A069497 -id:A069497 - cp's OEIS frontend

A287733 First differences of A069497.

6, 30, 30, 12, 42, 90, 66, 24, 78, 150, 102, 36, 114, 210, 138, 48, 150, 270, 174, 60, 186, 330, 210, 72, 222, 390, 246, 84, 258, 450, 282, 96, 294, 510, 318, 108, 330, 570, 354, 120, 366, 630, 390, 132, 402, 690, 426, 144, 438, 750, 462, 156, 474, 810, 498, 168, 510, 870, 534

Offset: 1

Greg Huber, May 30 2017

First differences of the subsequence of triangular numbers that are divisible by 6.

By definition, these numbers are themselves divisible by 6.

			The first triangular number divisible by 6 is 6, and the second triangular number divisible by 6 is 36.  Therefore a(2) = 36 - 6 = 30. (The zeroth triangular number divisible by 6 is taken to be 0.)

Cf. A069497, A108782, A154293.

S:= [seq(seq((12*i+j)*(12*i+j+1)/2, j=[0,3,8,11]), i=0..50)]:
S[2..-1]-S[1..-2]; # Robert Israel, May 30 2017

Differences@ Select[Array[# (# + 1)/2 &, 180, 0], Mod[#, 6] == 0 &] (* Robert G. Wilson v, May 30 2017 *)
Differences[Select[Accumulate[Range[0, 209]], Divisible[#, 6] &]] (* Alonso del Arte, May 31 2017 *)

G.f.: 6*(x^2+4*x+1)*(x^2-x+1)/((x-1)^2*(x^2+1)^2). - Robert Israel, May 30 2017

More terms from Robert G. Wilson v, May 30 2017

A154293 Integers of the form t/6, where t is a triangular number (A000217).

Original entry on oeis.org

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

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

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

			G.f. = x^2 + 6*x^3 + 11*x^4 + 13*x^5 + 20*x^6 + 35*x^7 + 46*x^8 + ...

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).

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

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

f:=n-> 8*(n-floor(n/4))+(-1)^n:seq((f(n)^2-1)/48,n=0..51); # Gary Detlefs, Mar 01 2010

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 *)

a(n)=n--;(8*(n-n\4)+(-1)^n)^2\48 \\ Charles R Greathouse IV, Jan 02 2012

{a(n) = (3*n^2 + if( (n+1)\2%2, -5*n+2,-n)) / 4}; /* Michael Somos, Feb 10 2015 */

{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 */

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)

Definition rewritten by M. F. Hasler, Dec 31 2012

A165721 Integers of the form k*(k+13)/12.

Original entry on oeis.org

4, 14, 22, 25, 35, 55, 69, 74, 90, 120, 140, 147, 169, 209, 235, 244, 272, 322, 354, 365, 399, 459, 497, 510, 550, 620, 664, 679, 725, 805, 855, 872, 924, 1014, 1070, 1089, 1147, 1247, 1309, 1330, 1394, 1504, 1572, 1595, 1665, 1785, 1859, 1884, 1960, 2090

Offset: 1

Vladimir Joseph Stephan Orlovsky, Sep 24 2009

Integers of the form k+k*(k+1)/12 = k+A000217(k)/6 (see A069497). - R. J. Mathar, Sep 25 2009
Are all terms composite numbers?
Contribution from Zak Seidov, Sep 25 2009: (Start)
Integers of form n(13+n)/12, n=0,1,2,...
Each four terms of the sequence are composite numbers of forms:
{(4+3 m) (1+4 m), (2+3 m) (7+4 m), (2+m) (11+12 m), m (13+12 m)}, m=0,1,2,...
m=0: {4,14,22,25}; m=1: {35,55,69,74}; m=2: {90,120,140,147}, etc. (End)

Index entries for linear recurrences with constant coefficients, signature (3,-5,7,-7,5,-3,1).

Cf. A000217, A069497, A165717, A165718, A165719, A165720.

q=6;s=0;lst={};Do[s+=((n+q)/q);If[IntegerQ[s],AppendTo[lst,s]],{n,6!}];lst
Select[Table[k (k+13)/12,{k,200}],IntegerQ] (* or *) LinearRecurrence[ {3,-5,7,-7,5,-3,1},{4,14,22,25,35,55,69},50] (* Harvey P. Dale, Jan 30 2013 *)

From R. J. Mathar, Sep 25 2009: (Start)
a(n) = 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).
G.f.: x*(-4-2*x-x^3+x^5)/((x^2+1)^2*(x-1)^3). (End)
Sum_{n>=1} 1/a(n) = 712/507 - (3 + 4*sqrt(3))*Pi/39. - Amiram Eldar, Jul 26 2024

Definition simplified by R. J. Mathar, Sep 25 2009

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A287733 First differences of A069497.

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A154293 Integers of the form t/6, where t is a triangular number (A000217).

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A165721 Integers of the form k*(k+13)/12.

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A287746 First differences of A154293.

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