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
Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..200 from T. D. Noe; terms 576 onward using A053624)
Zach Wissner-Gross, The Riddler, Solution to last week's Riddler Express, FiveThirtyEight, Feb 18 2022.
FORMULA
a(n) = A053624(i) for n in d(A053624(i-1))..d(A053624(i))-1, where d(x) is the number of divisors of x. - Michael S. Branicky, Feb 18 2022
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))
print(list(islice(agen(), 50))) # Michael S. Branicky, Feb 18 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Robin Garcia, Jul 27 2012
EXTENSIONS
Definition corrected by Jonathan Sondow, Feb 19 2014
STATUS
approved