A084990 a(n) = n*(n^2+3*n-1)/3.

0, 1, 6, 17, 36, 65, 106, 161, 232, 321, 430, 561, 716, 897, 1106, 1345, 1616, 1921, 2262, 2641, 3060, 3521, 4026, 4577, 5176, 5825, 6526, 7281, 8092, 8961, 9890, 10881, 11936, 13057, 14246, 15505, 16836, 18241, 19722, 21281, 22920, 24641, 26446, 28337, 30316

Offset

0,3

Comments

Row sums of triangle A131782 starting (1, 6, 17, 36, 65, 106, ...). - Gary W. Adamson, Jul 14 2007

a(n) is the number of triples (x,y,z) in {1,2,...,n}^3 with x <= y <= z or x >= y >= z. - Jack Kennedy, Mar 14 2009

a(2*n) is the difference between numbers of nonnegative multiples of 2*n+1 with even and odd digit sum in base 2*n in interval [0, 16*n^4). - Vladimir Shevelev, May 18 2012

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, A q-queens problem III. Partial queens, arXiv:1402.4886 [math.CO], 2014-2018. See Conjecture 4.4.

Vladimir Shevelev, On monotonic strengthening of Newman-like phenomenon on (2m+1)-multiples in base 2m, arXiv:0710.3177 [math.NT], 2007.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Formula

a(n) = 2*A000292(n-1) - 1 (notice offset=-1 in A000292!).

a(n) = (n-1)*(n+1)*(n+3)/3 + 1. - Reinhard Zumkeller, Aug 20 2007

a(n) = A077415(n+1) + 1 for n > 0; a(n) = A000290(n) + A007290(n); a(n+1) = Sum_{k=0..n} A028387(k). - Reinhard Zumkeller, Aug 20 2007

a(2*n) = Sum_{i=0..16*n^4, i==0 (mod 2*n+1)} (-1)^s_(2*n)(i), where s_k(n) is the digit sum of n in base k. - Vladimir Shevelev, May 18 2012

a(2*n) = (2/(2*n+1))*Sum_{i=1..n} tan^4(Pi*i/(2*n+1)). - Vladimir Shevelev, May 23 2012

a(n) = Sum_{i=1..n} i*(i+1)-1. - Wesley Ivan Hurt, Oct 19 2013

G.f.: x*(1+2*x-x^2)/(1-x)^4. - Vincenzo Librandi, Mar 28 2014

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. - Vincenzo Librandi, Mar 28 2014

a(n) = A064043(n)/3. - Alois P. Heinz, Jul 21 2017

E.g.f.: x*exp(x)*(x^2 + 6*x + 3)/3. - Stefano Spezia, Mar 06 2024

Example

Let n=2. Consider nonnegative multiples of 5 up to 16*2^4 - 1 = 255. There are 52 such numbers and from them only 8 (namely, 35, 50, 55, 115, 140, 200, 205, 220) have an odd digit sum in base 4. Therefore, a(4) = (52 - 8) - 8 = 36. - Vladimir Shevelev, May 18 2012

Maple

A084990:=n->n*(n^2+3*n-1)/3; seq(A084990(k), k=0..100); # Wesley Ivan Hurt, Oct 19 2013

Mathematica

Table[n*(n^2+3*n-1)/3, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Mar 08 2010 *)
CoefficientList[Series[(x + 2 x^2 - x^3)/((1 - x)^4), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 28 2014 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 1, 6, 17}, 50] (* Harvey P. Dale, Aug 18 2015 *)

Prog

(Magma) I:=[0, 1, 6, 17]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Mar 28 2014
(PARI) a(n) = n*(n^2+3*n-1)/3 \\ Charles R Greathouse IV, Sep 24 2015

Crossrefs

Cf. A000290, A000292, A028387, A064043, A077415, A131782.

Sequence in context: A012277 A358245 A307502 * A024181 A023663 A048208

Adjacent sequences: A084987 A084988 A084989 * A084991 A084992 A084993

Keyword

nonn,easy

Author

Gary W. Adamson, Jul 16 2003

Status

approved