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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 (list; graph; refs; listen; history; text; internal format)
 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 G. C. Greubel, Table of n, a(n) for n = 1..1000 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). 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 CROSSREFS Cf. A000217, A001318, A074378, A057569, A057570, A154292, A218171. Sequence in context: A315371 A315372 A315373 * A199717 A068308 A163871 Adjacent sequences:  A154290 A154291 A154292 * A154294 A154295 A154296 KEYWORD nonn,easy AUTHOR Vladimir Joseph Stephan Orlovsky, Jan 06 2009 EXTENSIONS Definition rewritten by M. F. Hasler, Dec 31 2012 STATUS approved

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Last modified July 6 21:17 EDT 2022. Contains 355114 sequences. (Running on oeis4.)