0,4

Ed Pegg Jr (ed(AT)mathpuzzle.com) conjectures that n^3 - n = k! has a solution iff n is 2, 3, 5 or 9 (when k is 3, 4, 5 and 6).

Three-dimensional promic (or oblong) numbers, cf. A002378 - Alexandre Wajnberg (alexandre.wajnberg(AT)skynet.be), Dec 29 2005

Doubled first differences of tritriangular numbers A050534(n) = (1/8)n(n + 1)(n - 1)(n - 2). a(n) = 2*(A050534(n+1) - A050534(n)). - Alexander Adamchuk (alex(AT)kolmogorov.com), Apr 11 2006

If Y is a 4-subset of an n-set X then, for n>=6, a(n-4) is the number of (n-5)-subsets of X having exactly two elements in common with Y. - Milan R. Janjic (agnus(AT)blic.net), Dec 28 2007

Convolution of A005843 with A008585. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Mar 07 2009]

a(n) = A000578(n) - A000567(n). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Sep 18 2009]

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

R. K. Guy, Unsolved Problems in Theory of Numbers, Section D25.

L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 40.

Index entries for two-way infinite sequences

Index entries for sequences related to linear recurrences with constant coefficients

Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets

Sum(Polygorial(3, i), i=1..n) - Daniel Dockery (peritus(AT)gmail.com) Jun 16, 2003

a(0) = a(1) = a(2) = 0, a(n) = 3a(n-1) - 3a(n-2) + a(n-3) + 6. - Zak Seidov, (zakseidov(AT)yahoo.com) Feb 09 2006.

G.f.: 6x^2/(1-x)^4. a(-n)=-a(n+2).

1/6 + 3/24 + 5/60 +...= 3/4 - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 21 2006

a(n)=numbperm(n,3). - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 26 2007

Other than the first two 0's this sequence is the same as the sequence a(n)=n^3-n. - Mohammad K. Azarian (azarian(AT)evansville.edu), Jul 26 2007

E.g.f.:x^3*exp(x) [From Geoffrey Critzer (critzer(AT)usd443.org), Feb 08 2009]

[seq(6*binomial(n, 3), n=0..41)]; - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 24 2006

a:=n->sum(numbperm (n, 2), j=0..n): seq(a(n), n=0..40); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 25 2007

seq(numbperm(n, 3), n=0..41); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 26 2007

a:=n->sum(sum(sum(1, j=0..n), k=1..n), m=2..n): seq(a(n), n=-1..40); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 18 2007

seq(sum(n^2-1, k=1..n), n=-1..39); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jan 28 2008

restart: G(x):=x^3*exp(x): f[0]:=G(x): for n from 1 to 41 do f[n]:=diff(f[n-1], x) od: x:=0: seq(f[n], n=0..41); # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 05 2009]

Table[n^3 - 3n^2 + 2n, {n, 0, 41}]

(PARI) a(n)=n*(n-1)*(n-2)

Cf. A002378.

Equals 6*A000292(n-1). Cf. A002378, A005563, A084939, A084940, A084941, A084942, A084943, A084944.

Cf. A007531.

Sequence in context: A086768 A160944 A160936 this_sequence A130669 A101854 A101877

Adjacent sequences: A007528 A007529 A007530 this_sequence A007532 A007533 A007534

nonn,easy

N. J. A. Sloane (njas(AT)research.att.com), Robert G. Wilson v (rgwv(AT)rgwv.com)