A214771 - OEIS

%I #32 Feb 18 2022 22:48:46

%S 3,9,15,45,45,105,105,225,315,315,315,945,945,945,945,1575,1575,2835,

%T 2835,3465,3465,3465,3465,10395,10395,10395,10395,10395,10395,10395,

%U 10395,17325,17325,17325,17325,31185,31185,31185,31185,45045,45045,45045,45045

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

%C 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.

%H Michael S. Branicky, <a href="/A214771/b214771.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..200 from T. D. Noe; terms 576 onward using A053624)

%H Zach Wissner-Gross, The Riddler, <a href="https://fivethirtyeight.com/features/can-you-find-the-luckiest-coin/">Solution to last week's Riddler Express</a>, FiveThirtyEight, Feb 18 2022.

%F 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

%e a(1) = 3 = 1+2;

%e a(2) = 9 = 4+5 = 2+3+4;

%e a(3) = 15 = 7+8 = 4+5+6 = 1+2+3+4+5;

%e 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.

%t 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 *)

%o (Python)

%o import heapq

%o from itertools import islice

%o def agen(): # generator of terms

%o     p = v = 3; h = [(v, 1, 2)]; nextcount = 3; oldv = ways = highways = 0

%o     while True:

%o         (v, s, l) = heapq.heappop(h)

%o         if v == oldv: ways += 1

%o         else:

%o             if ways > highways:

%o                 for n in range(highways+1, ways+1):

%o                     yield oldv

%o                 highways = ways

%o             ways = 1

%o         if v >= p:

%o             p += nextcount

%o             heapq.heappush(h, (p, 1, nextcount))

%o             nextcount += 1

%o         oldv = v

%o         v -= s; s += 1; l += 1; v += l

%o         heapq.heappush(h, (v, s, l))

%o print(list(islice(agen(), 50))) # _Michael S. Branicky_, Feb 18 2022

%Y Cf. A053624 (union of these terms), A057716 (not powers of 2).

%K nonn

%O 1,1

%A _Robin Garcia_, Jul 27 2012

%E Definition corrected by _Jonathan Sondow_, Feb 19 2014