A195037 17 times triangular numbers.

0, 17, 51, 102, 170, 255, 357, 476, 612, 765, 935, 1122, 1326, 1547, 1785, 2040, 2312, 2601, 2907, 3230, 3570, 3927, 4301, 4692, 5100, 5525, 5967, 6426, 6902, 7395, 7905, 8432, 8976, 9537, 10115, 10710, 11322, 11951, 12597, 13260, 13940, 14637, 15351, 16082, 16830

Offset: 0

Author

Omar E. Pol, Sep 12 2011

Keywords

• nonn
• easy

Comments

Related to the primitive Pythagorean triple [5, 12, 13].

Sequence found by reading the line from 0, in the direction 0, 17, ..., and the same line from 0, in the direction 0, 51, ..., in the Pythagorean spiral whose edges have length A195031 and whose vertices are the numbers A195032. This is the main diagonal of the square spiral.

Sum of the numbers from 8*n to 9*n. - Wesley Ivan Hurt, Dec 23 2015

Links

• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Bisection of A195032.

Cf. A000217, A195031.

Programs

Magma

[17*n*(n+1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Dec 23 2015

Maple

A195037:=n->17*n*(n+1)/2: seq(A195037(n), n=0..60); # Wesley Ivan Hurt, Dec 23 2015

Mathematica

17*Accumulate[Range[0,50]] (* Harvey P. Dale, May 31 2014 *)
Table[17*n*(n + 1)/2, {n, 0, 50}] (* Wesley Ivan Hurt, Dec 23 2015 *)

PARI

a(n)=17*n*(n+1)/2 \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = (17*n^2 + 17*n)/2 = 17*n*(n+1)/2 = 17*A000217(n).

From Wesley Ivan Hurt, Dec 23 2015: (Start)

G.f.: 17*x/(1-x)^3.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.

a(n) = Sum_{i=8*n..9*n} i. (End)

From Amiram Eldar, Feb 22 2023: (Start)

Sum_{n>=1} 1/a(n) = 2/17.

Sum_{n>=1} (-1)^(n+1)/a(n) = (4*log(2) - 2)/17.

Product_{n>=1} (1 - 1/a(n)) = -(17/(2*Pi))*cos(5*Pi/(2*sqrt(17))).

Product_{n>=1} (1 + 1/a(n)) = (17/(2*Pi))*cos(3*Pi/(2*sqrt(17))). (End)

From Elmo R. Oliveira, Dec 25 2024: (Start)

E.g.f.: 17*exp(x)*x*(2 + x)/2.

a(n) = A195032(2*n). (End)