OFFSET
0,1
COMMENTS
tri(n) = n*(n+1)/2 is the n-th triangular number, A000217(n).
a(n) is how many consecutive triangular numbers starting from tri(n) are needed to sum up to tri(x) for some x. The requirement a(n) > 1 is needed, because otherwise all a(n) = 1.
Because an oblong number (A002378) is twice a triangular number, this sequence is also the least k > 1 such that oblong(n) + ... + oblong(n+k-1) is an oblong number.
a(n) is least k > 1 such that 12*k^3 + 36*n*k^2 + 36*k*n^2 - 12*k + 9 is a perfect square. - Chai Wah Wu, Mar 01 2016
a(n) <= 3*n^2 - 3*n - 1 for n > 1, since 12*k^3 + 36*n*k^2 + 36*k*n^2 - 12*k + 9 is a square when k = 3*n^2 - 3*n - 1. - Robert Israel, Mar 03 2016
LINKS
Chai Wah Wu, Table of n, a(n) for n = 0..5000
EXAMPLE
0+1 = 1 is a triangular number, two summands, so a(0)=2.
1+3+6 = 10 is a triangular number, three summands, so a(1)=3.
3+6+10+15+21 = 55 is a triangular number, five summands, so a(2)=5.
Starting from Triangular(5)=15: 15+21=36 is a triangular number, two summands, so a(5)=2.
MAPLE
f:= proc(n) local k;
for k from 2 do if issqr(12*k^3+36*k^2*n+36*k*n^2-12*k+9) then return k fi od
end proc:
map(f, [$0..100]); # Robert Israel, Mar 03 2016
MATHEMATICA
triQ[n_] := IntegerQ[Sqrt[1+8*n]]; Table[k = n+1; s = k^2; While[! triQ[s], k++; s = s + k*(k+1)/2]; k - n + 1, {n, 0, 55}] (* T. D. Noe, Jul 26 2012 *)
PROG
(Python)
for n in range(77):
i = ti = n
sum = 0
tn_gte_sum = 0 # least oblong number >= sum
while i-n<=1 or tn_gte_sum!=sum:
sum += i*(i+1)
i+=1
while tn_gte_sum<sum:
tn_gte_sum = ti*(ti+1)
ti+=1
print i-n,
(Python)
from math import sqrt
def A214697(n):
k, a1, a2, m = 2, 36*n, 36*n**2 - 12, n*(72*n + 144) + 81
while int(round(sqrt(m)))**2 != m:
k += 1
m = k*(k*(12*k + a1) + a2) + 9
return k # Chai Wah Wu, Mar 01 2016
CROSSREFS
KEYWORD
nonn,look
AUTHOR
Alex Ratushnyak, Jul 26 2012
STATUS
approved