A064538 a(n) is the smallest positive integer such that a(n)*(1^n + 2^n + ... + x^n) is a polynomial in x with integer coefficients.

1, 2, 6, 4, 30, 12, 42, 24, 90, 20, 66, 24, 2730, 420, 90, 48, 510, 180, 3990, 840, 6930, 660, 690, 720, 13650, 1092, 378, 56, 870, 60, 14322, 7392, 117810, 7140, 210, 72, 1919190, 103740, 8190, 1680, 94710, 13860, 99330, 9240, 217350, 9660, 9870, 10080, 324870

Offset: 0

Author

Floor van Lamoen, Oct 08 2001

Keywords

• nonn
• nice
• look
• easy

Comments

a(n) is a multiple of n+1. - Vladimir Shevelev, Dec 20 2011

Let P_n(x) = 1^n + 2^n + ... + x^n = Sum_{i=1..n+1}c_i*x^i. Let P^*n(x) = Sum{i=1..n+1}(c_i/(i+1))*(x^(i+1)-x). Then b(n) = (n+1)*a(n+1)is the smallest positive integer such that b(n)*P^*n(x) is a polynomial with integer coefficients. Proof follows from the recursion P(n+1)(x) = x + (n+1)*P^*n(x). As a corollary, note that, if p is the maximal prime divisor of a(n), then p<=n+1. - _Vladimir Shevelev, Dec 21 2011

The recursion P_(n+1)(x) = x + (n+1)*P^*n(x) is due to Abramovich (1973); see also Shevelev (2007). - _Jonathan Sondow, Nov 16 2015

The sum S_m(n) = Sum_{k=0..n} k^m can be written as S_m(n) = n(n+1)(2n+1)P_m(n)/a(m) for even m>1, or S_m(n) = n^2*(n+1)^2*P_m(n)/a(m) for odd m>1, where a(m) is the LCM of the denominators of the coefficients of the polynomial P_m/a(m), i.e., the smallest integer such that P_m defined in this way has integer coefficients. (Cf. Michon link.) - M. F. Hasler, Mar 10 2013

a(n)/(n+1) is squarefree, by Faulhaber's formula and the von Staudt-Clausen theorem on the denominators of Bernoulli numbers. - Kieren MacMillan and Jonathan Sondow, Nov 20 2015

a(n) equals n+1 times the product of the primes p <= (n+2)/(2+(n mod 2)) such that the sum of the base-p digits of n+1 is at least p. - Bernd C. Kellner and Jonathan Sondow, May 24 2017

Examples

			1^3 + 2^3 + ... + x^3 = (x(x+1))^2/4 so a(3)=4.
1^4 + 2^4 + ... + x^4 = x(x+1)(2x+1)(3x^2+3x-1)/30, therefore a(4)=30.
		

References

• M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprints), p. 804, Eq. 23.1.4.

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..10000 (n = 0..1000 from T. D. Noe)
• V. S. Abramovich, Power sums of natural numbers, Kvant 5 (1973), 22-25. (in Russian)
• M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
• Bernd C. Kellner, On a product of certain primes, J. Number Theory, 179 (2017), 126-141; arXiv:1705.04303 [math.NT], 2017.
• Bernd C. Kellner, On the finiteness of Bernoulli polynomials whose derivative has only integral coefficients, J. Integer Seq. 27 (2024), Article 24.2.8, 11 pp.; arXiv:2310.01325 [math.NT], 2023.
• Bernd C. Kellner and Jonathan Sondow, Power-Sum Denominators, Amer. Math. Monthly, 124 (2017), 695-709; arXiv:1705.03857 [math.NT], 2017.
• Bernd C. Kellner and Jonathan Sondow, The denominators of power sums of arithmetic progressions, Integers 18 (2018), #A95, 17 pp.; arXiv:1705.05331 [math.NT], 2017.
• Bernd C. Kellner and Jonathan Sondow, On Carmichael and polygonal numbers, Bernoulli polynomials, and sums of base-p digits, Integers 21 (2021), #A52, 21 pp.; arXiv:1902.10672 [math.NT], 2019.
• Meng Fai Lim, On the p-divisibility of even K-groups of the ring of integers of a cyclotomic field, arXiv:2308.04099 [math.NT], 2023.
• Dr. Math, Summing n^k.
• R. Mestrovic, A congruence modulo n^3 involving two consecutive sums of powers and its applications, arXiv:1211.4570 [math.NT], 2012.
• R. Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.
• G. Michon, Faulhaber's Formula on NUMERICANA.com.
• E. S. Rowland, Sums of Consecutive Powers
• Vladimir Shevelev, A Short Proof of a Known Relation for Consecutive Power Sums, arXiv:0711.3692 [math.CA], 2007.
• Eric Weisstein's World of Mathematics, Power Sum
• Wikipedia, Faulhaber's Formula.

Crossrefs

Cf. A144845, A195441, A256581, A286516, A286762, A286763, A324369, A324370, A324371.

Programs

Maple

A064538 := n -> denom((bernoulli(n+1,x)-bernoulli(n+1))/(n+1)): # Peter Luschny, Aug 19 2011
# Formula of Kellner and Sondow (2017):
a := proc(n) local s; s := (p,n) -> add(i,i=convert(n,base,p));
select(isprime,[$2..(n+2)/(2+irem(n,2))]);
(n+1)*mul(i,i=select(p->s(p,n+1)>=p,%)) end: seq(a(n), n=0..48); # Peter Luschny, May 14 2017

Mathematica

A064538[n_] := Denominator[ Together[ (BernoulliB[n+1, x] - BernoulliB[n+1])/(n+1)]];
Table[A064538[n], {n, 0, 44}] (* Jean-François Alcover, Feb 21 2012, after Maple *)

PARI

a(n) = {my(vp = Vec(bernpol(n+1, x)-bernfrac(n+1))/(n+1)); lcm(vector(#vp, k, denominator(vp[k])));} \\ Michel Marcus, Feb 07 2016

Python

from _future_ import division
from sympy.ntheory.factor_ import digits, nextprime
def A064538(n):
    p, m = 2, n+1
    while p <= (n+2)//(2+ (n% 2)):
        if sum(d for d in digits(n+1,p)[1:]) >= p:
            m *= p
        p = nextprime(p)
    return m # Chai Wah Wu, Mar 07 2018

Sage

A064538 = lambda n: (n+1)*mul([p for p in (2..(n+2)//(2+n%2)) if is_prime(p) and sum((n+1).digits(base=p)) >= p])
print([A064538(n) for n in (0..48)]) # Peter Luschny, May 14 2017

Formula

a(n) = (n+1)*A195441(n). - Jonathan Sondow, Nov 12 2015

A001221(a(n)/(n+1)) = A001222(a(n)/(n+1)). - Kieren MacMillan and Jonathan Sondow, Nov 20 2015

rad(a(n)) = A007947(a(n)) = A144845(n) = A324369(n+1) * A324370(n+1) * A324371(n+1). - Bernd C. Kellner, Oct 12 2023