

A214771


a(n) is the smallest number that can be written as the sum of consecutive positive integers in at least n ways.


3



3, 9, 15, 45, 45, 105, 105, 225, 315, 315, 315, 945, 945, 945, 945, 1575, 1575, 2835, 2835, 3465, 3465, 3465, 3465, 10395, 10395, 10395, 10395, 10395, 10395, 10395, 10395, 17325, 17325, 17325, 17325, 31185, 31185, 31185, 31185, 45045, 45045, 45045, 45045
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OFFSET

1,1


COMMENTS

The number of forms of writing a number x with odd prime factors as distinct sums of at least two nonzero summands of consecutive positive integers is: d(2x)/2 1 = d(x)  1, where d(x) is the number of divisors of x.


LINKS



FORMULA



EXAMPLE

a(1) = 3 = 1+2;
a(2) = 9 = 4+5 = 2+3+4;
a(3) = 15 = 7+8 = 4+5+6 = 1+2+3+4+5;
a(4) = a(5) = 45 is the sum of 2,3,5,6 and 9 consecutive integers beginning with 22, 14, 7, 5 and 1 respectively.


MATHEMATICA

nn = 50000; t = Table[0, {nn}]; Do[tot = i; j = i; While[j++; tot = tot + j; tot <= nn, t[[tot]]++], {i, nn/2  1}]; Table[Position[t, _?(# >= n &), 1, 1][[1, 1]], {n, Max[t]}] (* T. D. Noe, Jul 28 2012 *)


PROG

(Python)
import heapq
from itertools import islice
def agen(): # generator of terms
p = v = 3; h = [(v, 1, 2)]; nextcount = 3; oldv = ways = highways = 0
while True:
(v, s, l) = heapq.heappop(h)
if v == oldv: ways += 1
else:
if ways > highways:
for n in range(highways+1, ways+1):
yield oldv
highways = ways
ways = 1
if v >= p:
p += nextcount
heapq.heappush(h, (p, 1, nextcount))
nextcount += 1
oldv = v
v = s; s += 1; l += 1; v += l
heapq.heappush(h, (v, s, l))


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



