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

Author

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Keywords

• nonn
• easy

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

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. [added Jan 21 2025: this is correct - see Berndt et al., Theorem 3.2.] - Peter Bala, Feb 04 2021

From Peter Bala, Dec 12 2024 (Start)

The sequence terms occur as exponents in the expansion of F(x)*Product_{n >= 1} (1 - x^n) = Product_{n >= 1} (1 - x^n)*(1 + x^(4*n))^2*(1 + x^(4*n-2))*(1 + x^(8*n-3))*(1 + x^(8*n-5)) = 1 - x - x^6 + x^11 + x^13 - x^20 - x^35 + x^46 + x^50 - - + + ..., where F(x) is the g.f. of A069910.

It appears that the sequence terms occur as exponents in the expansion 1/(1 - x) * ( - x^2 + Sum_{n >= 1} x^floor((3*n+1)/2) * 1/Product_{k = 1..n} (1 + x^k) ) = x^6 + x^11 - x^13 - x^20 + x^35 + x^46 - - + + .... (End)

It appears that the sequence terms occur as exponents in the expansion Sum_{n >= 0} x^(n+1)/Product_{k = 1..2*n+2} (1 + x^k) = x - x^6 - x^11 + x^13 + x^20 - x^35 - x^46 + + - - .... - Peter Bala, Jan 21 2025

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).

Crossrefs

Cf. A000217, A001318, A069497, A074378, A057569, A057570, A154292, A193828, A218171.

Programs

Magma

/* By definition: */ [t/6: n in [0..160] | IsIntegral(t/6) where t is n*(n+1)/2]; // Bruno Berselli, Mar 07 2016

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 *)

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 */

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

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;

a(4*n+2) = (3*n + 1)*(4*n + 1) = A033577(n); a(4*n+3) = (3*n + 2)*(4*n + 3) = A033578(n+1).

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