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A225844
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Least k>0 such that triangular(n) + k*(k+1) is a triangular number.
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0
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2, 1, 3, 5, 7, 2, 11, 13, 5, 17, 19, 3, 6, 25, 27, 9, 31, 33, 35, 4, 9, 41, 8, 45, 47, 10, 14, 53, 9, 5, 59, 61, 21, 18, 67, 69, 21, 73, 75, 14, 22, 6, 11, 13, 87, 15, 91, 26, 20, 34, 12, 101, 26, 105, 30, 7, 20, 33, 115, 117, 119, 34, 21, 125, 37, 129, 29, 133, 14, 137
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OFFSET
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0,1
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COMMENTS
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For n>0, a(n) <= 2*n-1, because n*(n+1)/2 + (2*n-1)*2*n = (9*n^2 - 3*n)/2 = 3*n*(3*n-1)/2 = triangular(3*n-1).
The subsequence with terms less than 2*n-1 begins: 2, 5, 3, 6, 9, 4, 9, 8, 10, 14, 9, 5, 21, 18, 21, 14, 22, 6, 11, 13, 15, ...
The sequence of n's such that a(n) < 2*n-1 begins: 5, 8, 11, 12, 15, 19, 20, 22, 25, 26, ...
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LINKS
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MATHEMATICA
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lktrno[n_]:=Module[{t=(n(n+1))/2, k=1}, While[!IntegerQ[(Sqrt[ 8(t+k(k+1))+1]-1)/2], k++]; k]; Array[lktrno, 70, 0] (* Harvey P. Dale, Aug 19 2014 *)
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PROG
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(Python)
def isTriangular(a):
sr = 1L << (long.bit_length(long(a)) >> 1)
a += a
while a < sr*(sr+1): sr>>=1
b = sr>>1
while b:
s = sr+b
if a >= s*(s+1): sr = s
b>>=1
return (a==sr*(sr+1))
n = tn = 0
while 1:
for m in range(1, 1000000000):
if isTriangular(tn + m*(m+1)): break
print str(m)+', ',
n += 1
tn += n
(PARI) a(n)=for(k=1, 2*n, t=n*(n+1)/2+k*(k+1); x=sqrtint(2*t); if(t==x*(x+1)/2, return(k))) /* from Ralf Stephan */
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CROSSREFS
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Cf. A101157 (least k>0 such that triangular(n) + k^2 is a triangular number).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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