). The row sums look like A198631(n)/A006519(n+1), n>=0.
  The e.g.f. for the alternating row sums is 2/(exp(x)*(exp(x)+1)). These sums look like (-1)^n*A143074(n)/ A006519(n+1).
  The e.g.f. for the a-sequence of this Sheffer array is 1. The z-sequence has e.g.f. (1-exp(x))/(2*x). This z-sequence is -1/(2*A000027(n))=-1/(2*(n+1)) (see the link under A006232 for the definition of a- and z-sequences). This leads to the recurrences given below.
The alternating power sums for the first n positive integers are given by sum((-1)^(n-j)*j^k,j=1..n) = (E(k, x=n+1)+(-1)^n*E(k, x=0))/2, k>=1, n>=1,with the row polynomials E(n, x)(see the Abramowitz-Stegun reference, p. 804, 23.1.4, and an addendum in the W. Lang link under A196837).
(End)

			n\m  0    1    2    3    4    5    6    7  8  ...
0:   1
1:  -1    1
2:   0   -1    1
3:   1    0   -3    1
4:   0    1    0   -2    1
5:  -1    0    5    0   -5    1
6:   0   -3    0    5    0   -3    1
7:  17    0  -21    0   35    0   -7    1
8:   0   17    0  -28    0   14    0   -4  1
...
The rational triangle a(n,m)/A060097(n,m) starts
n\m  0    1    2    3    4    5    6    7  8  ...
0:   1
1: -1/2   1
2:   0   -1    1
3:  1/4   0  -3/2   1
4:   0    1    0   -2    1
5: -1/2   0   5/2   0  -5/2   1
6:   0   -3    0    5    0   -3    1
7: 17/8   0 -21/2   0  35/4   0  -7/2   1
8:   0   17    0  -28    0   14    0   -4  1
...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 809.
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 20, equations 20:4:1 - 20:4:8 at pages 177-178.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Cf. A060097.

A060096 := proc(n,m) coeff(euler(n,x),x,m) ; numer(%) ;end proc:
seq(seq(A060096(n,m),m=0..n),n=0..12) ; # R. J. Mathar, Dec 21 2010

Numerator[Flatten[Table[CoefficientList[EulerE[n, x], x], {n, 0, 12}]]] (* Jean-François Alcover, Apr 29 2011 *)

E(n, x)= sum((a(n, m)/b(n, m))*x^m, m=0..n), denominators b(n, m)= A060097(n, m).
From Wolfdieter Lang, Oct 31 2011: (Start)
E.g.f. for E(n, x) is 2*exp(x*z)/(exp(z)+1).
E.g.f. of column no. m, m>=0, is 2*x^{m+1}/(m!*(exp(x)+1)).
Recurrences for E(n,m):=a(n,m)/A060097(n,m) from the Sheffer a-and z-sequence:
E(n,m)=(n/m)*E(n-1,m-1), n>=1,m>=1.
E(n,0)=-n*sum(E(n-1,j)/(2*(j+1)),j=0..n-1), n>=1, E(0,0)=1.
(see the Sheffer comments above).
(End)
E(n,m) = binomial(n,m)*sum(((-1)^j)*j!*S2(n-m,j)/2^j ,j=0..n-m), 0<=m<=n, with S2 given by A008277. From S. Roman, The umbral calculus, reference under A048854, eq. (4.2.10), p. 101, with a=1, and a misprint corrected: replace 1/k! by binomial(n,k) (also in the two preceding formulas). - Wolfdieter Lang, Nov 03 2011
The first (m=0) column of the rational triangle is conjectured to be E(n,0) = ((-1)^n)*A198631(n) / A006519(n+1). See also the first column shown in A209308 (different signs). - Wolfdieter Lang, Jun 15 2015

Table rewritten by Wolfdieter Lang, Oct 31 2011

A213268 Denominators of the Inverse semi-binomial transform of A001477(n) read downwards antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 4, 4, 1, 2, 2, 8, 2, 1, 1, 4, 1, 16, 16, 1, 2, 1, 8, 8, 32, 16, 1, 1, 4, 4, 16, 8, 64, 64, 1, 2, 2, 8, 4, 32, 32, 128, 16, 1, 1, 4, 2, 16, 16, 64, 8, 256, 256, 1, 2, 1, 8, 8, 32, 4, 128, 128, 512, 256, 1, 1, 4, 4, 16, 2, 64, 64, 256, 128, 1024, 1024

Offset: 0

Paul Curtz, Jun 08 2012

Starting from any sequence a(k) in the first row, define the array T(n,k) of the inverse semi-binomial transform  by T(0,k) = a(k), T(n,k) = T(n-1,k+1) -T(n-1,k)/2, n>=1.
Here, where the first row is the nonnegative integers, the array is
0        1      2      3    4      5      6      7     8   =A001477(n)
1       3/2     2    5/2    3    7/2      4    9/2     5   =A026741(n+2)/A000034(n)
1       5/4   3/2    7/4    2    9/4    5/2   11/4     3   =A060819(n+4)/A176895(n)
3/4     7/8     1    9/8  5/4   11/8    3/2   13/8   7/4   =A106609(n+6)/A205383(n+6)
1/2    9/16   5/8  11/16  3/4  13/16    7/8  15/16     1   =A106617(n+8)/TBD
5/16  11/32   3/8  13/32 7/16  15/32    1/2  17/32  9/16
3/16  13/64  7/32  15/64  1/4  17/64   9/32  19/64  5/16
7/64 15/128   1/8 17/128 9/64 19/128   5/32 21/128 11/64
1/16 17/256 9/128 19/256 5/64 21/256 11/128 23/256  3/32.
The first column contains 0, followed by fractions A000265/A084623, that is Oresme numbers n/2^n multiplied by 2 (see A209308).

			The array of denominators starts:
  1   1   1   1   1   1   1   1   1   1   1 ...
  1   2   1   2   1   2   1   2   1   2   1 ...
  1   4   2   4   1   4   2   4   1   4   2 ...
  4   8   1   8   4   8   2   8   4   8   1 ...
  2  16   8  16   4  16   8  16   1  16   8 ...
16  32   8  32  16  32   2  32  16  32   8 ...
16  64  32  64   4  64  32  64  16  64  32 ...
64 128   8 128  64 128  32 128  64 128  16 ...
16 256 128 256  64 256 128 256  32 256 128 ...
256 512 128 512 256 512  64 512 256 512 128 ...
All entries are powers of 2.

A213268frac := proc(n,k)
        if n = 0 then
                return k ;
        else
                return procname(n-1,k+1)-procname(n-1,k)/2 ;
        end if;
end proc:
A213268 := proc(n,k)
        denom(A213268frac(n,k)) ;
end proc: # R. J. Mathar, Jun 30 2012

T[0, k_] := k; T[n_, k_] := T[n, k] = T[n-1, k+1] - T[n-1, k]/2; Table[T[n-k, k] // Denominator, {n, 0, 11}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Sep 12 2014 *)

A255935 Triangle read by rows: a(n) = Pascal's triangle A007318(n) + A197870(n+1).

Original entry on oeis.org

0, 1, 2, 1, 2, 0, 1, 3, 3, 2, 1, 4, 6, 4, 0, 1, 5, 10, 10, 5, 2, 1, 6, 15, 20, 15, 6, 0, 1, 7, 21, 35, 35, 21, 7, 2, 1, 8, 28, 56, 70, 56, 28, 8, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 2, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 0

Offset: 0

Paul Curtz, Mar 11 2015

Consider the difference table of a sequence with A000004(n)=0's as main diagonal. (Example: A000045(n).) We call this sequence an autosequence of the first kind.
Based on Pascal's triangle, a(n) =
0,                   T1
1, 2,
1, 2, 0,
1, 3, 3, 2,
etc.
transforms every sequence s(n) in an autosequence of the first kind via the multiplication by the triangle
s0,                  T2
s0, s1,
s0, s1, s2,
s0, s1, s2, s3,
etc.
Examples.
1) s(n) = A198631(n)/A006519(n+1), the second fractional Euler numbers (see A209308). This yields 0*1, 1*1+2*1/2=2, 1*1+2*1/2+0*0=2, 1*1+3*1/2++3*0+2*(-1/4)=2, ... .
The autosequence is 0 followed by 2's or 2*(0,1,1,1,1,1,1,1,... = b(n)).
b(n), the basic autosequence of the first kind, is not in the OEIS (see A140575 and A054977).
2) s(n) = A164555(n)/A027642(n), the second Bernoulli numbers, yields 0,2,2,3,4,5,6,7,... = A254667(n).
Row sums of T1: A062510(n) = 3*A001045(n).
Antidiagonal sums of T1: A111573(n).
With 0's instead of the spaces, every column, i.e.,
0, 1, 1, 1, 1,  1,  1,  1,  1,  1,   1, ...
0, 2, 2, 3, 4,  5,  6,  7,  8,  9,  10, ... = A001477(n) with 0 instead of 1 = A254667(n)
0, 0, 0, 3, 6, 10, 15, 21, 28, 36,  45, ... = A161680(n) with 0 instead of 1
0, 0, 0, 2, 4, 10, 20, 35, 56, 84, 120, ...
etc., is an autosequence of the first kind.
With T(0,0) = 1, it is (1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, -2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 24 2015

			Triangle starts:
0;
1, 2;
1, 2, 0;
1, 3, 3, 2;
1, 4, 6, 4, 0;
1, 5, 10, 10, 5, 2;
1, 6, 15, 20, 15, 6, 0;
...

Cf. A001045, A001477, A010673, A011973, A054977, A062510, A007318, A197870, A006519, A198631, A140575, A209308, A164555, A027642, A111573, A161680, A254667.

a[n_, k_] := If[k == n, 2*Mod[n, 2], Binomial[n, k]]; Table[a[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 23 2015 *)

a(n) = Pascal's triangle A007318(n) with main diagonal A010673(n) (= period 2: repeat 0, 2) instead of 1's=A000012(n).
a(n) = reversal abs(A140575(n)).
a(n) = A007318(n) + A197870(n+1).
T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = 0, T(1,0) = 1, T(1,1) = 2, T(n,k) = 0 if k>n or if k<0 . - Philippe Deléham, May 24 2015
G.f.: (-1-2*x*y+x^2*y+x^2*y^2)/((x*y+1)*(x*y+x-1)) - 1. - R. J. Mathar, Aug 12 2015

A227577 Square array read by antidiagonals, A(n,k) the numerators of the elements of the difference table of the Euler polynomials evaluated at x=1, for n>=0, k>=0.

Original entry on oeis.org

1, -1, 1, 0, -1, 0, 1, 1, -1, -1, 0, 1, 1, 1, 0, -1, -1, -1, 1, 1, 1, 0, -1, -1, -5, -1, -1, 0, 17, 17, 13, 5, -5, -13, -17, -17, 0, 17, 17, 47, 13, 47, 17, 17, 0, -31, -31, -107, -73, -13, 13, 73, 107, 31, 31, 0, -31, -31, -355

Offset: 0

Paul Curtz, Jul 16 2013

sign

The difference table of the Euler polynomials evaluated at x=1:
    1,   1/2,     0,    -1/4,     0,     1/2,      0,      -17/8, ...
  -1/2, -1/2,   -1/4,    1/4,    1/2,   -1/2,   -17/8,      17/8, ...
    0,   1/4,    1/2,    1/4;    -1,   -13/8,    17/4,     107/8, ...
   1/4,  1/4,   -1/4,   -5/4,   -5/8,   47/8,    73/8,    -355/8, ...
    0,  -1/2,    -1,     5/8    13/2,   13/4,  -107/2,    -655/8, ...
  -1/2, -1/2,   13/8,   47/8,  -13/4, -227/4,  -227/8,    5687/8, ...
    0,  17/8,   17/4,  -73/8, -107/2,  227/8,  2957/4,    2957/8, ...
  17/8, 17/8, -107/8, -355/8,  655/8, 5687/8, -2957/8, -107125/8, ...
To compute the difference table, take
    1,   1/2;
  -1/2;
The next term is always half of the sum of the antidiagonals. Hence (-1/2 + 1/2 = 0)
    1,   1/2,   0;
  -1/2, -1/2;
    0;
The first column (inverse binomial transform) lists the numbers (1, -1/2, 0, 1/4, ..., not in the OEIS; corresponds to A027641/A027642). See A209308 and A060096.
A198631(n)/A006519(n+1) is an autosequence. See A181722.
Note the main diagonal: 1, -1/2, 1/2, -5/4, 13/2, -227/4, 2957/4, -107125/8, .... (See A212196/A181131.)
This twice the first upper diagonal. The autosequence is of the second kind.
From 0, -1, the algorithm gives A226158(n), full Genocchi numbers, autosequence of the first kind.
The difference table of the Bernoulli polynomials evaluated at x=1 is (apart from signs) A085737/A085738 and its analysis by Ludwig Seidel was discussed in the Luschny link. - Peter Luschny, Jul 18 2013

			Read by antidiagonals:
    1;
  -1/2,  1/2;
    0,  -1/2,   0;
   1/4,  1/4, -1/4, -1/4;
    0,   1/4,  1/2,  1/4,   0;
  -1/2, -1/2, -1/4,  1/4,  1/2,  1/2;
    0,  -1/2, - 1,  -5/4,  -1,  -1/2,   0;
  ...
Row sums: 1, 0, -1/2, 0, 1, 0, -17/4, 0, ... = 2*A198631(n+1)/A006519(n+2).
Denominators: 1, 1, 2, 1, 1, 1, 4, 1, ... = A160467(n+2)?

Peter Luschny, The computation and asymptotics of the Bernoulli numbers.
OEIS Wiki, Autosequence

Cf. A164555/A027642 in A190339.

DifferenceTableEulerPolynomials := proc(n) local A,m,k,x;
A := array(0..n,0..n); x := 1;
for m from 0 to n do for k from 0 to n do A[m,k]:= 0 od od;
for m from 0 to n do A[m,0] := euler(m,x);
   for k from m-1 by -1 to 0 do
      A[k,m-k] := A[k+1,m-k-1] - A[k,m-k-1] od od;
LinearAlgebra[Transpose](convert(A, Matrix)) end:
DifferenceTableEulerPolynomials(7);  # Peter Luschny, Jul 18 2013

t[0, 0] = 1; t[0, k_] := EulerE[k, 1]; t[n_, 0] := -t[0, n]; t[n_, k_] := t[n, k] = t[n-1, k+1] - t[n-1, k]; Table[t[n-k, k] // Numerator, {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 18 2013 *)

def DifferenceTableEulerPolynomialsEvaluatedAt1(n) :
    @CachedFunction
    def ep1(n):          # Euler polynomial at x=1
        if n < 2: return 1 - n/2
        s = add(binomial(n,k)*ep1(k) for k in (0..n-1))
        return 1 - s/2
    T = matrix(QQ, n)
    for m in range(n) :  # Compute difference table
        T[m,0] = ep1(m)
        for k in range(m-1,-1,-1) :
            T[k,m-k] = T[k+1,m-k-1] - T[k,m-k-1]
    return T
def A227577_list(m):
    D = DifferenceTableEulerPolynomialsEvaluatedAt1(m)
    return [D[k,n-k].numerator() for n in range(m) for k in (0..n)]
A227577_list(12)  # Peter Luschny, Jul 18 2013

Corrected by Jean-François Alcover, Jul 17 2013

A242971 Alternate n+1, 2^n.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 4, 8, 5, 16, 6, 32, 7, 64, 8, 128, 9, 256, 10, 512, 11, 1024, 12, 2048, 13, 4096, 14, 8192, 15, 16384, 16, 32768, 17, 65536, 18, 131072, 19, 262144, 20, 524288, 21, 1048576, 22, 2097152, 23, 4194304, 24, 8388608, 25, 16777216, 26, 33554432

Offset: 0

Paul Curtz, Jun 22 2014

nonn

The offset 0 is a choice. Another sequence could begin with A001477 instead of A000027. The Akiyama-Tanigawa transform applied to 1/(n+1) and 1/2^n are the second Bernoulli numbers A164555(n)/A027642(n) and the second (fractional) Euler numbers A198631(n)/A006519(n+1). (The first Euler numbers are not in the OEIS). Hence a(n).
a(2n+1) - a(2n) = 2^n -n -1 = 0, 0, 1, 4, 11,... = A000295(n) (Eulerian numbers).
a(2n+1) + a(2n) = 2^n +n +1 = A005126(n).

			G.f. = 1 + x + 2*x^2 + 2*x^3 + 3*x^4 + 4*x^5 + 4*x^6 + 8*x^7 + 5*x^8 + ...

Index entries for linear recurrences with constant coefficients, signature (0,4,0,-5,0,2).

Cf. A000027, A000079, A209308, A227577.

A242971:=n->((n+1) mod 2)*(n/2 + 1) + (n mod 2) * 2^((n-1)/2); seq(A242971(n), n=0..50); # Wesley Ivan Hurt, Jun 29 2014

Table[Mod[n + 1, 2] (n/2 + 1) + Mod[n, 2] 2^((n - 1)/2), {n, 0, 50}] (* Wesley Ivan Hurt, Jun 29 2014 *)

a(n) = ((n+1) mod 2) * (n/2 + 1) + (n mod 2) * 2^((n-1)/2). - Wesley Ivan Hurt, Jun 29 2014

G.f.: (1 + x - x^2) * (1 - x^2 - x^3) / ((1 - x^2)^2 * (1 - 2*x^2)). - Michael Somos, Jun 30 2014

A273692 a(n) is the denominator of 2*O(n+1) - O(n+2) where O(n) = n/2^n, the n-th Oresme number.

Original entry on oeis.org

2, 8, 2, 32, 32, 128, 64, 512, 512, 2048, 128, 8192, 8192, 32768, 16384, 131072, 131072, 524288, 131072, 2097152, 2097152, 8388608, 4194304, 33554432, 33554432, 134217728, 16777216, 536870912, 536870912, 2147483648, 1073741824, 8589934592, 8589934592, 34359738368

Offset: 0

Paul Curtz, May 28 2016

O(n) is the Horadam notation.
O(n) or Oresme(n) = n/2^n = 0, 1/2, 1/2, 3/8, 1/4, ... . The positive Oresme numbers are O(n+1) = A000265(n+1)/A075101(n+1). See A209308. Consider Oco(n) = 2*O(n+1) - O(n+2) = 1/2, 5/8, 1/2, 11/32, 7/32, ... = A075677(n+1)/a(n). (See Coll(n) in A209308.)
Oco(n) = 1/2, 5/8, 1/2, 11/32, 7/32, 17/128, 5/64, 23/512, 13/512, 29/2048, 1/128, 35/8192, 19/8192, ... . Compare to (2+3*n)/2^(n+2).
Differences table of Oco(n):
   1/2,    5/8,    1/2,   11/32,    7/32,  17/128, 5/64, ...
   1/8,   -1/8,   -5/32,  -1/8,   -11/128, -7/128, ...
  -1/4,   -1/32,   1/32,   5/128,   1/32, ...
   7/32,   1/16,   1/128, -1/128, ...
  -5/32,  -7/128, -1/64, ...
  13/128,  5/128, ...
  -1/16, ... .
First column: Io(n) = 1/2 followed by (-1)^n* A067745(n)/(8, 4, 32, 32, ...).
1) Alternated Oco(2n) + Io(2n) and Oco(2n+1) - Io(2n+1) gives 2^n.
2) Alternated Oco(2n) - Io(2n) and Oco(2n+1) + Io(2n+1) gives 3*O(n)/2.
   (1/2 - 1/2 = 0, 5/8 + 1/8 = 3/4, 1/2 + 1/4 = 3/4, 11/32 + 7/32 = 9/16, ...)

M. R. Bacon and C. K. Cook, Some properties of Oresme numbers and convolutions ..., Fib. Q., 62:3 (2024), 233-240.

Seiichi Manyama, Table of n, a(n) for n = 0..3320
A. F. Horadam, Oresme numbers, Fib. Quart., 12 (1974), 267-271.

Cf. A000079, A000265, A000302, A004171 (main diagonal), A006519, A016789, A067745, A075101, A075677, A209308.

Or(n) = n/2^n;
a(n) = denominator(2*Or(n+1) - Or(n+2)); \\ Michel Marcus, May 28 2016

a(n) = denominator of (2+3*n)/2^(n+2).
a(2n+1) = 8*4^n.
a(2n+2) = a(2n+1)/(4, 1, 2, 1, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, ..., shifted A006519?).

More terms from Michel Marcus, May 28 2016

A278331 Shifted sequence of second differences of Genocchi numbers.

Original entry on oeis.org

0, -2, -2, 6, 14, -34, -138, 310, 1918, -4146, -36154, 76454, 891342, -1859138, -27891050, 57641238, 1080832286, -2219305810, -50833628826, 103886563462, 2853207760750, -5810302084962, -188424521441482, 382659344967926, 14464296482284734, -29311252309537394, -1277229462293249018

Offset: 0

Jean-François Alcover and Paul Curtz, Nov 18 2016

sign

This is an autosequence of the first kind (array of successive differences shows typical zero diagonal).
Last digits are apparently of period 20.
From A226158(n) for the continuity of autosequences of the first kind.
b(n) = 0, 1, -1, 0, 1, 0, -3, 0, 17, ... = A226158(n) with 1 as second term instead of -1.
c(n) = 0, 0, -1, 0, 1, 0, -3, 0, 17, ... = A226158(n) with 0 as second term instead of -1.
Respective difference tables:
0, -1, -1,  0,  1,  0, -3,   0,   17, ...
-1, 0,  1,  1, -1, -3 , 3,  17,  -17, ...
1,  1,  0, -2, -2,  6, 14, -34, -138, ...
etc,
0,  1, -1,  0,  1,  0, -3,   0,   17, ... = 0 followed by A036968(n+1)
1, -2,  1,  1, -1, -3,  3,  17,  -17, ...
-3, 3,  0, -2, -2,  6, 14, -34, -138, ...
etc,
0,  0, -1,  0,  1,  0, -3,   0,   17, ...
0, -1,  1,  1, -1, -3,  3,  17,  -17, ...
-1, 2,  0, -2, -2,  6, 14, -34, -138, ...
etc.
Since it is in the three tables, a(n) is the core of the Genocchi numbers.

Eric Weisstein's MathWorld, Genocchi Number.
Wikipedia, Genocchi number

Cf. A001469, A014781, A036968, A005439 (a(n) second and third diagonals), A164555/A027642, A209308, A226158, A240581(n)/A239315(n) (core of Bernoulli numbers).

g[0] = 0; g[1] = -1; g[n_] := n*EulerE[n-1, 0]; G = Table[g[n], {n, 0, 30}]; Drop[Differences[G, 2], 2]
(* or, from Seidel's triangle A014781: *)
max = 26; T[1, 1] = 1; T[n_, k_] /; 1 <= k <= (n + 1)/2 := T[n, k] = If[EvenQ[n], Sum[T[n - 1, i], {i, k, max}], Sum[T[n - 1, i], {i, 1, k}]]; T[, ] = 0; a[n_] := With[{k = Floor[(n - 1)/2] + 1}, (-1)^k*T[n + 3, k]]; Table[a[n], {n, 0, max}]

a(n) = (n+2)*E(n+1, 0) - 2*(n+3)*E(n+2, 0) + (n+4)*E(n+3, 0), where E(n,x) is the n-th Euler polynomial.

a(n) = -2*(2^(n+2)-1)*B(n+2) + 4*(2^(n+3)-1)*B(n+3) - 2*(2^(n+4)-1)*B(n+4), where B(n) is the n-th Bernoulli number.

A281500 Reduced denominators of f(n) = (n+1)/(2^(2+n)-2) with A026741(n+1) as numerators.

Original entry on oeis.org

2, 3, 14, 15, 62, 63, 254, 255, 1022, 1023, 4094, 4095, 16382, 16383, 65534, 65535, 262142, 262143, 1048574, 1048575, 4194302, 4194303, 16777214, 16777215, 67108862, 67108863, 268435454, 268435455, 1073741822, 1073741823, 4294967294, 4294967295, 17179869182, 17179869183

Offset: 0

Paul Curtz, Jan 23 2017

f(n) = (n+1)/A000918(n+2) = 1/2, 2/6, 3/14, 4/30, 5/62, 6/126, 7/254, 8/510, 9/1022, 10/2046, 11/4094, 12/8190, ... .
Partial reduction: 1/2, 1/3, 3/14, 2/15, 5/62, 3/63, 7/254, 4/255, 9/1022, 5/1023, 11/4094, 6/4095, ... = A026741(n+1)/a(n).
Full reduction: 1/2, 1/3, 3/14, 2/15, 5/62, 1/21, 7/254, ... = A111701(n+1)/(2, 3, 14, 15, 62, 21, ... )
A164555(n+1)/A027642(n) = 1/2, 1/6, 0, -1/30, 0, 1/42, ... = f(n) * A198631(n)/A006519(n+1) = 1, 1/2, 0, -1/4, 0, 1/2, ... .).
Via f(n), we go from the second fractional Euler numbers to the second Bernoulli numbers.
a(n) mod 10: periodic sequence of length 4: repeat [2, 3, 4, 5].
a(n) differences table:
.  2,   3, 14,  15,  62,   63, 254,  255, ...
.  1,  11,  1,  47,   1,  191,   1,  767, ... see A198693
. 10, -10, 46, -46, 190, -190, 766, -766, ... see A096045, from Bernoulli(2n).
Extension of a(n): a(-2) = -1, a(-1) = 0.

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

Cf. A000027, A000918, A001477, A006519, A026741, A027642, A096045, A111701, A117856, A164555, A198631, A198693, A209308, A267921.

a[n_] := (3+(-1)^n)*(2^(n+1)-1)/2; (* or *) a[n_] := If[EvenQ[n], 4^(n/2+1)-2, 4^((n+1)/2)-1]; Table[a[n], {n, 0, 35}] (* Jean-François Alcover, Jan 24 2017 *)

Vec((2 + 3*x + 4*x^2) / ((1 - x)*(1 + x)*(1 - 2*x)*(1 + 2*x)) + O(x^50)) \\ Colin Barker, Jan 24 2017

From Colin Barker, Jan 24 2017: (Start)
G.f.: (2 + 3*x + 4*x^2) / ((1 - x)*(1 + x)*(1 - 2*x)*(1 + 2*x)).
a(n) = 5*a(n-2) - 4*a(n-4) for n>3. (End)
From Jean-François Alcover, Jan 24 2017: (Start)
a(n) = (3 + (-1)^n)*(2^(n + 1) - 1)/2.
a(n) = 4^((n + 1 + ((n + 1) mod 2))/2) - 1 - ((n + 1) mod 2). (End)
a(n) = a(n-2) + A117856(n+1) for n>1.
a(2*k) = 4^(k + 1) - 2, a(2*k+1) = a(2*k) + 1 = 4^(k+1) - 1.
a(2*k) + a(2*k+1) = A267921(k+1).

cp's OEIS Frontend

A000265 Remove all factors of 2 from n; or largest odd divisor of n; or odd part of n.

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A060096 Numerator of coefficients of Euler polynomials (rising powers).

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A213268 Denominators of the Inverse semi-binomial transform of A001477(n) read downwards antidiagonals.

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A255935 Triangle read by rows: a(n) = Pascal's triangle A007318(n) + A197870(n+1).

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A227577 Square array read by antidiagonals, A(n,k) the numerators of the elements of the difference table of the Euler polynomials evaluated at x=1, for n>=0, k>=0.

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A242971 Alternate n+1, 2^n.

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