A003683 -id:A003683 - cp's OEIS frontend

A334908 Area/6 of primitive Pythagorean triangles generated by {{2, 0}, {1, -1}}^n * {{2}, {1}}, for n >= 0.

1, 10, 220, 3080, 52976, 818720, 13333440, 211474560, 3398520576, 54257082880, 869067996160, 13897453373440, 222420341682176, 3558236809994240, 56935698394234880, 910939899548958720, 14575288593717067776, 233202615903456460800

Offset: 0

Ralf Steiner, May 16 2020

Matrix {{2, 0}, {1, -1}} is [g_{-2}] given by Firstov in eq. (24).
These primitive Pythagorean triples are also given by Lee Price as (M_2)^n (3,4,5)^T (T for transposed), with M_2 = {{2, 1, 1}, {2, -2, 2}, {2, -1, 3}}.
For a primitive Pythagorean triangle (x, y, z) = (u^2-v^2, 2*u*v, u^2+v^2) the area is A = x*y/2 = u*v*(u^2 - v^2) = z*h/2 with altitude h, and h is an irreducible fraction. Here:
x(n) = A084175(n+2).
y(n) = 4*(A084175(n+1) - A084175(n)) = A054881(n+2).
     = 2*A192382(n+1) = 4*A003683(n+1).
z(n) = A084175(n+2) + 2*A084175(n+1) - 4*A084175(n).
     = A108924(n+2)/2 = A084175(n+2) + 2*A139818(n+1).
     = A000302(n+1) + A139818(n+1).
u(n) = A000079(n+1) = 2^(n+1).
v(n) = A001045(n+1) = (2^(n+1) + (-1)^n)/3.
For the area A(n): Limit_{n -> oo} (3^3/(2^(4*n+7)))*A(n) = 1. See the formula section. - Wolfdieter Lang, Jun 14 2020

			a(0) = 3*4/12 = 1 for the triangle (3, 4, 5).

G. C. Greubel, Table of n, a(n) for n = 0..825
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).
H. Lee Price, The Pythagorean Tree: A New Species, arXiv:0809.4324 [math.HO], 2008-2011
R. Steiner, Spezielle Folge primitiver pythagoräischer Dreiecke, researchgate.net, 2020
Index entries for linear recurrences with constant coefficients, signature (10,120,-320,-1024).

Cf. A000079, A000302, A001045, A003683, A054881, A084175, A108924, A139818, A192382.

[(2^(2*n+1)*(2^(2*n+5) -3) +(-2)^n*(3*2^(2*n+3) -1))/81: n in [0..40]]; // G. C. Greubel, Feb 18 2023

Table[(2^(2*n+1)*(2^(2*n+5) -3) + (-2)^n*(3*2^(2*n+3) -1))/3^4, {n,0,40}]

[(2^(2*n+1)*(2^(2*n+5) -3) +(-2)^n*(3*2^(2*n+3) -1))/81 for n in range(41)] # G. C. Greubel, Feb 18 2023

a(n) = ( 2^(4*n+6) - 3*2^(2*n+1) - 3*(-2)^(3*n+3) - (-2)^n )/3^4.
G.f.: 1 / ((1 + 2*x)*(1 - 4*x)*(1 + 8*x)*(1 - 16*x)). - Colin Barker, Jun 11 2020
E.g.f.: (1/81)*(24*exp(-8*x) - exp(-2*x) - 6*exp(4*x) + 64*exp(16*x)). - G. C. Greubel, Feb 18 2023

A368043 Triangle read by rows: T(n, k) = 2^(n + k).

Original entry on oeis.org

1, 2, 4, 4, 8, 16, 8, 16, 32, 64, 16, 32, 64, 128, 256, 32, 64, 128, 256, 512, 1024, 64, 128, 256, 512, 1024, 2048, 4096, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144

Offset: 0

Peter Luschny, Dec 09 2023

			[0]  [  1]
[1]  [  2,   4]
[2]  [  4,   8,  16]
[3]  [  8,  16,  32,    64]
[4]  [ 16,  32,  64,   128,  256]
[5]  [ 32,  64,  128,  256,  512, 1024]
[6]  [ 64, 128,  256,  512, 1024, 2048,  4096]
[7]  [128, 256,  512, 1024, 2048, 4096,  8192, 16384]
[8]  [256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536]

Cf. A000079 (T(n,0)), A004171 (T(n,n-1)), A000302 (T(n,n)), A171476 (row sums), A003683 (alternating row sums), A134353 (antidiagonal sums), A001018 (T(2n, n)), A094014 (T(n, n/2)), A002697.

Array[2^Range[#,2#]&,10,0] (* Paolo Xausa, Dec 09 2023 *)

from functools import cache
@cache
def T_row(n: int) -> list[int]:
    if n == 0: return [1]
    row = T_row(n - 1) + [0]
    for k in range(n): row[k] *= 2
    row[n] = row[n - 1] * 2
    return row
for n in range(11): print(T_row(n))

G.f.: 1/((1 - 2*x)*(1 - 4*x*y)). - Stefano Spezia, Dec 09 2023

A374098 a(n) = A112387(n)^2.

Original entry on oeis.org

1, 1, 4, 1, 16, 9, 64, 25, 256, 121, 1024, 441, 4096, 1849, 16384, 7225, 65536, 29241, 262144, 116281, 1048576, 466489, 4194304, 1863225, 16777216, 7458361, 67108864, 29822521, 268435456, 119311929, 1073741824, 477204025, 4294967296, 1908903481, 17179869184

Offset: 0

Paul Curtz, Jun 28 2024

Index entries for linear recurrences with constant coefficients, signature (0,3,0,6,0,-8).

Cf. A000302, A003683, A084175, A112387, A139818.

LinearRecurrence[{0, 3, 0, 6, 0, -8}, {1, 1, 4, 1, 16, 9}, 35] (* Amiram Eldar, Jul 01 2024 *)

a(2*n) = A000302(n); a(2*n+1) = A139818(n+1).
(a(2*n) + a(2*n-1))^2 = A084175(n+1)^2 + 16*A003683(n)^2, for n >= 1. - Thomas Scheuerle, Jun 28 2024
G.f. ( 1+x+x^2-2*x^3-2*x^4 ) / ( (x-1)*(2*x+1)*(2*x-1)*(1+x)*(2*x^2+1) ). - R. J. Mathar, Aug 02 2024

A110953 Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 9 times the bottom to get the new top.

Original entry on oeis.org

2, 12, 40, 176, 672, 2752, 10880, 43776, 174592, 699392, 2795520, 11186176, 44736512, 178962432, 715816960, 2863333376, 11453202432, 45813071872, 183251763200, 733008101376, 2932030308352, 11728125427712, 46912493322240, 187649990066176, 750599926710272

Offset: 1

Cino Hilliard, Sep 25 2005

The limit of the sequence of fractions used to generate this sequence is 3.

Essentially the same as A003683. - R. J. Mathar, May 25 2009

Prime Obsession, John Derbyshire, Joseph Henry Press, April 2004, p. 16.

Index entries for linear recurrences with constant coefficients, signature (2,8).

g(n,k,typ) = /* typ = 1 numerator, 2 denominator, k = multiple of denom */ { local(a,b,x,tmp); a=1;b=1; for(x=1,n, tmp=b; b=a+b; a=k*tmp+a; if(typ==1,print1(a","),print1(b",")) ); print(); print(a/b+.) }

from itertools import islice
def A110953_gen(): # generator of terms
    a, b = 1, 1
    while True:
        a, b = a+9*b, a+b
        yield b
A110953_list = list(islice(A110953_gen(),30)) # Chai Wah Wu, Apr 15 2025

Given a(0)=1, b(0)=1 then for i>=1, a(i)/b(i) = (a(i-1)+ 9*b(i-1)) / (a(i-1) + b(i-1)).
From Chai Wah Wu, Apr 15 2025: (Start)
a(n) = 2*a(n-1) + 8*a(n-2) for n > 2.
G.f.: x*(-8*x - 2)/((2*x + 1)*(4*x - 1)). (End)

a(21)-a(25) from Chai Wah Wu, Apr 15 2025

A292847 a(n) is the smallest odd prime of the form ((1 + sqrt(2n))^k - (1 - sqrt(2n))^k)/(2sqrt(2n)).

Original entry on oeis.org

5, 7, 101, 11, 13, 269, 17, 19, 509, 23, 709, 821, 29, 31, 46957, 55399, 37, 168846239, 41, 43, 9177868096974864412935432937651459122761, 47, 485329129, 2789, 53, 3229, 3461, 59, 61, 1563353111, 139237612541, 67, 5021, 71, 73, 484639, 6221, 79, 6869, 83, 7549

Offset: 1

XU Pingya, Sep 24 2017

nonn

			For k = {1, 2, 3, 4, 5}, ((1 + sqrt(6))^k - (1 - sqrt(6))^k)/(2*sqrt(6)) = {1, 2, 9, 28, 101}. 101 is odd prime, so a(3) = 101.

Cf. A000129, A002605, A015518, A063727, A002532, A083099, A015519, A003683, A002534, A083102, A015520, A091914, A079773, A161007, A099134.

g[n_, k_] := ((1 + Sqrt[n])^k - (1 - Sqrt[n])^k)/(2Sqrt[n]);
Table[k = 3; While[! PrimeQ[Expand@g[2n, k]], k++]; Expand@g[2n, k], {n, 41}]

g(n,k) = ([0,1;2*n-1,2]^k*[0;1])[1,1]
a(n) = for(k=3,oo,if(ispseudoprime(g(n,k)),return(g(n,k)))) \\ Jason Yuen, Apr 12 2025

When 2*n + 3 = p is prime, a(n) = p.

A352692 a(n) + a(n+1) = 2^n for n >= 0 with a(0) = 4.

Original entry on oeis.org

4, -3, 5, -1, 9, 7, 25, 39, 89, 167, 345, 679, 1369, 2727, 5465, 10919, 21849, 43687, 87385, 174759, 349529, 699047, 1398105, 2796199, 5592409, 11184807, 22369625, 44739239, 89478489, 178956967, 357913945, 715827879, 1431655769, 2863311527, 5726623065, 11453246119, 22906492249

Offset: 0

Paul Curtz, Mar 29 2022

Difference table D(n,k) = D(n-1,k+1) - D(n-1,k), D(0,k) = a(k):
    4,  -3,   5,  -1,   9,   7, 25, ...
   -7,   8,  -6,  10,  -2,  18, 14,  50, ...
   15, -14,  16, -12,  20,  -4, 36,  28, 100, ...
  -29,  30, -28,  32, -24,  40, -8,  72,  56, 200, ...
   59, -58,  60, -56,  64, -48, 80, -16, 144, 112, 400, ...
  ...
The diagonals are given by D(n,n+k) = a(k)*2^n.
D(n,1) = -(-1)^n* A340627(n).
a(n)   - a(n) =  0,  0,  0,  0,   0, ...       (trivially)
a(n+1) + a(n) =  1,  2,  4,  8,  16, ... = 2^n (by definition)
a(n+2) - a(n) =  1,  2,  4,  8,  16, ... = 2^n
a(n+3) + a(n) =  3,  6, 12, 24,  48, ... = 2^n*3
a(n+4) - a(n) =  5, 10, 20, 40,  80, ... = 2^n*5
a(n+5) + a(n) = 11, 22, 44, 88, 176, ... = 2^n*11
     (...)
This table is given by T(r,n) = A001045(r)*2^n with r, n >= 0.
Sums of antidiagonals are A045883(n).
Main diagonal: A192382(n).
First upper diagonal: A054881(n+1).
First subdiagonal: A003683(n+1).
Second subdiagonal: A246036(n).
Now consider the array from c(n) = (-1)^n*a(n) with its difference table:
   4,   3,   5,    1,    9,   -7,   25,   -39, ...  = c(n)
  -1,   2,  -4,    8,  -16,   32,  -64,   128, ...  = -A122803(n)
   3,  -6,  12,  -24,   48,  -96,  192,  -384, ...  =
  -9,  18, -36,   72, -144,  288, -576,  1152, ...
  27, -54, 108, -216,  432, -864, 1728, -3456, ...
...
The first subdiagonal is -A000400(n). The second is A169604(n).

Index entries for linear recurrences with constant coefficients, signature (1,2).

If a(0) = k then A001045 (k=0), A078008 (k=1), A140966 (k=2), A154879 (k=3), this sequence (k=4).
Essentially the same as A115335.
Cf. A000079, A002450, A340627.
Cf. A020714, A175805.
Cf. A045883, A192382, A003683, A246036.
Cf. A054881, A000400, A169604, A122803.
Cf. A132805, A082311, A082365.
Cf. A024495, A132804, A163868.

a := proc(n) option remember; ifelse(n = 0, 4, 2^(n-1) - a(n-1)) end: # Peter Luschny, Mar 29 2022
A352691 := proc(n)
    (11*(-1)^n + 2^n)/3
end proc: # R. J. Mathar, Apr 26 2022

LinearRecurrence[{1, 2}, {4, -3}, 40] (* Amiram Eldar, Mar 29 2022 *)

a(n) = (11*(-1)^n + 2^n)/3; \\ Thomas Scheuerle, Mar 29 2022

abs(a(n)) = A115335(n-1) for n >= 1.
a(3*n) - (-1)^n*4 = A132805(n).
a(3*n+1) + (-1)^n*4 = A082311(n).
a(3*n+2) - (-1)^n*4 = A082365(n).
From Thomas Scheuerle, Mar 29 2022: (Start)
G.f.: (-4 + 7*x)/(-1 + x + 2*x^2).
Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*a(m + 2*n-k) = a(m)*2^n.
Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*a(1 + n-k) = -(-1)^n*A340627(n).
a(n) = (11*(-1)^n + 2^n)/3.
a(n + 2*m) = a(n) + A002450(m)*2^n.
a(2*n) = A192382(n+1) + (-1)^n*a(n).
a(n) = ( A045883(n) - Sum_{k=0..n-1}(-1)^k*a(k) )/n, for n > 0. (End)
a(n) = A001045(n) + 4*(-1)^n.
a(n+1) = 2*a(n) -11*(-1)^n.
a(n+2) = a(n) + 2^n.
a(n+4) = a(n) + A020714(n).
a(n+6) = a(n) + A175805(n).
a(2*n) = A163868(n).
a(2*n+1) = (2^(2*n+1) - 11)/3.

Warning: The DATA is correct, but there may be errors in the COMMENTS, which should be rechecked. - Editors of OEIS, Apr 26 2022

Edited by M. F. Hasler, Apr 26 2022.

A370627 a(n) = 2^(n - 1)((-1)^(n + 1) + 72^n)/3 = 2^(n - 1)*A062092(n).

Original entry on oeis.org

1, 5, 18, 76, 296, 1200, 4768, 19136, 76416, 305920, 1223168, 4893696, 19572736, 78295040, 313171968, 1252704256, 5010784256, 20043202560, 80172679168, 320690978816, 1282763390976, 5131054612480, 20524216352768, 82096869605376, 328387470032896, 1313549896908800, 5254199554080768

Offset: 0

Paul Curtz, Jul 03 2024

Index entries for linear recurrences with constant coefficients, signature (2,8).

Cf. A003683, A122803, A133125, A255470, A255471.

Cf. A108982, A062092.

LinearRecurrence[{2, 8}, {1, 5}, 27] (* Amiram Eldar, Jul 03 2024 *)

a(n) = 2^(n-1)*((-1)^(n+1) + 7*2^n)/3 \\ Thomas Scheuerle, Jul 03 2024

Binomial transform of A133125.
G.f.: (1 + 3*x)/(1 - 2*x - 8*x^2).
E.g.f.: (1/3)*exp(x)*(3*exp(3*x) + sinh(3*x)).
a(n) = 2*a(n-1) + 8*a(n-2), for n > 1.
a(n) = 4*a(n-1) + (-2)^n, for n > 0.
a(n) = (a(n+2) - 2*a(n+1))/8.
From Thomas Scheuerle, Jul 03 2024: (Start)
a(n) = 2^(n - 1)*((-1)^(n + 1) + 7*2^n)/3.
a(n) = A003683(n) + 4^n.
a(n) = A255470(2^n - 1) - A255470(2^(n-1) - 1) = A255471(n) - A255471(n-1), for n > 0. (End)
Binomial transform: A108982.

cp's OEIS Frontend

A334908 Area/6 of primitive Pythagorean triangles generated by {{2, 0}, {1, -1}}^n * {{2}, {1}}, for n >= 0.

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A368043 Triangle read by rows: T(n, k) = 2^(n + k).

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A374098 a(n) = A112387(n)^2.

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A110953 Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 9 times the bottom to get the new top.

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A292847 a(n) is the smallest odd prime of the form ((1 + sqrt(2*n))^k - (1 - sqrt(2*n))^k)/(2*sqrt(2*n)).

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A352692 a(n) + a(n+1) = 2^n for n >= 0 with a(0) = 4.

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A370627 a(n) = 2^(n - 1)*((-1)^(n + 1) + 7*2^n)/3 = 2^(n - 1)*A062092(n).

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A292847 a(n) is the smallest odd prime of the form ((1 + sqrt(2n))^k - (1 - sqrt(2n))^k)/(2sqrt(2n)).

A370627 a(n) = 2^(n - 1)((-1)^(n + 1) + 72^n)/3 = 2^(n - 1)*A062092(n).