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A033994 a(n) = n*(n+1)*(5*n+1)/6. 15
2, 11, 32, 70, 130, 217, 336, 492, 690, 935, 1232, 1586, 2002, 2485, 3040, 3672, 4386, 5187, 6080, 7070, 8162, 9361, 10672, 12100, 13650, 15327, 17136, 19082, 21170, 23405, 25792, 28336, 31042, 33915, 36960, 40182, 43586, 47177, 50960, 54940 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Partial sums of A005476.

REFERENCES

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

LINKS

Harvey P. Dale, Table of n, a(n) for n = 1..1000

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

FORMULA

G.f.: x*(2+3*x)/(1-x)^4.

a(n) = A132121(n,1). - Reinhard Zumkeller, Aug 12 2007

a(n) = A000292(n) + A002412(n) = A000330(n) + A002411(n). - Omar E. Pol, Jan 11 2013

a(n) = Sum_{i=1..n} Sum_{j=1..n} i+min(i,j). - Enrique Pérez Herrero, Jan 15 2013

a(n) = Sum_{i=1..n} i*(n+i). - Charlie Marion, Apr 10 2013

Sum_{n>=1} 1/a(n) = 36 - 3*Pi*5^(3/4)*phi^(3/2)/4 - 15*sqrt(5)*log(phi)/4 - 75*log(5)/8 = 0.66131826232008423794478..., where phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 01 2018

MAPLE

[n*(n+1)*(5*n+1)/6$n=1..40]; # Muniru A Asiru, Jan 01 2019

MATHEMATICA

Table[Range[x].Range[x+1, 2x], {x, 40}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {2, 11, 32, 70}, 40] (* Harvey P. Dale, Jun 01 2018 *)

PROG

(PARI) a(n) = n*(n+1)*(5*n+1)/6;

(MAGMA) [n*(n+1)*(5*n+1)/6 : n in [1..40]]; // Vincenzo Librandi, Jan 01 2019

(GAP) a:=List([1..40], n->n*(n+1)*(5*n+1)/6);; Print(a); # Muniru A Asiru, Jan 01 2019

CROSSREFS

Cf. A005476, A016873, A000330, A132124, A132112, A050409.

Sequence in context: A190261 A000755 A183460 * A023659 A094792 A173707

Adjacent sequences:  A033991 A033992 A033993 * A033995 A033996 A033997

KEYWORD

easy,nonn

AUTHOR

Barry E. Williams, Dec 16 1999

EXTENSIONS

More terms from James A. Sellers, Jan 19 2000

STATUS

approved

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Last modified December 13 06:26 EST 2019. Contains 329968 sequences. (Running on oeis4.)