### Citations

351 | Tableaux, with Applications to Representation Theory and Geometry - Young - 1997 |

190 | Advanced combinatorics: The art of finite and infinite expansions - Comtet - 1974 |

5 | Hyper-sums of powers of integers and the Akiyama-Tanigawa matrix,
- Inaba
- 2005
(Show Context)
Citation Context ...of integers put forward recently, we develop an iterative procedure which allows us to express a hypersum of arbitrary order in terms of ordinary (zeroth order) power sums. Then, we derive the coefficients of the hypersum polynomial as a function of the Bernoulli numbers and the Stirling numbers of the first kind. 1 Introduction For every integer m, m ≥ 1, the hypersums of powers of integers are defined recursively as follows: P (m) k (n) = ∑n j=1 P (m−1) k (j), where P (0) k (n) is the sum of the first n positive integers each raised to the integer power k ≥ 0, P (0) k (n) = 1 k+2k+ · · ·+nk [1, 2]. The latter is given by a polynomial in n of degree k+1 with zero constant term. Hence, inductively P (m) k (n) is given by a polynomial in n of degree k +m+ 1 with zero constant term: P (m) k (n) = k+m+1 ∑ r=1 crk,mn r. (1) An explicit formula for the coefficients crk,m involving the Stirling numbers of the first and second kinds has been given by the author [3]. In this paper (Section 2), by an iterative procedure, we obtain a new representation of the m-th order hypersum P (m) k (n) in terms of 1 ordinary (zeroth order) power sums. Specifically, we will show that P (m) k (n) can be express... |

2 | Johann Faulhaber and sums of powers,
- Knuth
- 1993
(Show Context)
Citation Context ...of integers put forward recently, we develop an iterative procedure which allows us to express a hypersum of arbitrary order in terms of ordinary (zeroth order) power sums. Then, we derive the coefficients of the hypersum polynomial as a function of the Bernoulli numbers and the Stirling numbers of the first kind. 1 Introduction For every integer m, m ≥ 1, the hypersums of powers of integers are defined recursively as follows: P (m) k (n) = ∑n j=1 P (m−1) k (j), where P (0) k (n) is the sum of the first n positive integers each raised to the integer power k ≥ 0, P (0) k (n) = 1 k+2k+ · · ·+nk [1, 2]. The latter is given by a polynomial in n of degree k+1 with zero constant term. Hence, inductively P (m) k (n) is given by a polynomial in n of degree k +m+ 1 with zero constant term: P (m) k (n) = k+m+1 ∑ r=1 crk,mn r. (1) An explicit formula for the coefficients crk,m involving the Stirling numbers of the first and second kinds has been given by the author [3]. In this paper (Section 2), by an iterative procedure, we obtain a new representation of the m-th order hypersum P (m) k (n) in terms of 1 ordinary (zeroth order) power sums. Specifically, we will show that P (m) k (n) can be express... |

1 | Generalized Akiyama-Tanigawa algorithm for hypersums of powers of integers,
- Cereceda
- 2013
(Show Context)
Citation Context ...ums of powers of integers are defined recursively as follows: P (m) k (n) = ∑n j=1 P (m−1) k (j), where P (0) k (n) is the sum of the first n positive integers each raised to the integer power k ≥ 0, P (0) k (n) = 1 k+2k+ · · ·+nk [1, 2]. The latter is given by a polynomial in n of degree k+1 with zero constant term. Hence, inductively P (m) k (n) is given by a polynomial in n of degree k +m+ 1 with zero constant term: P (m) k (n) = k+m+1 ∑ r=1 crk,mn r. (1) An explicit formula for the coefficients crk,m involving the Stirling numbers of the first and second kinds has been given by the author [3]. In this paper (Section 2), by an iterative procedure, we obtain a new representation of the m-th order hypersum P (m) k (n) in terms of 1 ordinary (zeroth order) power sums. Specifically, we will show that P (m) k (n) can be expressed as a linear combination of P (0) k (n), P (0) k+1(n), . . . , P (0) k+m(n), as follows: P (m) k (n) = m ∑ i=0 (−1)iqm,i(n) m! P (0) k+i(n), (2) where qm,i(n) is a polynomial in n of degree m− i. (Note that formula (2) holds for m = 0 if we set q0,0(n) = 1.) In Section 3, we determine the explicit form of the coefficients of qm,i(n). Then, using (2), we obtain t... |