cp's OEIS Frontend

A143815 Let A(0)=1, B(0)=0 and C(0)=0. Let B(n+1) = Sum_{k = 0..n} binomial(n,k)*A(k), C(n+1) = Sum_{k = 0..n} binomial(n,k)*B(k) and A(n+1) = Sum_{k = 0..n} binomial(n,k)*C(k). This entry gives the sequence A(n).

%I A143815 #36 Aug 13 2025 10:18:55
%S A143815 1,0,0,1,6,25,91,322,1232,5672,32202,209143,1432454,9942517,69363840,
%T A143815 490303335,3565609732,27118060170,218183781871,1861370544934,
%U A143815 16729411124821,156706028787827,1514442896327792,14999698898942772,151838974745743228,1571513300578303070
%N A143815 Let A(0)=1, B(0)=0 and C(0)=0. Let B(n+1) = Sum_{k = 0..n} binomial(n,k)*A(k), C(n+1) = Sum_{k = 0..n} binomial(n,k)*B(k) and A(n+1) = Sum_{k = 0..n} binomial(n,k)*C(k). This entry gives the sequence A(n).
%C A143815 Compare with A024429 and A024430.
%C A143815 This sequence and its companion sequences B(n) = A143816(n) and C(n) = A143817(n) may be viewed as generalizations of the Bell numbers A000110. Define a sequence R(n) of real numbers by R(n) = Sum_{k >= 0} (3*k)^n/(3*k)! for n = 0, 1, 2, .... It is easy to verify that this sequence satisfies the recurrence relation u(n+3) = 3*u(n+2) - 2*u(n+1) + Sum_{i = 0..n} binomial(n, i)*3^(n-i)*u(i). Hence R(n) is an integral linear combination of R(0), R(1) and R(2). Some examples are given below.
%C A143815 To find the precise form of the linear relation define two other sequences of real numbers by S(n) = Sum_{k >= 0} (3*k+1)^n/(3*k+1)! and T(n) = Sum_{k >= 0} (3*k+2)^n/(3*k+2)! for n = 0, 1, 2, .... Both S(n) and T(n) satisfy the above recurrence. Then by means of the identities S(n+1) = Sum_{i = 0..n} binomial(n, i)*R(i), T(n+1) = Sum_{i = 0..n} binomial(n, i)*S(i) and R(n+1) = Sum_{i = 0..n} binomial(n, i)*T(i) we obtain the result R(n) = A(n)*R(0) + (B(n) - C(n))*R(1) + C(n)*R(2) = A(n)*R(0) + B(n)*R(1) + C(n)*(R(2) - R(1)) (with corresponding expressions for S(n) and T(n)). This generalizes the Dobinski's relation for the Bell numbers Sum_{k >= 0} k^n/k! = A000110(n)*exp(1).
%C A143815 Some examples of R(n) as a linear combination of R(0), R(1) and R(2) - R(1) are given below. The decimal expansions of R(0) = 1 + 1/3! + 1/6! + 1/9! + ..., R(2) - R(1) = 1/1! + 1/4! + 1/7! + ... and R(1) = 1/2! + 1/5! + 1/8! + ... may be found in A143819, A143820 and A143821 respectively. Compare with A143628 through A143631.
%C A143815 For n > 0, the number of partitions of {1,2,...,n} into 3,6,9,... classes. - _Geoffrey Critzer_, Mar 05 2010
%H A143815 Alois P. Heinz, <a href="/A143815/b143815.txt">Table of n, a(n) for n = 0..576</a>
%H A143815 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BellPolynomial.html">Bell Polynomial</a>.
%F A143815 a(n) = Sum_{k = 0..floor(n/3)} Stirling2(n, 3*k).
%F A143815 Let w = exp(2*Pi*i/3) and set F(x) = (exp(x) + exp(w*x) + exp(w^2*x))/3 = 1 + x^3/3! + x^6/6! + ... . Then the e.g.f. for the sequence is F(exp(x) - 1).
%F A143815 A143815(n) + A143816(n) + A143817(n) = Bell(n).
%F A143815 E.g.f. is B(A(x)) where A(x) = exp(x) - 1 and B(x) = (1/3)*(exp(x) + 2*exp(-x/2)*cos(sqrt(3)*x/2)). - _Geoffrey Critzer_, Mar 05 2010
%F A143815 a(n) = ( Bell_n(1) + Bell_n(w) + Bell_n(w^2) )/3, where Bell_n(x) is n-th Bell polynomial and w = exp(2*Pi*i/3). - _Seiichi Manyama_, Oct 13 2022
%F A143815 a(n) ~ n^n / (3 * (LambertW(n))^n * exp(n+1-n/LambertW(n)) * sqrt(1+LambertW(n))). - _Vaclav Kotesovec_, Jun 10 2025
%e A143815 R(n) as a linear combination of R(i), i = 0..2.
%e A143815   ==================================
%e A143815   R(n)  |     R(0)    R(1)    R(2)
%e A143815   ==================================
%e A143815   R(3)  |       1      -2       3
%e A143815   R(4)  |       6      -5       7
%e A143815   R(5)  |      25      -5      16
%e A143815   R(6)  |      91      20      46
%e A143815   R(7)  |     322     149     203
%e A143815   R(8)  |    1232     552    1178
%e A143815   R(9)  |    5672     991    7242
%e A143815   R(10) |   32202   -3799   43786
%e A143815   ...
%e A143815 Column 2 of the above table is A143818.
%e A143815 R(n) as a linear combination of R(0),R(1) and R(2) - R(1).
%e A143815   =====================================
%e A143815   R(n)  |     R(0)     R(1)   R(2)-R(1)
%e A143815   =====================================
%e A143815   R(3)  |       1        1        3
%e A143815   R(4)  |       6        2        7
%e A143815   R(5)  |      25       11       16
%e A143815   R(6)  |      91       66       46
%e A143815   R(7)  |     322      352      203
%e A143815   R(8)  |    1232     1730     1178
%e A143815   R(9)  |    5672     8233     7242
%e A143815   R(10) |   32202    39987    43786
%e A143815   ...
%p A143815 # (1)
%p A143815 M:=24: a:=array(0..100): b:=array(0..100): c:=array(0..100):
%p A143815 a[0]:=1: b[0]:=0: c[0]:=0:
%p A143815 for n from 1 to M do
%p A143815 b[n]:=add(binomial(n-1,k)*a[k], k=0..n-1);
%p A143815 c[n]:=add(binomial(n-1,k)*b[k], k=0..n-1);
%p A143815 a[n]:=add(binomial(n-1,k)*c[k], k=0..n-1);
%p A143815 end do:
%p A143815 A143815:=[seq(a[n], n=0..M)];
%p A143815 # (2)
%p A143815 seq(add(Stirling2(n,3*i),i = 0..floor(n/3)), n = 0..24);
%p A143815 # third Maple program:
%p A143815 b:= proc(n, t) option remember; `if`(n=0, irem(t, 2),
%p A143815       add(b(n-j, irem(t+1, 3))*binomial(n-1, j-1), j=1..n))
%p A143815     end:
%p A143815 a:= n-> b(n, 1):
%p A143815 seq(a(n), n=0..25);  # _Alois P. Heinz_, Feb 20 2018
%t A143815 a = Exp[x] - 1; f[x_] := 1/3 (E^x + 2 E^(-x/2) Cos[(Sqrt[3] x)/2]); CoefficientList[Series[f[a], {x, 0, 25}], x]*Table[n!, {n, 0, 25}] (* _Geoffrey Critzer_, Mar 05 2010 *)
%o A143815 (PARI) Bell_poly(n, x) = exp(-x)*suminf(k=0, k^n*x^k/k!);
%o A143815 a(n) = my(w=(-1+sqrt(3)*I)/2); round(Bell_poly(n, 1)+Bell_poly(n, w)+Bell_poly(n, w^2))/3; \\ _Seiichi Manyama_, Oct 13 2022
%Y A143815 Cf. A000110, A024429, A024430, A143628, A143629, A143630, A143631, A143816, A143817, A143818, A143819, A143820, A143821.
%K A143815 easy,nonn
%O A143815 0,5
%A A143815 _Peter Bala_, Sep 03 2008