A171977 -id:A171977 - cp's OEIS frontend

A224270 Absolute values of the numerators of the third column of ( 0 followed by (interleave 0 , A001803(n))/A060818(n) ) and its successive differences.

1, 1, 5, 11, 95, 203, 861, 1815, 30459, 63635, 264979, 550069, 4555915, 9412543, 38816525, 79898895, 2627302995, 5392044675, 22104436695, 45256266825, 370241638305, 756514878405, 3088866211275, 6300861570705, 102746354288175, 209286947903319

Offset: 0

Paul Curtz, Apr 02 2013

The array is
  0,          0,      1,     0,   3/2,     0,  15/8, 0,...
  0,          1,     -1,   3/2,  -3/2,  15/8, -15/8,...
  1,         -2,    5/2,    -3,  27/8, -15/4,...
  -3,       9/2,  -11/2,  51/8, -57/8,...
  15/2,     -10,   95/8, -27/2,...
  -35/2,  175/8, -203/8,...
  315/8, -189/4,...
  -693/8,...
Note A001803 in the first column and a variant of A206771(n) in the second column.
Now consider a(n)/A046161(n) and its differences:
  1,            1/2,     5/8,  11/16,  95/128, 203/256, 861/1024,...
  -1/2,         1/8,    1/16,  7/128,  13/256, 49/1024,...   =b(n)/A046161(n)
  5/8,        -1/16,  -1/128, -1/256, -3/1024,...
  -11/16,     7/128,   1/256, 1/1024,...
  95/128,   -13/256, -3/1024,...
  -203/256, 49/1024,...
  861/1024,...
This an autosequence of second kind. The first column is the signed sequence.
(Its companion, the corresponding autosequence of first kind, is 0, 1, 1, 9/8, 5/4,... in A206771).
Main diagonal: 1, 1/8, -1/128,... = A002596(n)/A061549(n) ?
b(n) = a(n+1) - A171977*a(n). Also for two successive rows (with shifted A171977).

			a(n)=numerators of 0+1=1, 0+1/2=1/2, 1/4+3/8=5/8, 3/8+5/16=11/16, 15/32+35/128=95/128,... .

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A001790, A001803, A002596, A046161, A061549, A098597, A171977, A206771.

nmax = 25; t1 = Table[ Numerator[ (2*n+1)*(Binomial[2*n, n]/4^n)] / Denominator[ Binomial[2*n, n]/4^n], {n, 0, Ceiling[nmax/2]}]; t2 = Join[{0}, Table[ If[ OddQ[n], 0, t1[[n/2]] ], {n, 1, nmax+2}] ]; t3 = Table[ Differences[t2, n], {n, 0, nmax}]; t3[[All, 3]] // Numerator // Abs (* Jean-François Alcover, Apr 02 2013 *)

Numerators of (0, 0 followed by A001803(n)/(4*A046161(n))) + A001790(n)/A046161(n).

More terms from Jean-François Alcover, Apr 02 2013

A292791 The numerator of the real part of E(2n-1, i), where E(n, x) is the Euler polynomial.

Original entry on oeis.org

-1, 7, -11, 199, -361, 8017, -63311, 10775663, -37120861, 2572609327, -54738555011, 11225458402189, -170606509547761, 24269619087650437, -998364772178081111, 1505193846304099711711, -10065529459831250937061, 2427246234079407797537347, -163790353311268893725697611

Offset: 1

Robert G. Wilson v, Sep 23 2017

sign

The imaginary part is +-i.
The denominators are powers of two; A171977(n) = 2^A001511(n).
For E(2n, i) see A292792.
a(4n) == +-1 (mod 6),
a(4n+1) == 5 (mod 6),
a(4n+2) == 1 (mod 6),
a(4n+3) == 1 (mod 6).
Inspired by A291897.

			a(3) = -11 since E(5, i) = -11/2 + i.

Robert G. Wilson v, Table of n, a(n) for n = 1..276

Cf. A292792.

f[n_] := Numerator[ EulerE[2n -1, I] - I^(2n -1)]; Array[f, 19]

A349102 Increase the odd part of n to the next greater odd number.

Original entry on oeis.org

3, 6, 5, 12, 7, 10, 9, 24, 11, 14, 13, 20, 15, 18, 17, 48, 19, 22, 21, 28, 23, 26, 25, 40, 27, 30, 29, 36, 31, 34, 33, 96, 35, 38, 37, 44, 39, 42, 41, 56, 43, 46, 45, 52, 47, 50, 49, 80, 51, 54, 53, 60, 55, 58, 57, 72, 59, 62, 61, 68, 63, 66, 65, 192, 67, 70, 69

Offset: 1

Kevin Ryde, Mar 26 2022

This is +2 at the bit position of the odd part of n, that being the least significant 1-bit.
The least significant run of 1-bits changes from 0111..111 in n to 1000..001 in a(n).
Arrays A054582 and A135764 arrange terms into rows having the same number of trailing 0 bits. a(n) is the term to the right of n, i.e., next in its row.

			n    = 3448 = binary 1101 0111 1 000
a(n) = 3464 = binary 1101 1000 1 000

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A000265 (odd part), A171977 (2 at odd part), A285326.

Arrays: A054582, A135764.

Array[# + 2^(IntegerExponent[#, 2] + 1) &, 67] (* Michael De Vlieger, Mar 27 2022 *)

PARI
```
a(n) = n + 2<
				
```

a(n) = n + A171977(n).
a(2*n) = 2*a(n).
a(2*n+1) = 2*n + 3.

A360156 a(n) is the sum of the even unitary divisors of 2*n.

Original entry on oeis.org

2, 4, 8, 8, 12, 16, 16, 16, 20, 24, 24, 32, 28, 32, 48, 32, 36, 40, 40, 48, 64, 48, 48, 64, 52, 56, 56, 64, 60, 96, 64, 64, 96, 72, 96, 80, 76, 80, 112, 96, 84, 128, 88, 96, 120, 96, 96, 128, 100, 104, 144, 112, 108, 112, 144, 128, 160, 120, 120, 192, 124, 128

Offset: 1

Amiram Eldar, Jan 28 2023

a(n) is the unitary analog of A146076(2*n).

Amiram Eldar, Table of n, a(n) for n = 1..10000
Octavio A. Agustín-Aquino, Wang-Sun Formula in GL(Z/2kZ), INTEGERS, Vol. 23 (2023), #A37; arXiv preprint, arXiv:2207.14495 [math.NT], 2022.

Cf. A002117, A034448, A100008, A146076, A171977, A192066, A328258.

usigma[n_] := Times @@ (1 + Power @@@ FactorInteger[n]); usigma[1] = 1; a[n_] := Module[{e = IntegerExponent[n, 2]}, 2^(e + 1) * usigma[n/2^e]]; Array[a, 100]

usigma(n) = {my(f = factor(n)); prod(i=1, #f~, f[i, 1]^f[i, 2] + 1)} ;
a(n) = {my(e = valuation(n, 2)); (1 << (e+1)) * usigma(n >> e); }

a(n) = Sum_{even d|(2*n), gcd(d, 2*n/d)=1} d.
a(n) = A034448(2*n) - A192066(2*n).
a(n) = A192066(2*n) - A328258(2*n).
a(n) = A171977(n) * A192066(n).
Sum_{k=1..n} a(k) ~ Pi^2 * n^2 / (7*zeta(3)).
Dirichlet g.f. of b(n): (zeta(s)*zeta(s-1)/zeta(2*s-1))*(2^(s+1)-2)/(2^(2*s)-2), where b(n) is the sum of the even unitary divisors of n: b(n) = a(n/2) if n is even and 0 otherwise.

A377336 Square array read by antidiagonals: T(n,k) is the number of fully symmetric, k-celled, n-dimensional polyhypercubes; n >= 0, k >= 1.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Pontus von Brömssen, Oct 25 2024

To be included, a polyhypercube should have all the 2^n*n! symmetries of the n-dimensional hypercube.

Let n >= 1 and m = A171977(n). Then T(n,k) = 0 if neither k nor k-1 is a multiple of m. Also, there exists a number K such that T(n,k) > 0 if k >= K and either k or k-1 is a multiple of m. In particular, T(n,k) > 0 for all sufficiently large k if and only if n is odd. Sketch of proof: Assume that the point of rotation of the symmetries is in the origin and that the center of each cell have integer or half-integer coordinates, depending on whether the point of rotation is in the center of a cell or at the common corner of 2^n cells. The number of cells that are equivalent to a given cell c is n!/(x_0!*x_1!*...)*2^(n-x_0), where x_1, x_2, ... are the frequencies of the absolute values of the nonzero coordinates of c and x_0 is the number of zero coordinates of c. It can be proved that this number is divisible by m unless c is the cell at the origin (in which case x_0 = n and there are no other equivalent cells). (It is sufficient to check the case where all nonzero coordinates have the same absolute value, i.e., that all numbers except 1 in the n-th row of A013609 are divisible by m; the other numbers are multiples of these.) Since either none or all of the cells equivalent to a given cell must be part of the polyhypercube, this proves the first part. For the second part, say that a cell where the absolute values of all coordinates are equal and nonzero is a corner cell, and that a cell with a single nonzero coordinate is a spike cell. Corner cells and spike cells come in sets of 2^n and 2*n equivalent cells, respectively, and the GCD of 2^n and 2*n is already equal to m. Assume that n >= 3 (the case n <= 2 is easily handled), that k >= (4*n-1)^n, and that either k or k-1 is a multiple of m. Start with a solid cube made up of (4*n-1)^n cells. Remove the central cell if k is even, so that the number of remaining cells is congruent to k (mod m). Since GCD(2^n,2*n) = m, we can remove at most 2*n-1 sets of 2^n equivalent corner cells each, until the number of remaining cells is congruent to k (mod 2*n). The resulting polyhypercube is still connected. Then add sets of 2*n spike cells until the total number of cells is equal to k. This proves the second part. The bound k >= (4*n-1)^n resulting from this construction is far from optimal.

			Array begins:
  n\k| 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
  ---+-----------------------------------------------------------
   0 | 1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
   1 | 1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1
   2 | 1  0  0  1  1  0  0  1  2  0  0  3  2  0  0  5  4  0  0 12
   3 | 1  0  0  0  0  0  1  1  0  0  0  0  1  0  0  0  0  1  2  1
   4 | 1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  1  1  0  0  0
   5 | 1  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0
   6 | 1  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0
   7 | 1  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0
   8 | 1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0
   9 | 1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0
  10 | 1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0

Pontus von Brömssen, Table of n, a(n) for n = 0..5049 (first 100 antidiagonals)
Index entries for sequences related to polyominoes.

Cf. A013609, A142886 (2nd row), A171977, A330891, A376791 (3rd row).

A185342 Triangle of successive recurrences in columns of A117317(n).

Original entry on oeis.org

2, 4, -4, 6, -12, 8, 8, -24, 32, -16, 10, -40, 80, -80, 32, 12, -60, 160, -240, 192, -64, 14, -84, 280, -560, 672, -448, 128, 16, -112, 448, -1120, 1792, -1792, 1024, -256, 18, -144, 672, -2016, 4032, -5376, 4608, -2304, 512, 20, -180, 960, -3360, 8064

Offset: 0

Paul Curtz, Jan 26 2012

A117317 (A):
1
2    1
4    5   1
8   16   9   1
16  44  41  14   1
32 112 146  85  20  1
64 272 456 377 155 27 1
have for their columns successive signatures
(2) (4,-4) (6,-12,8) (8,-24, 32, -16) (10,-40,80,-80,32) i.e. a(n).
Take based on abs(A133156) (B):
1
2   0
4   1  0
8   4  0 0
16 12  1 0 0
32 32  6 0 0 0
64 80 24 1 0 0 0.
The recurrences of successive columns are also a(n). a(n) columns: A005843(n+1), A046092(n+1), A130809, A130810, A130811, A130812, A130813.
A053220 + A001787 = A014480.

			Triangle T(n,k),for 1<=k<=n, begins :
2                                         (1)
4    -4                                   (2)
6   -12   8                               (3)
8   -24  32   -16                         (4)
10  -40  80   -80   32                    (5)
12  -60 160  -240  192   -64              (6)
14  -84 280  -560  672  -448  128         (7)
16 -112 448 -1120 1792 -1792 1024 -256    (8)
Successive rows can be divided by A171977.

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. For (A): A053220, A056243. For (B): A000079, A001787, A001788, A001789. For A193862: A115068 (a Coxeter group). For (2): A014480 (also (3),(4),(5),..); also A053220 and A001787.

Cf. A007318.

Table[(-1)*Binomial[n, k]*(-2)^k, {n, 1, 20}, {k, 1, n}] // Flatten (* G. C. Greubel, Jun 27 2017 *)

for(n=1,20, for(k=1,n, print1((-2)^(k+1)*binomial(n,k)/2, ", "))) \\ G. C. Greubel, Jun 27 2017

Take A133156(n) without 1's or -1's double triangle (C)=
2
4
8      -4
16    -12
32    -32   6
64    -80  24
128  -192  80   -8
256  -448 240  -40
512 -1024 672 -160 10;
a(n) is increasing odd diagonals and increasing (sign changed) even diagonals. Rows sum of (C) = A201629 (?) Another link between Chebyshev polynomials and cos( ).
Absolute values: A013609(n) without 1's. Also 2*A193862 = (2*A002260)*A135278.
T(n,k) = T(n-1,k) - 2*T(n-1,k-1) for k>1, T(n,1) = 2*n = 2*T(n-1,1) - T(n-2,1). - Philippe Deléham, Feb 11 2012
T(n,k) = (-1)* Binomial(n,k)*(-2)^k, 1<=k<=n. - Philippe Deléham, Feb 11 2012

A237425 Denominators of A164555(n)/A027642(n) + A198631(n)/A006519(n+1).

Original entry on oeis.org

1, 1, 6, 4, 30, 2, 42, 8, 30, 2, 66, 4, 2730, 2, 6, 16, 510, 2, 798, 4, 330, 2, 138, 8, 2730, 2, 6, 4, 870, 2, 14322, 32, 510, 2, 6, 4, 1919190, 2, 6, 8, 13530, 2, 1806, 4, 690, 2, 282, 16, 46410, 2, 66, 4, 1590, 2, 798, 8

Offset: 0

Paul Curtz, Feb 07 2014

nonn

An autosequence is a sequence which has its inverse binomial transform equal to the signed sequence. There are two possibilities. For the first kind, the main diagonal is 0's=A000004, the first two following diagonals being the same (generally not A000004). Integers example: A000045(n).
For the second kind, the main diagonal is the double of the following diagonal. Example: the companion to A000045(n) is A000032(n)=2, 1, 3, ... .
A000032(n)/2 is also a possibility. Here a(n) is the denominator of the sum of two autosequences of second kind involving (fractional) Euler and Bernoulli numbers. The corresponding fractional sequence is also an autosequence of the second kind: 2, 1, 1/6, -1/4, -1/30, 1/2, 1/42, -17/8, -1/30, 31/2, 5/66, -691/4, -691/2730,... . It could be divided by 2.

Cf. 1/n in A003506, A194767, A218289, A168516/A168426, A224270/A046161.

a[n_] := BernoulliB[n] + EulerE[n, 1]/2^IntegerExponent[n, 2]; a[0] = 2; a[1] = 1; Table[a[n] // Denominator, {n, 0, 55}] (* Jean-François Alcover, Feb 11 2014 *)

a(2n) = A002445(n). a(2n+2) = A171977(n+2).

A371402 a(n) = gcd(2n, 4^n)^(2n + 1) mod (2^(2*n + 1) - 1)^2.

Original entry on oeis.org

0, 8, 63, 128, 1534, 2048, 16383, 32768, 524285, 524288, 4194303, 8388608, 100663294, 134217728, 1073741823, 2147483648, 42949672956, 34359738368, 274877906943, 549755813888, 6597069766654, 8796093022208, 70368744177663, 140737488355328, 2251799813685245

Offset: 0

Peter Luschny, Mar 26 2024

nonn

Cf. A004171, A001511, A013709, A085058.

Cf. A171977, A089080, A123725.

a := n -> modp(igcd(2*n, 4^n)^(2*n + 1), (2^(2*n + 1) - 1)^2):
seq(a(n), n = 0..19);

a(n) = lift(Mod(gcd(2*n, 4^n),(2^(2*n + 1) - 1)^2)^(2*n + 1)); \\ Michel Marcus, Mar 27 2024

def A371402(n): return ((~n & n-1).bit_length()+2<<(n<<1) if n&1 else ((m:=(~n & n-1).bit_length())+1<<(n<<1)+1)-m) if n else 0 # Chai Wah Wu, Mar 27 2024

def v2(n): return valuation(2*n, 2)
def a(n):
    if n == 0: return 0
    return 4^n*(v2(n) + 1) if n % 2 else 2*4^n*v2(n) - v2(n//2)
print([a(n) for n in range(0, 25)])

a(2*n) = 2*4^(2*n)*A001511(2*n) - A001511(n) for n >= 1.

a(2*n+1) = 4^(2*n + 1)*(A001511(2*n + 1) + 1) for n >= 1.

A292531 a(n) = 0 if n is a power of 2. Otherwise, product of 2 numbers nearest n that have more 2's in their prime factorization than n.

Original entry on oeis.org

0, 0, 8, 0, 24, 32, 48, 0, 80, 96, 120, 128, 168, 192, 224, 0, 288, 320, 360, 384, 440, 480, 528, 512, 624, 672, 728, 768, 840, 896, 960, 0, 1088, 1152, 1224, 1280, 1368, 1440, 1520, 1536, 1680, 1760, 1848, 1920, 2024, 2112, 2208, 2048, 2400, 2496, 2600, 2688

Offset: 1

J. Lowell, Sep 18 2017

1) For all odd n, a(n) = n^2 - 1.
2) All numbers in sequence are divisible by 8.
3) a(n) is not divisible by 16 if and only if n = 8k+3 or n = 8k+5.
Proposition 2) is true. Proof: 0 mod 2 = 0, so the conjecture trivially holds when n is a power of 2. For n not a power of 2, a(n) has by definition the repeated prime factor 2^2 and so is divisible by 8 when a(n) > 4. - Felix Fröhlich, Sep 19 2017

			a(40) = 1536 because 40 has three 2's in its prime factorization, and the closest integers to 40 that have at least four 2's are 32 and 48, and 32 times 48 = 1536.

Cf. A000290.

a[n_] := Block[{p = 2 2^IntegerExponent[n, 2]}, Floor[n/p] Ceiling[n/p] p^2]; Array[a, 60] (* Giovanni Resta, Sep 19 2017 *)

a(n) = p^2 * ceiling(n/p) * floor(n/p), where p = A171977(n). - Giovanni Resta, Sep 19 2017

More terms from Giovanni Resta, Sep 19 2017

cp's OEIS Frontend

A224270 Absolute values of the numerators of the third column of ( 0 followed by (interleave 0 , A001803(n))/A060818(n) ) and its successive differences.

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A292791 The numerator of the real part of E(2n-1, i), where E(n, x) is the Euler polynomial.

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A349102 Increase the odd part of n to the next greater odd number.

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A360156 a(n) is the sum of the even unitary divisors of 2*n.

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A377336 Square array read by antidiagonals: T(n,k) is the number of fully symmetric, k-celled, n-dimensional polyhypercubes; n >= 0, k >= 1.

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A185342 Triangle of successive recurrences in columns of A117317(n).

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A237425 Denominators of A164555(n)/A027642(n) + A198631(n)/A006519(n+1).

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A371402 a(n) = gcd(2*n, 4^n)^(2*n + 1) mod (2^(2*n + 1) - 1)^2.

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A371402 a(n) = gcd(2n, 4^n)^(2n + 1) mod (2^(2*n + 1) - 1)^2.