A267140 Least m>0 for which m + n^2 is a square and m + triangular(n) is a triangular number (A000217).

1, 35, 12, 72, 180, 336, 45, 792, 1092, 208, 1836, 2280, 112, 315, 3900, 4536, 644, 5952, 6732, 7560, 225, 715, 10332, 627, 12420, 13536, 924, 1575, 17172, 840, 396, 21240, 22692, 3267, 2565, 27336, 28980, 3392, 32412, 34200, 1881, 3795, 637, 1400, 1785, 45936, 2240

Offset

0,2

Comments

For n>1, a(n) <= b(n), where b(n) = 24*n^2 - 60*n + 36 = A143698(n-1), because b(n) + n^2 = (5*n-6)^2, and b(n) + n*(n+1)/2 = (7*n-9)*(7*n-8)/2 = triangular(7*n-9).

Links

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Example

12 + 2^2 = 16 is a square, and 12 + 2*3/2 = 15 is a triangular number, and 12 is the least such integer, so a(2)=12.

Mathematica

lmst[n_]:=Module[{m=1, n2=n^2, nt=(n(n+1))/2}, While[ !IntegerQ[Sqrt[ m+n2]] || !OddQ[Sqrt[1+8(m+nt)]], m++]; m]; Join[{1}, Array[lmst, 50]] (* Harvey P. Dale, Aug 15 2021 *)

Prog

(PARI) a(n) = {m = 1; while (! (issquare(m+n^2) && ispolygonal(m+n*(n+1)/2, 3)), m++); m; } \\ Michel Marcus, Jan 11 2016
(Python)
from math import sqrt
def A267140(n):
    u, r, k, m = 2*n+1, 4*n*(n+1)+1, 0, 2*n+1
    while True:
        if int(sqrt(8*m+r))**2 == 8*m+r:
            return m
        k += 2
        m += u + k # Chai Wah Wu, Jan 13 2016

Crossrefs

Cf. A000217, A000290, A143698, A267077.

Sequence in context: A248477 A051050 A267198 * A348893 A088830 A033355

Adjacent sequences: A267137 A267138 A267139 * A267141 A267142 A267143

Keyword

nonn

Author

Alex Ratushnyak, Jan 10 2016

Status

approved