A132337 Sum of the integers from 1 to n, excluding the perfect sixth powers.

0, 2, 5, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, 324, 350, 377, 405, 434, 464, 495, 527, 560, 594, 629, 665, 702, 740, 779, 819, 860, 902, 945, 989, 1034, 1080, 1127, 1175, 1224, 1274, 1325, 1377, 1430, 1484

Offset

1,2

Links

T. D. Noe, Table of n, a(n) for n = 1..10000

Formula

Let r = floor(n^(1/6)). Then a(n) = n(n+1)/2 - (r^7/7 + r^6/2 + r^5/2 - r^3/6 + r/42) = A000217(n) - A000540(r).

a(n) = A000217(n) - A000540(A178489(n)). - M. F. Hasler, Oct 09 2010

Maple

A132337 := proc(n) r := floor(n^(1/6)) ; A000217(n)-A000540(r); end proc: seq(A132337(n), n=1..40) ; # R. J. Mathar

Mathematica

Accumulate[Table[If[IntegerQ[Surd[n, 6]], 0, n], {n, 60}]] (* Harvey P. Dale, Jun 01 2022 *)

Prog

(PARI) g6(n)=for(x=1, n, r=floor(x^(1/6)); sum6=r^7/7+r^6/2+r^5/2-r^3/6+r/ 42; sn=x* (x+1)/2; print1(sn-sum6", "))
(PARI) A132337(n)=n*(n+1)/2-(1+n=floor(sqrtn(n+.5, 6)))*(2*n+1)*((n^3+2*n^2-1)*n*3+1)*n/42 \\ M. F. Hasler, Oct 09 2010

Crossrefs

Different from A000096.

Cf. A132336, A178489.

Sequence in context: A272370 A212342 A080956 * A000096 A134189 A109470

Adjacent sequences: A132334 A132335 A132336 * A132338 A132339 A132340

Keyword

nonn,easy

Author

Cino Hilliard, Nov 07 2007

Extensions

Incorrect formula deleted by Jon E. Schoenfield, Jun 12 2010

Incorrect program replaced by R. J. Mathar, Oct 08 2010

Edited by the Assoc. Editors of the OEIS, Oct 12 2010. Thanks to Daniel Mondot for pointing out that the sequence needed editing.

Incorrect linear recurrence removed by Georg Fischer, Apr 11 2019

Status

approved