cp's OEIS Frontend

Compare with A024429 and A024430.

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.

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).

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.

For n > 0, the number of partitions of {1,2,...,n} into 3,6,9,... classes. - Geoffrey Critzer, Mar 05 2010

Compare with A024429 and A024430.

This sequence and its companion sequences A(n) = A143815 and B(n) = A143816 may be viewed as generalizations of the Bell numbers A000110. Define R(n) = Sum_{k >= 0} (3k)^n/(3k)! for n = 0,1,2,.... Then the real number R(n) is an integral linear combination of R(0) = 1 + 1/3! + 1/6! + ...., R(2) - R(1) = 1/1! + 1/4! + 1/7! + ... and R(1) = 1/2! + 1/5! + 1/8! + ... . Some examples are given below. The precise result is R(n) = A(n)*R(0) + B(n)*R(1) + C(n)*(R(2)-R(1)). This generalizes the Dobinski relation for the Bell numbers: Sum_{k >= 0} k^n/k! = A000110(n)*exp(1). See A143815 for more details. Compare with A143628 through A143631. The decimal expansions of R(0), R(2) - R(1) and R(1) may be found in A143819, A143820 and A143821 respectively.

a(n) equals (-1)^n times the determinant of the (3n+1) X (3n+1) matrix with consecutive integers from 1 to 3n+1 along the main diagonal, consecutive integers from 2 to 3n+1 along the superdiagonal, consecutive integers from 1 to 3n along the subdiagonal, and 1's everywhere else (see Mathematica code below). - John M. Campbell, Jul 12 2011

Compare with A024429 and A024430.

This sequence and its companion sequences A(n) = A143815 and C(n) = A143817 may be viewed as generalizations of the Bell numbers A000110. Define R(n) = Sum_{k >= 0} (3k)^n/(3k)! for n = 0,1,2,.... Then the real number R(n) is an integral linear combination of R(0) = 1 + 1/3! + 1/6! + ...., R(2) - R(1) = 1/1! + 1/4! + 1/7! + ... and R(1) = 1/2! + 1/5! + 1/8! + .... Some examples are given below. The precise result is R(n) = A(n)*R(0) + B(n)*R(1) + C(n)*(R(2)-R(1)). This generalizes the Dobinski relation for the Bell numbers: Sum_{k >= 0} k^n/k! = A000110(n)*exp(1). See A143815 for more details. Compare with A143628 through A143631. The decimal expansions of R(0), R(2) - R(1) and R(1) may be found in A143819, A143820 and A143821 respectively.

Define a sequence R(n) of real numbers by R(n) := Sum_{k >= 0} (3*k)^n/(3*k)! for n = 0,1,2,... . This constant is R(2) - R(1); the decimal expansions of R(0) = 1 + 1/3! + 1/6! + ... and R(1) = 1/2! + 1/5! + 1/8! + ... may be found in A143819 and A143821. It is easy to verify that the sequence R(n) 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) and so also an integral linear combination of R(0), R(1) and R(2) - R(1).

R(n) as a linear combination of R(0), R(1) and R(2) - R(1).

========================================

| linear combination of

R(n) | R(0) R(1) R(2) - R(1)

========================================

R(3) | 1 1 3

R(4) | 6 2 7

R(5) | 25 11 16

R(6) | 91 66 46

R(7) | 322 352 203

R(8) | 1232 1730 1178

R(9) | 5672 8233 7242

R(10) | 32202 39987 43786

...

The column entries are from A143815, A143816 and A143817.

The Abraham Ungar 1982 article defines H_{n,r}(z) = Sum_{k>=0} z^(nk+r)/(nk+r)! as equation (1). The constant is H_{3,1}(1). In equation (13) H_{3,1}(x) = (exp(x) + 2 * exp(-x/2) * cos(sqrt(3)/2*x - 2*Pi/3))/3. In equation (12) the expression H_{3,1}(x) = (e^x + q_2 e^{q_1 x} + q_1 e^{q_2 x})/3 where q_1 = (-1 + I sqrt(3))/2 and q_2 = (-1 - I sqrt(3))/2 is given for H_{3,2}(x) instead. - Michael Somos, Nov 01 2024

For q integer >= 1, Sum_{m>=0} 1/(q*m)! = (1/q) * Sum_{k=1..q} exp(X_k) where X_1, X_2, ..., X_q are the q-th roots of unity.

Define a sequence of real numbers R(n) by R(n) := Sum_{k >= 0} (3*k)^n/(3*k)! for n = 0,1,2... . This constant is R(1); the decimal expansions of R(0) = 1 + 1/3!+ 1/6! + 1/9! + ... and R(2) - R(1) = 1/1! + 1/4! + 1/7! + ... may be found in A143819 and A143820. It is easy to verify that the sequence R(n) 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) and so also an integral linear combination of R(0), R(1) and R(2) - R(1). Some examples are given below.

From Vaclav Kotesovec, Feb 24 2016: (Start)

Sum_{k>=0} 1/k! = A001113 = exp(1).

Sum_{k>=0} 1/(2k)! = A073743 = cosh(1).

Sum_{k>=0} 1/(3k)! = A143819 = (2*cos(sqrt(3)/2)*exp(-1/2) + exp(1))/3.

Sum_{k>=0} 1/(4k)! = (cos(1) + cosh(1))/2 = 1.0416914703416917479394211141...

(End)

For q integer >= 1, Sum_{m>=0} 1/(q*m)! = (1/q) * Sum_{k=1..q} exp(X_k) where X_1, X_2, ..., X_q are the q-th roots of unity. - Bernard Schott, Mar 02 2020

Continued fraction: 1 + 1/(120 - 120/(30241 - 30240/(360361 - ... - P(n-1)/((P(n) + 1) - ... )))), where P(n) = (5*n)*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4) for n >= 1. Cf. A346441. - Peter Bala, Feb 22 2024

For q integer >= 1, Sum_{m>=0} 1/(q*m)! = (1/q) * Sum_{k=1..q} exp(X_k) where X_1, X_2, ..., X_q are the q-th roots of unity.