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A133480 Left 3-step factorial (n,-3)! : signed A008544. 1
1, -2, 10, -80, 880, -12320, 209440, -4188800, 96342400, -2504902400, 72642169600 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

To motivate the definition, consider c(t) = column vector(1, t, t^2, t^3, t^4, t^5,...), T = A094638 and the list of integers.

Starting at 1 and sampling every integer to the right, we obtain (1,2,3,4,5,...) from which factorials may be formed. It's true that

T * c(1) = (1, 1*2, 1*2*3, 1*2*3*4,...), giving n! for n>0. Call this sequence the right 1-step factorial (n,+1)! .

Starting at 1 and sampling every integer to the left, we obtain (1,0,-1,-2,-3,-4,-5,...). And,

T * c(-1) = (1, 1*0, 1*0*-1, 1*0*-1*-2,...) = (1,0,0,0,...) . Call this the left 1-step factorial (n,-1)! .

Sampling every other integer to the right, we obtain (1,3,5,7,9,...).

T * c(2) = (1, 1*3, 1*3*5, ...) = (1,3,15,105,945,...), giving A001147 for n>0, the right 2-step factorial, (n,+2)! .

Sampling every other integer to the left, we obtain (1,-1,-3,-5,-7...).

T * c(-2) = (1, 1*-1, 1*-1*-3, 1*-1*-3*-5,...) = (1,-1,3,-15,105,-945,...) = signed A001147, the left 2-step factorial, (n,-2)! .

Sampling every 3 steps to the right, we obtain (1,4,7,10,...).

T * c(3) = (1, 1*4, 1*4*7,...) = (1,4,28,280,...), giving A007559 for n>0, the right 3-step factorial, (n,+3)! .

Sampling every 3 steps to the left, we obtain (1,-2,-5,-8,-11,...), giving

T * c(-3) = (1, 1*-2, 1*-2*-5, 1*-2*-5*-8,...) = (1,-2,10,-80,880,...) = signed A008544 = the left 3-step factorial, (n,-3)! .

The list partition transform A133314 of [1,T * c(t)] gives signed [1,T *c(-t)] . For example:

LPT[1,T*c(1)] = LPT[1,(n,+1)! ] = LPT[A000142] = (1,-1,0,0,0,...) = signed [1,(n,-1)! ]

LPT[1,T*c(2)] = LPT[1,(n,+2)! ] = LPT[A001147] = (1,-1,-1,-3,-15,-105,-945,...) = (1,-A001147) = signed [1,(n,-2)! ]

LPT[1,T*c(3)] = LPT[1,(n,+3)! ] = LPT[A007559] = (1,-1,-2,-10,-80,-880,...) = (1,-A008544) = signed [1,(n,-3)! ]

LPT[1,T*c(-3)] = LPT[1,(n,-3)! ] = LPT[A000001] = signed A007559 = signed [1,(n,+3)! ] .

And, e.g.f.[1,T * c(m)] = e.g.f.[1,(n,m)! ] = (1-m*x)^(-1/m) .

Also with P(n,t) = sum(k=0,...,n-1) T(n,k+1) * t^k = 1*(1+t)*(1+2t)...(1+(n-1)*t) and P(0,t)=1, exp[P(.,t)*x] = (1-tx)^(-1/t) .

T(n,k+1) = (1/k!) (D_t)^k (D_x)^n [ (1-tx)^(-1/t) - 1 ] eval. at t=x=0 .

And, (1-tx)^(-1/t) - 1 is the e.g.f. for plane increasing m-ary trees when t = (m-1), discussed by Bergeron et al. in "Varieties of Increasing Trees" and the book Combinatorial Species and Tree-Like Structures, cited in the OEIS.

The above relations reveal the intimate connections, through T or LPT or sampling, between the right and left step factorials, (n,+m)! and (n,-m)!. The pairs have conjugate interpretations as trees, ignoring signs, which Callan and Lang have noted in several of the OEIS entries above. Also note unsigned (n,-2)! is the diagonal of A001498 and (n,+2)!, the first subdiagonal.

LINKS

Table of n, a(n) for n=0..10.

FORMULA

a(n) = b(0)*b(1)...b(n) where b = (1,-2,-5,-8,-11,...) .

a(n)=3^(n+1)*sum(k=1..n+1, stirling1(n+1,k)/3^k), [From Vladimir Kruchinin, Jul 02 2011]

G.f.: (1/(Q(0)-1)/x where Q(k) = 1 + x*(3*k-1)/( 1 + x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 22 2013

G.f.: G(0)/(2*x) -1/x, where G(k)= 1 + 1/(1 - x*(3*k-1)/(x*(3*k-1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013

CROSSREFS

Sequence in context: A220112 A048286 A227463 * A008544 A227464 A269353

Adjacent sequences:  A133477 A133478 A133479 * A133481 A133482 A133483

KEYWORD

sign

AUTHOR

Tom Copeland, Dec 23 2007

STATUS

approved

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Last modified March 24 06:56 EDT 2019. Contains 321444 sequences. (Running on oeis4.)