A268539 Numbers k such that 48*k + 25 is a perfect square.

0, 2, 3, 7, 17, 25, 28, 38, 58, 72, 77, 93, 123, 143, 150, 172, 212, 238, 247, 275, 325, 357, 368, 402, 462, 500, 513, 553, 623, 667, 682, 728, 808, 858, 875, 927, 1017, 1073, 1092, 1150, 1250, 1312, 1333, 1397, 1507, 1575, 1598, 1668, 1788, 1862, 1887, 1963, 2093, 2173

Offset: 1

Author

N. J. A. Sloane, Feb 24 2016

Keywords

• nonn
• easy

Comments

Equivalently, integers of the form (h+5)*(h-5)/48, where h must be odd, h = 2*m+1, thus also integers of the form (m+3)*(m-2)/12, with m = 2, 5, 6, 9, 14, 17, 18, ... = {2, 5, 6, 9} + 12 N. - M. F. Hasler, Mar 02 2016

The sequence terms are the exponents in the expansion of Product_{n >= 1} (1 - q^(8*n))*(1 + q^(8*n-1))*(1 + q^(8*n-7))/(1 + q^n) = Sum_{n >= 0} q^(2*n*(n+1)) * Product_{k >= 2*n+2} 1 - q^k = 1 - q^2 - q^3 + q^7 + q^17 - q^25 - q^28 + + - - ... (by the quintuple product identity and Mc Laughlin et al., S.38, p 16). - Peter Bala, Dec 30 2024

Conjecture: the sequence terms are also the exponents in the expansion of Sum_{n >= 0} q^n/(Product_{k = 1..2*n+1} 1 + q^k) = 1 + q^2 - q^3 - q^7 + q^17 + q^25 - - + + .... - Peter Bala, Jan 15 2025

Links

• Zak Seidov, Table of n, a(n) for n = 1..10000
• J. Mc Laughlin, A. V. Sills and P. Zimmer, Rogers-Ramanujan-Slater Type Identities, Electronic J. Combinatorics, DS15, 1-59, May 31, 2008.
• Mircea Merca, The bisectional pentagonal number theorem, Journal of Number Theory, Volume 157, December 2015, Pages 223-232, see Corollary 4.4.
• Index entries for linear recurrences with constant coefficients, signature (3,-5,7,-7,5,-3,1).

Crossrefs

Cf. A078405, A154293.

Subsequence of A011865.

Programs

Magma

[n: n in [0..2200] | IsSquare(48*n+25)]; // Vincenzo Librandi, Feb 25 2016

Maple

L := [5, 11, 13, 19, 29, 35, 37, 43]:
seq(seq(((L[i]+48*j)^2-25)/48, i=1..8), j=0..10); # Robert Israel, Feb 29 2016

Mathematica

Select[Range[0, 2500], IntegerQ[Sqrt[48 # + 25]] &] (* Vincenzo Librandi, Feb 25 2016 *)
Table[(3 (n - 1) n + (2 n - 1) (-1)^((n - 2) (n - 1)/2) - 1)/4, {n, 1, 60}] (* Bruno Berselli, Feb 29 2016 *)
LinearRecurrence[{3, -5, 7, -7, 5, -3, 1}, {0, 2, 3, 7, 17, 25, 28}, 48] (* Robert G. Wilson v, Mar 05 2016 *)
CoefficientList[ Series[ x*(2 - 3x + 8x^2 - 3x^3 + 2x^4)/((1 - x)^3*(1 + x^2)^2), {x, 0, 47}], x] (* Robert G. Wilson v, Mar 05 2016 *)

PARI

isok(n) = issquare(48*n+25); \\ Michel Marcus, Feb 25 2016

PARI

A268539(n)={my(m=n\4*12+[-3,2,5,6][n%4+1]);(3+m)*(m-2)/12} \\ M. F. Hasler, Mar 03 2016

Python

from gmpy2 import is_square
[k for k in range(2200) if is_square(48*k+25)] # Bruno Berselli, Dec 05 2016

Sage

[n for n in (0..2200) if is_square(48*n+25)] # Bruno Berselli, Feb 29 2016

Formula

For n>25, a(n) = 3*( a(n-8)-a(n-16) ) + a(n-24). - Zak Seidov, Feb 28 2016

From Robert Israel, Feb 29 2016: (Start)

Let L = [5, 11, 13, 19, 29, 35, 37, 43].

Then a(i + 8*j) = ( (L(i) + 48*j)^2 - 25 )/48 for i = 1..8, j >= 0. (End)

From Bruno Berselli, Feb 29 2016: (Start)

G.f.: x^2*(2 - 3*x + 8*x^2 - 3*x^3 + 2*x^4)/((1 - x)^3*(1 + x^2)^2).

a(n) = a(-n+1) = 3*a(n-1) - 5*a(n-2) + 7*a(n-3) - 7*a(n-4) + 5*a(n-5) - 3*a(n-6) + a(n-7) for n>6.

a(n) = (3*(n-1)*n + (2*n-1)*(-1)^((n-2)*(n-1)/2) - 1)/4. Therefore:

a(4*k) = k*(12*k -5),

a(4*k+1) = k*(12*k +5),

a(4*k+2) = k*(12*k+11)+2 = (3*k+2)*(4*k+1),

a(4*k+3) = k*(12*k+13)+3 = (3*k+1)*(4*k+3).

From the previous formulas follows that 2, 3, 7 and 17 are the only primes of the sequence. (End)

Sum_{n>=2} 1/a(n) = 12/25 + (4/sqrt(3)-1)*Pi/5. - Amiram Eldar, Jul 30 2024

Extensions

More terms from Michel Marcus, Feb 25 2016