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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A226508 a(n) = Sum_{i=3^n..3^(n+1)-1} i.

Original entry on oeis.org

3, 33, 315, 2889, 26163, 235953, 2125035, 19129689, 172180323, 1549662273, 13947078555, 125524061289, 1129717614483, 10167461718993, 91507165036875, 823564514029689, 7412080712360643, 66708726669526113, 600378540800575995, 5403406869529706889
Offset: 0

Views

Author

Michel Marcus, Jun 10 2013

Keywords

Comments

Partial sums give 3, 36, 351, 3240, 29403,...: A026121.
a(n) is the sum of all integers having n+1 digits in their ternary expansion (without leading zeros). - Jonathan Vos Post, Sep 07 2006

Examples

			a(0) = 1+2 = 3,
a(1) = 3+4+5+6+7+8 = 33.

Links

Crossrefs

Cf. A010035, A010036 (base 2), A026121, A101291 (base 10).
Cf. A007089 (numbers in base 3).

Programs

  • Mathematica
    Table[3^(n - 1) (4 3^(n + 1) - 3), {n, 0, 20}] (* Bruno Berselli, Jun 11 2013 *)
LinearRecurrence[{12,-27},{3,33},30] (* Harvey P. Dale, Jun 19 2013 *)
  • PARI
    a(n) = sum(i=3^n, 3^(n+1)-1, i) \\ Michel Marcus, Jun 11 2013

Formula

G.f.: 3*(1-x)/(1-12*x+27*x^2). [Bruno Berselli, Jun 11 2013]
a(n) = 3^(n-1)*(4*3^(n+1)-3). [Bruno Berselli, Jun 11 2013]
a(0)=3, a(1)=33, a(n)=12*a(n-1)-27*a(n-2). - Harvey P. Dale, Jun 19 2013

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.

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

Views

Author

Ralf Steiner, May 07 2020

Keywords

Comments

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)

Examples

			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

Links

Crossrefs

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

Programs

  • Mathematica
    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
  • PARI
    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

Formula

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)

Extensions

Edited, and corrected proportion by Wolfdieter Lang, Jun 13 2020
Minor grammatical edits. - N. J. A. Sloane, Sep 12 2020
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