A336535 a(n) = (m(n)^2 + 3)*(m(n)^2 + 7)/32, where m(n) = 2*n - 1.

1, 6, 28, 91, 231, 496, 946, 1653, 2701, 4186, 6216, 8911, 12403, 16836, 22366, 29161, 37401, 47278, 58996, 72771, 88831, 107416, 128778, 153181, 180901, 212226, 247456, 286903, 330891, 379756, 433846, 493521, 559153, 631126, 709836, 795691, 889111, 990528, 1100386, 1219141, 1347261, 1485226, 1633528, 1792671

Offset

1,2

Comments

For m(n) = 3,5,11, and 181, the perfect numbers (A000396), 6, 28, 496, and 33550336 are produced, respectively. 3,5,11, and 181 are the numbers m(n) such that (m(n)^2+7) is a power of 2. cf A038198.

The unique primitive Pythagorean triple whose inradius T(n) and its long leg and hypotenuse are consecutive natural numbers is (2*T(n)+1, 2*T(n)*(T(n)+1), 2*T(n)*(T(n)+1)+1) and its semiperimeter is (T(n)+1)*(2*T(n)+1) where T(n) = A000217(n). - Miguel-Ángel Pérez García-Ortega, May 16 2025

References

David M. Burton, Elementary Number Theory, McGraw-Hill (2011), 25.

Links

Table of n, a(n) for n=1..44.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

From Stefano Spezia, Jul 25 2020: (Start)

O.g.f.: x*(1 + x + 8*x^2 + x^3 + x^4)/(1 - x)^5.

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

a(n) = A002061(n)*A014206(n-1)/2.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5. (End)

a(n) = (A000217(n-1)+1)*(2*A000217(n-1)+1). - Miguel-Ángel Pérez García-Ortega, May 16 2025

Example

m(2) = 2*2-1 = 3 and (3^2+3)*(3^2+7)/32 = 6, so 6 is in the sequence.

Mathematica

Table[((n^2+3)*(n^2+7))/32, {n, 1, 100, 2}]
a=Table[(n(n+1))/2, {n, 0, 40}]; Apply[Join, Map[{(#+1)(2#+1)}&, a]] (* Miguel-Ángel Pérez García-Ortega, May 16 2025 *)

Prog

(PARI) a(n)=(1-n+n^2)*(2-n+n^2)/2 \\ Charles R Greathouse IV, Oct 21 2022

Crossrefs

Cf. A000396, A038198.

Cf. A002061, A014206.

Cf. A000217, A058919, A383833, A383834.

Sequence in context: A141255 A091321 A125310 * A342380 A138874 A172141

Adjacent sequences: A336532 A336533 A336534 * A336536 A336537 A336538

Keyword

nonn,easy

Author

Jeff Brown, Jul 24 2020

Status

approved