A232179 Least k >= 0 such that n^2 + triangular(k) is a triangular number.
0, 0, 3, 1, 15, 2, 0, 3, 63, 4, 8, 5, 11, 6, 20, 3, 255, 8, 1, 9, 3, 10, 38, 11, 59, 12, 45, 13, 8, 14, 2, 15, 1023, 16, 59, 0, 24, 18, 66, 19, 51, 20, 3, 21, 44, 10, 80, 23, 251, 24, 42, 25, 68, 26, 4, 27, 39, 28, 101, 29, 10, 30, 108, 8, 4095, 32, 5, 33, 128
Offset: 0
Keywords
Links
- Chai Wah Wu, Table of n, a(n) for n = 0..10000
Crossrefs
Cf. A082183 (least k>0 such that triangular(n) + triangular(k) is a triangular number).
Cf. A232177 (least k>0 such that triangular(n) + triangular(k) is a square).
Cf. A232176 (least k>0 such that n^2 + triangular(k) is a square).
Cf. A101157 (least k>0 such that triangular(n) + k^2 is a triangular number).
Cf. A232178 (least k>=0 such that triangular(n) + k^2 is a square).
Programs
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Mathematica
TriangularQ[n_] := IntegerQ[Sqrt[1 + 8*n]]; Table[k = 0; While[! TriangularQ[n^2 + k*(k + 1)/2], k++]; k, {n, 0, 68}] (* T. D. Noe, Nov 21 2013 *)
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PARI
a(n) = {my(k = 0); while (! ispolygonal(n^2 + k*(k+1)/2, 3), k++); k;} \\ Michel Marcus, Sep 15 2017
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Python
from _future_ import division from sympy import divisors def A232179(n): if n == 0: return 0 t = 2*n**2 ds = divisors(t) for i in range(len(ds)//2-1,-1,-1): x = ds[i] y = t//x a, b = divmod(y-x,2) if b: return a return -1 # Chai Wah Wu, Sep 12 2017
Formula
a(A001109(n)) = 0.
Comments