A214771 a(n) is the smallest number that can be written as the sum of consecutive positive integers in at least n ways.
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
Offset: 1
Keywords
Examples
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.
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.
Programs
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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 *)
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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
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
Extensions
Definition corrected by Jonathan Sondow, Feb 19 2014
Comments