A154293 Integers of the form t/6, where t is a triangular number (A000217).
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
Examples
G.f. = x^2 + 6*x^3 + 11*x^4 + 13*x^5 + 20*x^6 + 35*x^7 + 46*x^8 + ...
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- B. C. Berndt, B. Kim, and A. J. Yee, Ramanujan's lost notebook: Combinatorial proofs of identities associated with Heine's transformation or partial theta functions, J. Comb. Thy. Ser. A, 117 (2010), 957-973.
- Mircea Merca, The bisectional pentagonal number theorem, Journal of Number Theory, Volume 157 (December 2015), Pages 223-232.
- Index entries for linear recurrences with constant coefficients, signature (3,-5,7,-7,5,-3,1).
Programs
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Magma
/* By definition: */ [t/6: n in [0..160] | IsIntegral(t/6) where t is n*(n+1)/2]; // Bruno Berselli, Mar 07 2016
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Maple
f:=n-> 8*(n-floor(n/4))+(-1)^n:seq((f(n)^2-1)/48,n=0..51); # Gary Detlefs, Mar 01 2010
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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 *)
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PARI
a(n)=n--;(8*(n-n\4)+(-1)^n)^2\48 \\ Charles R Greathouse IV, Jan 02 2012
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PARI
{a(n) = (3*n^2 + if( (n+1)\2%2, -5*n+2,-n)) / 4}; /* Michael Somos, Feb 10 2015 */
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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 */
Formula
From R. J. Mathar, Jan 07 2009: (Start)
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
Sum_{n>=2} 1/a(n) = 12 - (1+4/sqrt(3))*Pi. - Amiram Eldar, Mar 18 2022
a(n) = A069497(n)/6. - Hugo Pfoertner, Nov 19 2024
From Peter Bala, Jan 21 2025: (Start)
a(4*n) = 12*n^2 - n; a(4*n+1) = 12*n^2 + n;
Let T(n) = n*(n + 1)/2 denote the n-th triangular number. Then
a(4*n) = (1/6) * T(12*n-1); a(4*n+1) = (1/6) * T(12*n);
a(4*n+2) = (1/6) * T(12*n+3); a(4*n+3) = (1/6) * T(12*n+8). (End)
Extensions
Definition rewritten by M. F. Hasler, Dec 31 2012
Comments