A000079 -id:A000079 - cp's OEIS frontend

A095362 Ratio A095110(n)/A000079(n-2) rounded to nearest integer.

Original entry on oeis.org

2, 6, 12, 20, 38, 53, 106, 200, 369, 657, 1227, 2191, 4268, 8076, 15291, 28932, 55125, 105983, 202482, 391506

Offset: 1

Antti Karttunen, Jun 12 2004

nonn

See Comments on A095361, where the same ratios are given rounded down.

Cf. A000079, A095110.

a(1) = 2, a(n) = round(A095110(n)/A000079(n-2)) for n > 1.

A162459 A002321*A000079.

Original entry on oeis.org

1, 0, -4, -8, -32, -32, -128, -256, -512, -512, -2048, -4096, -12288, -16384, -16384, -32768, -131072, -262144, -786432, -1572864, -2097152, -2097152, -8388608, -16777216, -33554432, -33554432, -67108864, -134217728, -536870912, -1610612736, -4294967296

Offset: 1

Mats Granvik, Jul 04 2009

sign

Appears to be the determinant of n X n (-1,1) matrix defined by A(i,j)=1 if j=1 or i divides j else -1.

Appears also to be the determinant of n X n (-i/j,i/j) matrix defined by A(i,j)=i/j if j=1 or i divides j else -i/j.

Cf. A002321, A000079.

from functools import lru_cache
@lru_cache(maxsize=None)
def A162459(n):
    if n == 0:
        return 0
    c, j = n, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*A162459(k1)//2**(k1-1)
        j, k1 = j2, n//j2
    return 2**(n-1)*(j-c) # Chai Wah Wu, Mar 30 2021

a(n) = 2^(n-1)*A002321(n). - Chai Wah Wu, Mar 30 2021

A171559 Powers of 2 (cf. A000079) with 1 replaced by 3.

Original entry on oeis.org

3, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 4294967296, 8589934592

Offset: 0

Paul Curtz, Dec 11 2009

First differences of A171449.

Index entries for linear recurrences with constant coefficients, signature (2).

a(n+1)-a(n) = A171449(n).
a(n) + A171449(n) = A000079(n+1) = 2*A000079(n).
G.f.: (3-4*x)/(1-2*x).

Edited by R. J. Mathar and N. J. A. Sloane, Dec 15 2009

A180338 Triangle by rows, A033452 convolved with A000079.

Original entry on oeis.org

1, 2, 5, 4, 10, 22, 8, 20, 44, 99, 16, 40, 88, 198, 471, 32, 80, 176, 396, 942, 2386, 64, 160, 352, 792, 1884, 4772, 12867, 128, 324, 704, 1584, 3768, 9544, 25734, 73681

Offset: 0

Gary W. Adamson, Aug 28 2010

Row sums = A058681: (1, 7, 36, 171, 813, 4012, 20891,...).

			First few rows of the triangle =
1;
2, 5;
4, 10, 22;
8, 20, 44, 99;
16, 40, 88, 198, 471;
32, 80, 176, 396, 942, 2386;
64, 160, 352, 792, 1884, 4772, 12867;
128, 324, 704, 1584, 3768, 9544, 25734, 73681;
...

Cf. A033452, A058681

Antidiagonals of a multiplication table generated from A033452 * powers of 2.

A181175 The "Fi1" sums of the powers-of-2 triangle A000079.

Original entry on oeis.org

1, 2, 40, 320, 21504, 688128, 178257920, 22817013760, 23433341566976, 11997870882291712, 49179307930885816320, 100719222642454151823360, 1650485975228872480767606784, 13520781109074923362448234774528

Offset: 0

Johannes W. Meijer, Oct 10 2010

The a(n) represent the Fi1(n) sums, see A180662, of the powers-of-2 triangle A000079. We observe that Fi2(2*n) = Fi1(2*n) and Fi2(2*n+1) = 2*Fi1(2*n+1).

nmax:=13: a(0):=1: a(1):=2: for n from 0 to nmax/2 do a(2*n+1):= 2^(2*n+1)*a(2*n): a(2*n+2):=2^(2*n+2)*((4^(n+2) - 1)/(4^(n+1) - 1))*a(2*n+1): od: seq(a(n),n=0..nmax);

a(2*n+1) = 2^(2*n+1)*a(2*n) and a(2*n+2) = 2^(2*n+2)*((4^(n+2) - 1)/(4^(n+1) - 1))*a(2*n+1) with a(0) = 1 and a(1) =2.

A334638 Three-column array pPT read by rows: subsequence of primitive Pythagorean triples (x, y, z) with x = A153893^2 - A000079^2, y = 2A153893A000079, z = A153893^2 + A000079^2, ordered by increasing z.

Original entry on oeis.org

3, 4, 5, 21, 20, 29, 105, 88, 137, 465, 368, 593, 1953, 1504, 2465, 8001, 6080, 10049, 32385, 24448, 40577, 130305, 98048, 163073, 522753, 392704, 653825, 2094081, 1571840, 2618369, 8382465, 6289408, 10479617, 33542145, 25161728, 41930753, 134193153, 100655104, 167747585, 536821761, 402636800, 671039489, 2147385345, 1610579968, 2684256257

Offset: 0

Ralf Steiner, May 07 2020

Let [h21] = {{1, 3}, {0, 2}} be the matrix [h_2]*[h_1] in Firstov's notation, from eqs. (24) and (39). Then primitive Pythagorean triples (pPT) (x(n), y(n), z(n)) = (u(n)^2 - v(n)^2, 2*u(n)*v(n), u(n)^2 + v(n)^2), with u(n) and v(n) of different parity, gcd(u(n), v(n)) = 1, and u(n) > v(n) > 0, are generated by (u(n), v(n))^T = [h21]^n*(2,1)^T (T for transpose).
For n > 0: (x(n), y(n), z(n)) = (1, 0, 1) (mod 4). Thus some z are Pythagorean primes (A002144).
The triples converge to the proportion (4:3:5) with:
lim_{n->infinity} x(n)/y(n) = 4/3, lim_{n->infinity} y(n)/z(n) = 3/5.
Altitude h(n) = x(n)*y(n)/z(n) is an irreducible fraction because of primitivity.
From Wolfdieter Lang, Jun 13 2020: (Start)
[h21]^n = sqrt(2)^n*(S(n, 3/sqrt(2))*[1_3] + S(n-1, 3/sqrt(2))*(1/sqrt(2))*([h21] - 3*[1_3])) with the Chebyshev S polynomials (A049310).
u(n) = sqrt(2)^n*(2*S(n, 3/sqrt(2)) - (1/sqrt(2))*S(n-1, 3/sqrt(2)))
     = A153893(n),
v(n) = sqrt(2)^n*(S(n, 3/sqrt(2)) - (1/sqrt(2))*S(n-1, 3/sqrt(2)))
     = A000079(n). Proof from the recurrence, using the Cayley-Hamilton theorem.
With the monic Chebyshev T polynomials, called R in A127672:
x(n)/3 = 2^(n+1)*(R(2*(n+1), 3/sqrt(2)) - (sqrt(2)/3)*R(2*n+1,3/sqrt(2)) - 1) = A171477(n),
y(n)/4 = 3*2^(n-1)*(sqrt(2)*R(2*n+1,3/sqrt(2)) - R(2*n,3/sqrt(2)) - 1/3)
= A010036(n),
z(n) = 3*2^(n+1)*((3/sqrt(2))*R(2*n+1, 3/sqrt(2)) - (4/3)*R(2*n,3/sqrt(2)) - 1).
Using 2^n*Rnx(2*n, 3/sqrt(2)) = A052539(n) = 2^(2*n) + 1, and
  2^(n)*(sqrt(2)/3)*Rnx(2*n+1, 3/sqrt(2)) = A007583(n) = (2^(2*n + 1) + 1)/3,
produces the explicit formulas given by the author in the formula section.
G.f.s for {x(n)} G0(x) = 3/((1 - 4*x)*(1 - 2*x)*(1 - x)), for {y(n)} G1(x) =  4*(1-x)/((1 - 4*x)*(1 - 2*x)), and for {z(n)} = (5 - 6*x + 4*x^2)/((1 - 4*x)*(1 - 2*x)*(1 - x)). This produces the g.f. for the array, read as sequence {a(n)}: G(x) = G0(x^3) + x*G1(x^3) + x^2*G2(x^3) given in the formula section by Colin Barker.
(End)

			The three-column array pPT(n,k) begins:
n\k        0        1         2
-------------------------------
0:         3        4         5
1:        21       20        29
2:       105       88       137
3:       465      368       593
4:      1953     1504      2465
5:      8001     6080     10049
6:     32385    24448     40577
7:    130305    98048    163073
8:    522753   392704    653825
9:   2094081  1571840   2618369
10:  8382465  6289408  10479617
... - _Wolfdieter Lang_, Jun 13 2020

V. E. Firstov, A Special Matrix Transformation Semigroup of Primitive Pairs and the Genealogy of Pythagorean Triples; Mathematical Notes, volume 84, number 2, August 2008, pages 263-279; Link of the page (for the Russian article).
Ralf Steiner, Weitere spezielle Folge primitiver pythagoraischer Dreiecke, ResearchGate.
Wikipedia, Tree of primitive Pythagorean triples.
Index entries for linear recurrences with constant coefficients, signature (0,0,7,0,0,-14,0,0,8).

Cf. A000079, A010035, A049310, A083329, A093357, A010036, A134057, A153893, A171477.

h21={{1, 3}, {0, 2}}; l = {}; Do[v = MatrixPower[h21, n, {2, 1}]; p = v[[1]]; q = v[[2]];
a = p^2 - q^2; b = 2 p q; c = p^2 + q^2; l = AppendTo[l, {a, b, c}], {n, 0, 14}]; l // Flatten

Vec((3 + 4*x + 5*x^2 - 8*x^4 - 6*x^5 + 4*x^7 + 4*x^8) / ((1 - x)*(1 + x + x^2)*(1 - 2*x^3)*(1 - 4*x^3)) + O(x^35)) \\ Colin Barker, Jun 12 2020

The three-column array PT(n, k) is for k = 0, 1, 2:  x(n), y(n), z(n), for n >= 0, with
x(n) = a(3*n + 0) = A153893(n)^2 - A000079(n)^2 = 1 - 3*2^(n+1) + 2^(2*n+3) = binomial(2^(n+2) - 1, 2) = 3*A171477(n),
y(n) = a(3*n + 1) = 2*A153893(n)*A000079(n) = 2^(n+1)*(-1 + 3*2^n) = 4*A010036(n),
z(n) = a(3*n + 2) = A153893(n)^2 + A000079(n)^2 = 1 - 6*2^n + 10*2^(2*n).
From Colin Barker, May 08 2020: (Start)
G.f. (read as sequence {a(n)}): (3 + 4*x + 5*x^2 - 8*x^4 - 6*x^5 + 4*x^7 + 4*x^8) / ((1 - x)*(1 + x + x^2)*(1 - 2*x^3)*(1 - 4*x^3)).
a(n) = 7*a(n-3) - 14*a(n-6) + 8*a(n-9), for n > 8.
(End)

Edited, and corrected proportion by Wolfdieter Lang, Jun 13 2020

Minor grammatical edits. - N. J. A. Sloane, Sep 12 2020

A340660 A000079 is the first row. For the second row, subtract A001045. For the third row, subtract A001045 from the second one, etc. The resulting array is read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 1, 4, 1, 0, 3, 8, 1, -1, 2, 5, 16, 1, -2, 1, 2, 11, 32, 1, -3, 0, -1, 6, 21, 64, 1, -4, -1, -4, 1, 10, 43, 128, 1, -5, -2, -7, -4, -1, 22, 85, 256, 1, -6, -3, -10, -9, -12, 1, 42, 171, 512, 1, -7, -4, -13, -14, -23, -20, -1, 86, 341, 1024

Offset: 0

Paul Curtz, Jan 15 2021

Every row has the signature (1,2).
(Among consequences: a(n) read by antidiagonals is
1,
1,  2,
1,  1, 4,
1,  0, 3,  8,
1, -1, 2,  5, 16
1, -2, 1,  2, 11, 32,
1, -3, 0, -1,  6, 21, 64,
... .
The row sums and their first two difference table terms are
1, 3, 6, 12, 23, 45, 88, ... = A086445(n+1) - 1
2, 3, 6, 11, 22, 43, 86, ... = A005578(n+2)
1, 3, 5, 11, 21, 43, 85, ... = A001045(n+2).
The antidiagonal sums are
b(n) = 1, 1, 3, 2, 5, 3, 9, 4, 15, 5, 27, 6, 49, 7, ... .)

			Square array:
1,  2,  4,   8,  16,  32,  64,  128, ... = A000079(n)
1,  1,  3,   5,  11,  21,  43,   85, ... = A001045(n+1)
1,  0,  2,   2,   6,  10,  22,   42, ... = A078008(n)
1, -1,  1,  -1,   1,  -1,   1,   -1, ... = A033999(n)
1, -2,  0,  -4,  -4, -12, -20,  -44, ... = -A084247(n)
1, -3, -1,  -7,  -9, -23, -41,  -87, ... = (-1)^n*A140966(n+1)
1, -4, -2, -10, -14, -34, -62, -130, ... = -A135440(n)
1, -5, -3, -13, -19, -45, -83, -173, ... = -A155980(n+3) or -A171382(n+1)
...

Cf. A000079, A001045, A033999, A078008, A084247, A135440, A140966, A155980, A171382.

Cf. A005578, A086445.

A:= (n, k)-> (<<0|1>, <2|1>>^k. <<1, 2-n>>)[1$2]:
seq(seq(A(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Jan 21 2021

A340660[m_, n_] := LinearRecurrence[{1, 2}, {1, m}, {n}]; Table[Reverse[Table[A340660[m, n + m - 2] // First, {m, 2, -n + 3, -1}]], {n, 1, 11}] // Flatten (* Robert P. P. McKone, Jan 28 2021 *)

T(n, k) = 2^k - n*(2^k - (-1)^k)/3;
matrix(10,10,n,k,T(n-1,k-1)) \\ Michel Marcus, Jan 19 2021

A(n,k) = 2^k - n*round(2^k/3).

A365207 The number of divisors d of n such that gcd(d, n/d) is a power of 2 (A000079).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 2, 4, 2, 6, 2, 4, 4, 5, 2, 4, 2, 6, 4, 4, 2, 8, 2, 4, 2, 6, 2, 8, 2, 6, 4, 4, 4, 6, 2, 4, 4, 8, 2, 8, 2, 6, 4, 4, 2, 10, 2, 4, 4, 6, 2, 4, 4, 8, 4, 4, 2, 12, 2, 4, 4, 7, 4, 8, 2, 6, 4, 8, 2, 8, 2, 4, 4, 6, 4, 8, 2, 10, 2, 4, 2, 12, 4, 4

Offset: 1

Amiram Eldar, Aug 26 2023

The sum of these divisors is A107749(n).

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

Cf. A000005, A000079, A000265, A007814, A034444, A038838, A042968, A107749, A122132.

Cf. A001620, A013661, A073002.

f[p_, e_] := If[p == 2, e + 1, 2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,1] == 2, f[i,2]+1, 2));}

Multiplicative with a(2^e) = e+1 and a(p^e) = 2 for an odd prime p.
a(n) <= A000005(n), with equality if and only if n is in A122132 (or equivalently, n is not in A038838).
a(n) >= A034444(n), with equality if and only if n is not divisible by 4 (A042968).
a(n) = A000005(A006519(n)) * A034444(A000265(n)).
a(n) = A034444(n) * (A007814(n)+1)/2^(1 - (n mod 2)).
Dirichlet g.f.: (4^s/(4^s-1)) * zeta(s)^2/zeta(2*s).
Sum_{k==1..n} a(k) ~ (8/Pi^2)*n*(log(n) + 2*gamma - 2*log(2)/3 - 2*zeta'(2)/zeta(2) - 1), where gamma is Euler's constant (A001620).

A072134 Second generation sequence in which each number is skipped that can be written as sum of distinct previous entries. To make the first generation we start with all natural numbers: this gives the powers of 2 (A000079). For the second generation we start with the natural numbers from which are removed the numbers of the first generation.

Original entry on oeis.org

3, 5, 6, 7, 17, 19, 53, 55, 161, 163, 485, 487, 1457, 1459, 4373, 4375, 13121, 13123, 39365, 39367, 118097, 118099, 354293, 354295, 1062881, 1062883, 3188645, 3188647, 9565937, 9565939, 28697813, 28697815, 86093441, 86093443, 258280325

Offset: 0

Floor van Lamoen, Jun 26 2002

nonn

			4 is not in the sequence because it is in A000079.
14 is not in the sequence because 14 = 3+5+6.

a_n is the union of {3, 6}; the numbers {v_n} where v_0 = 5, v_n=3*v_{n-1}+2; and the numbers u_n = 2 + v_n.
Conjectures from Colin Barker, Sep 09 2017: (Start)
G.f.: (3 + 8*x + 2*x^2 - 11*x^3 - 9*x^4 - 3*x^5) / ((1 + x)*(1 - 3*x^2)).
a(n) = -(-1)^n + 3^((n-1)/2)*(1-(-1)^n + sqrt(3) + (-1)^n*sqrt(3)) for n>2.
a(n) = -a(n-1) + 3*a(n-2) + 3*a(n-3) for n>5.
(End)

A093304 ((Cumulative sum A000045) + (A000079)) - A092176.

Original entry on oeis.org

1, 2, 3, 6, 11, 23, 45, 89, 172, 331, 641, 1239, 2405, 4685, 9156, 17955, 35323, 69687, 137804, 273055, 541963, 1077194, 2143479, 4269285, 8509953

Offset: 1

David Glenn Williams (davwill24(AT)aol.com), Apr 24 2004

			((0, 1, 2, 4, 7) + (1, 2, 4, 8, 16)) - (0, 1, 3, 6, 12)

Cf. A000045, A000079, A092176.

cp's OEIS Frontend

A095362 Ratio A095110(n)/A000079(n-2) rounded to nearest integer.

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A162459 A002321*A000079.

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A171559 Powers of 2 (cf. A000079) with 1 replaced by 3.

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A180338 Triangle by rows, A033452 convolved with A000079.

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A181175 The "Fi1" sums of the powers-of-2 triangle A000079.

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A334638 Three-column array pPT read by rows: subsequence of primitive Pythagorean triples (x, y, z) with x = A153893^2 - A000079^2, y = 2*A153893*A000079, z = A153893^2 + A000079^2, ordered by increasing z.

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A340660 A000079 is the first row. For the second row, subtract A001045. For the third row, subtract A001045 from the second one, etc. The resulting array is read by ascending antidiagonals.

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A365207 The number of divisors d of n such that gcd(d, n/d) is a power of 2 (A000079).

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A334638 Three-column array pPT read by rows: subsequence of primitive Pythagorean triples (x, y, z) with x = A153893^2 - A000079^2, y = 2A153893A000079, z = A153893^2 + A000079^2, ordered by increasing z.