A078405 -id:A078405 - cp's OEIS frontend

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

N. J. A. Sloane, Feb 24 2016

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

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).

Cf. A078405, A154293.

Subsequence of A011865.

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

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

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 *)

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

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

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

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

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

More terms from Michel Marcus, Feb 25 2016

A078413 Smallest positive integer than cannot be obtained from exactly n copies of n using parentheses and the operations +, -, /, *, ^ and concatenation.

Original entry on oeis.org

2, 2, 5, 13, 18, 89, 243, 475, 1257

Offset: 1

N. J. A. Sloane, Dec 28 2002

a(7) >= 192. - Rick L. Shepherd, May 08 2003
From Lars Blomberg, Apr 08 2018: (Start)
The terms a(6)-a(9) have been computed with these rules:
Division by 0 or exponentiation 0^0 is not allowed.
Concatenation where at least one operand is a fraction or where the second operand is negative are skipped.
Exponentiations yielding > 100000 digits or with exponent > 32-bit signed integer are skipped. (End)

			With three 3's one can form 1=(3/3)^3, 2=3-3/3, 3=3+3-3, 4=3+3/3, but not 5, so a(3)=5.
With four 4's one can get 1=44/44, 2=4/4+4/4, 3=4-(4/4)^4, 4=4+(4-4)^4, 5=4+(4/4)^4, 6=(4+4)/4+4, 7=44/4-4, 8=4+4+4-4, 9=4+4+4/4, 10=(44-4)/4, 11=(4/4) | (4/4), 12=(44+4)/4, but not 13, so a(4)=13 (| denotes concatenation).
With five 5's one can get 1 = (((5 | 5) | 5)^(5 - 5)), 2 = (((5 | 5) - 5) / (5 * 5)), 3 = (((5 * 5) + 5) / (5 + 5)), 4 = (((5 / 5) * 5) - (5 / 5)), 5 = (((5 | 5) * 5) / (5 | 5)), 6 = (((5 | 5) + 5) / (5 + 5)),
7 = (((5 / 5) + 5) + (5 / 5)), 8 = (( 5 + 5) - ((5 + 5) / 5)), 9 = (( 5 + 5) - ((5 / 5)^5)), 10 = (((5 | 5) / 5) - (5 / 5)), 11 = (((5 | 5) / 5)^(5 / 5)), 12 = (((5 | 5) / 5) + (5 / 5)),
13 = (((5 | 5) + (5 + 5)) / 5), 14 = (((5 / 5) | 5) - (5 / 5)), 15 = (((5 / 5) | 5)^(5 / 5)), 16 = (((5 / 5) | 5) + (5 / 5)), 17 = (( 5 | (5^5)) / (5^5)), but not 18, so a(5) = 18.

Index entries for similar sequences

Cf. A078405.

a(5) from Frank Ellermann, Dec 30 2002, who finds that a(6) >= 89

a(6)-a(9) from Lars Blomberg, Apr 08 2018

cp's OEIS Frontend

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

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