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A154293
Integers of the form t/6, where t is a triangular number (A000217).
24
0, 1, 6, 11, 13, 20, 35, 46, 50, 63, 88, 105, 111, 130, 165, 188, 196, 221, 266, 295, 305, 336, 391, 426, 438, 475, 540, 581, 595, 638, 713, 760, 776, 825, 910, 963, 981, 1036, 1131, 1190, 1210, 1271, 1376, 1441, 1463, 1530, 1645, 1716, 1740, 1813, 1938, 2015
OFFSET
1,3
COMMENTS
Old definition was "Integers of the form: 1/6+2/6+3/6+4/6+5/6+...".
1/6 + 2/6 + 3/6 = 1, 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 + 7/6 + 8/6 = 6, ...
a(n) is the set of all integers k such that 48k+1 is a perfect square. The square roots of 48*a(n) + 1 = 1, 7, 17, 23, 25, ... = 8*(n-floor(n/4)) + (-1)^n. - Gary Detlefs, Mar 01 2010
Conjecture: A193828 divided by 2. - Omar E. Pol, Aug 19 2011
The above conjecture is correct. - Charles R Greathouse IV, Jan 02 2012
Quasipolynomial of order 4. - Charles R Greathouse IV, Jan 02 2012
LINKS
Mircea Merca, The bisectional pentagonal number theorem, Journal of Number Theory, Volume 157 (December 2015), Pages 223-232.
FORMULA
From R. J. Mathar, Jan 07 2009: (Start)
a(n) = A000217(A108752(n))/6.
G.f.: x^2*(x^2-x+1)*(x^2+4*x+1)/((1+x^2)^2*(1-x)^3) (conjectured). (End)
The conjectured g.f. is correct. - Charles R Greathouse IV, Jan 02 2012
a(n) = (f(n)^2-1)/48 where f(n) = 8*(n-floor(n/4))+(-1)^n, with offset 0, a(0)=0. - Gary Detlefs, Mar 01 2010
a(n) = a(1-n) for all n in Z. - Michael Somos, Oct 27 2012
G.f.: x^2 * (1 + 4*x + x^2) * (1 - x^2) * (1 - x^6) / ((1 - x)^2 * (1 - x^3) * (1 - x^4)^2). - Michael Somos, Feb 10 2015
It appears that the sequence terms occur as exponents in the expansion Sum_{n >= 0} x^n/Product_{k = 1..2*n} (1 + x^k) = 1 + x - x^6 - x^11 + x^13 + x^20 - x^35 - x^46 + + - - .... Cf. A218171. - Peter Bala, Feb 04 2021
Sum_{n>=2} 1/a(n) = 12 - (1+4/sqrt(3))*Pi. - Amiram Eldar, Mar 18 2022
EXAMPLE
G.f. = x^2 + 6*x^3 + 11*x^4 + 13*x^5 + 20*x^6 + 35*x^7 + 46*x^8 + ...
MAPLE
f:=n-> 8*(n-floor(n/4))+(-1)^n:seq((f(n)^2-1)/48, n=0..51); # Gary Detlefs, Mar 01 2010
MATHEMATICA
lst={}; s=0; Do[s+=n/6; If[Floor[s]==s, AppendTo[lst, s]], {n, 0, 7!}]; lst (* Orlovsky *)
Join[{0}, Select[Table[Plus@@Range[n]/6, {n, 200}], IntegerQ]] (* Alonso del Arte, Jan 20 2012 *)
LinearRecurrence[{3, -5, 7, -7, 5, -3, 1}, {0, 1, 6, 11, 13, 20, 35}, 60] (* Charles R Greathouse IV, Jan 20 2012 *)
a[ n_] := (3 n^2 + If[ OddQ[ Quotient[ n + 1, 2]], -5 n + 2, -n]) / 4; (* Michael Somos, Feb 10 2015 *)
a[ n_] := Module[{m = n}, If[ n < 1, m = 1 - n]; SeriesCoefficient[ x^2 (1 + 4 x + x^2) (1 - x^2) (1 - x^6) / ((1 - x)^2 (1 - x^3) (1 - x^4)^2), {x, 0, m}]]; (* Michael Somos, Feb 10 2015 *)
PROG
(PARI) a(n)=n--; (8*(n-n\4)+(-1)^n)^2\48 \\ Charles R Greathouse IV, Jan 02 2012
(PARI) {a(n) = (3*n^2 + if( (n+1)\2%2, -5*n+2, -n)) / 4}; /* Michael Somos, Feb 10 2015 */
(PARI) {a(n) = if( n<1, n = 1-n); polcoeff( x^2 * (1 + 4*x + x^2) * (1 - x^2) * (1 - x^6) / ((1 - x)^2 * (1 - x^3) * (1 - x^4)^2) + x * O(x^n), n)}; /* Michael Somos, Feb 10 2015 */
(Magma) /* By definition: */ [t/6: n in [0..160] | IsIntegral(t/6) where t is n*(n+1)/2]; // Bruno Berselli, Mar 07 2016
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Definition rewritten by M. F. Hasler, Dec 31 2012
STATUS
approved