A266883 Numbers of the form m*(4*m+1)+1, where m = 0,-1,1,-2,2,-3,3,...

1, 4, 6, 15, 19, 34, 40, 61, 69, 96, 106, 139, 151, 190, 204, 249, 265, 316, 334, 391, 411, 474, 496, 565, 589, 664, 690, 771, 799, 886, 916, 1009, 1041, 1140, 1174, 1279, 1315, 1426, 1464, 1581, 1621, 1744, 1786, 1915, 1959, 2094, 2140, 2281, 2329, 2476, 2526

Offset: 0

Author

Bruno Berselli, Jan 05 2016

Keywords

• nonn
• easy

Comments

Also, numbers m such that 16*m-15 is a square. Therefore, the terms 1 and 4 are the only squares in this sequence.

Conjecture: the sequence terms are the exponents in the expansion of Sum_{n >= 1} q^n * (Product_{k >= 2*n-1} 1 - q^k) = q + q^4 + q^6 + q^15 + q^19 + q^34 + .... Cf. A174114. - Peter Bala, May 10 2025

Links

• Bruno Berselli, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Cf. A001057, A049486, A127365, A130472.

Cf. A002061: m*(4*m+2)+1 for m = 0,0,-1,1,-2,2,-3,3, ...

Cf. A174114: m*(4*m+3)+1 for m = 0,-1,1,-2,2,-3,3,-4,4, ...

Cf. A054556: m*(4*m+1)+1 for nonpositive m.

Cf. A054567: m*(4*m+1)+1 for nonnegative m.

Cf. A074378: numbers m such that 16*m+1 is a square.

Programs

Magma

[n*(n+1)+1-((2*n+1)*(-1)^n-1)/4: n in [0..50]];

Magma

I:=[1,4,6,15,19]; [n le 5 select I[n] else Self(n-1) + 2*Self(n-2) -2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..60]]; // Vincenzo Librandi, Jan 06 2016

Mathematica

Table[n (n + 1) + 1 - ((2 n + 1) (-1)^n - 1)/4, {n, 0, 50}]
LinearRecurrence[{1, 2, -2, -1, 1}, {1, 4, 6, 15, 19}, 60] (* Vincenzo Librandi, Jan 06 2016 *)

PARI

vector(50, n, n--; n*(n+1)+1-((2*n+1)*(-1)^n-1)/4)

PARI

Vec((1+3*x+3*x^3+x^4)/((1+x)^2*(1-x)^3) + O(x^100)) \\ Altug Alkan, Jan 06 2016

Python

[n*(n+1)+1-((2*n+1)*(-1)**n-1)/4 for n in range(60)]

Sage

[n*(n+1)+1-((2*n+1)*(-1)^n-1)/4 for n in range(50)]

Formula

O.g.f.: (1 + 3*x + 3*x^3 + x^4)/((1 + x)^2*(1 - x)^3).

E.g.f.: (5 + 8*x + 4*x^2)*exp(x)/4 -(1 - 2*x)*exp(-x)/4.

a(n) = a(-n-1) = n*(n + 1) + 1 - ((2*n + 1)*(-1)^n - 1)/4 = (2*n + 1)*floor((n + 1)/2) + 1.

a(n) = A002061(n+1) + A001057(n) = A074378(n)+1.

a(n+1) + a(n+2) = A049486(n+3).