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Left 3-step factorial (n,-3)!: a(n) = (-1)^n * A008544(n).
3

%I #36 Sep 24 2024 09:20:43

%S 1,-2,10,-80,880,-12320,209440,-4188800,96342400,-2504902400,

%T 72642169600,-2324549427200,81359229952000,-3091650738176000,

%U 126757680265216000,-5577337931669504000,262134882788466688000,-13106744139423334400000,694657439389436723200000,-38900816605808456499200000

%N Left 3-step factorial (n,-3)!: a(n) = (-1)^n * A008544(n).

%C 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.

%C 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

%C 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)!.

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

%C 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)!.

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

%C 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)!.

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

%C 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)!.

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

%C 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)!.

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

%C 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)!.

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

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

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

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

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

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

%C 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).

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

%C 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.

%C 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.

%H G. C. Greubel, <a href="/A133480/b133480.txt">Table of n, a(n) for n = 0..375</a>

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

%F a(n) = 3^(n+1)*Sum_{k=1..n+1} stirling1(n+1,k)/3^k. - _Vladimir Kruchinin_, Jul 02 2011

%F 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

%F 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

%F From _G. C. Greubel_, Mar 31 2019: (Start)

%F G.f.: Hypergeometric2F0(1,2/3; -; -3*x).

%F E.g.f.: (1+3*x)^(-2/3).

%F a(n) = (-3)^n*Pochhammer(2/3, n) = (-3)^n*(Gamma(n+2/3)/Gamma(2/3)). (End)

%F D-finite with recurrence: a(n) +(3*n-1)*a(n-1)=0. - _R. J. Mathar_, Jan 20 2020

%t Table[(-3)^n*Pochhammer[2/3, n], {n,0,20}] (* _G. C. Greubel_, Mar 31 2019 *)

%o (PARI) vector(20, n, n--; round((-3)^n*gamma(n+2/3)/gamma(2/3))) \\ _G. C. Greubel_, Mar 31 2019

%o (Magma) [Round((-3)^n*Gamma(n+2/3)/Gamma(2/3)): n in [0..20]]; // _G. C. Greubel_, Mar 31 2019

%o (Sage) [(-3)^n*rising_factorial(2/3,n) for n in (0..20)] # _G. C. Greubel_, Mar 31 2019

%Y Cf. A000142, A001147, A001498 , A007559, A008544, A133314 .

%K sign,easy

%O 0,2

%A _Tom Copeland_, Dec 23 2007

%E Terms a(11) onward added by _G. C. Greubel_, Mar 31 2019