A281026 a(n) = floor(3*n*(n+1)/4).

0, 1, 4, 9, 15, 22, 31, 42, 54, 67, 82, 99, 117, 136, 157, 180, 204, 229, 256, 285, 315, 346, 379, 414, 450, 487, 526, 567, 609, 652, 697, 744, 792, 841, 892, 945, 999, 1054, 1111, 1170, 1230, 1291, 1354, 1419, 1485, 1552, 1621, 1692, 1764, 1837, 1912, 1989, 2067, 2146

Offset

0,3

Links

Bruno Berselli, Table of n, a(n) for n = 0..1000

Bruno Berselli, Illustration of the initial terms.

Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Formula

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

E.g.f.: -(1 - 6*x - 3*x^2)*exp(x)/4 - (1 + i)*(i - exp(2*i*x))*exp(-i*x)/8, where i=sqrt(-1).

a(n) = a(-n-1) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) = a(n-4) + 6*n - 9.

a(n) = 3*n*(n+1)/4 + (i^(n*(n+1)) - 1)/4. Therefore:

a(4*k+r) = 12*k^2 + 3*(2*r+1)*k + r^2, where 0 <= r <= 3.

a(n) = n^2 - floor((n-1)*(n-2)/4).

a(n) = A011865(3*n+2).

Maple

A281026:=n->floor(3*n*(n+1)/4): seq(A281026(n), n=0..100); # Wesley Ivan Hurt, Jan 13 2017

Mathematica

Table[Floor[3 n (n + 1)/4], {n, 0, 60}]
LinearRecurrence[{3, -4, 4, -3, 1}, {0, 1, 4, 9, 15}, 60] (* Harvey P. Dale, Jun 04 2023 *)

Prog

(PARI) vector(60, n, n--; floor(3*n*(n+1)/4))
(Python) [int(3*n*(n+1)/4) for n in range(60)]
(Sage) [floor(3*n*(n+1)/4) for n in range(60)]
(Maxima) makelist(floor(3*n*(n+1)/4), n, 0, 60);
(Magma) [3*n*(n+1) div 4: n in [0..60]];

Crossrefs

Subsequence of A214068.

Partial sums of A047273.

Cf. A011865, A045943, A274757 (subsequence).

Cf. sequences with formula floor(k*n*(n+1)/4): A011848 (k=1), A000217 (k=2), this sequence (k=3), A002378 (k=4).

Cf. sequences with formula floor(k*n*(n+1)/(k+1)): A000217 (k=1), A143978 (k=2), this sequence (k=3), A281151 (k=4), A194275 (k=5).

Sequence in context: A022945 A022948 A022443 * A079423 A285283 A243536

Adjacent sequences: A281023 A281024 A281025 * A281027 A281028 A281029

Keyword

nonn,easy

Author

Bruno Berselli, Jan 13 2017

Status

approved