login
Integers of the form (k^2 - 49) / 120.
2

%I #30 Dec 29 2024 17:31:02

%S 0,1,2,4,11,15,18,23,37,44,49,57,78,88,95,106,134,147,156,170,205,221,

%T 232,249,291,310,323,343,392,414,429,452,508,533,550,576,639,667,686,

%U 715,785,816,837,869,946,980,1003,1038,1122,1159,1184,1222,1313,1353

%N Integers of the form (k^2 - 49) / 120.

%C From _Peter Bala_, Dec 26 2024: (Start)

%C The sequence terms are the exponents in the expansion of Product_{n >= 1} (1 - q^n)/( (1 - q^(10*n-4))*(1 - q^(10*n-6)) ) = 1 - q - q^2 + q^4 + q^11 - q^15 - q^18 + + - - ... (by the quintuple product identity).

%C The numbers 2*a(n) are the exponents in the expansion of Sum_{n >= 0} q^(n*(n+2)) * Product_{k >= 2*n+2} 1 - q^k = 1 - q^2 - q^4 + q^8 + q^22 - q^30 - q^36 + + - - .... See Andrews et al., p. 591, Exercise 6(b) (but note that the left side of the identity should be Sum_{n >= 0} q^(n^2+2*n)/(q; q)_(2*n+1)). (End)

%D George E. Andrews, Richard Askey, Ranjan Roy, Special Functions, Cambridge University Press, 1999.

%H G. C. Greubel, <a href="/A214429/b214429.txt">Table of n, a(n) for n = 0..5000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/QuintupleProductIdentity.html">Quintuple Product Identity</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,2,-2,0,0,-1,1).

%F G.f.: x * (1 + x + 2*x^2 + 7*x^3 + 2*x^4 + x^5 + x^6) / ((1 - x) * (1 - x^4)^2).

%F a(n) = 2*a(n-4) - a(n-8) + 15 = a(-1 - n).

%F From _Peter Bala_, Dec 26 2024: (Start)

%F a(n) is quasi-polynomial in n:

%F a(4*n) = n*(15*n + 7)/2; a(4*n+1) = (3*n + 2)*(5*n + 1)/2;

%F a(4*n+2) = (3*n + 1)*(5*n + 4)/2; a(4*n+3) = (n + 1)*(15*n + 8)/2.

%F For 0 <= k <= 3, a(4*n+k) = (N_k(n)^2 - 49)/120, where N_0(n) = 30*n + 7, N_1(n) = 30*n + 13, N_2(n) = 30*n + 17 and N_3(n) = 30*n + 23. (End)

%p A214429 := proc(q) local n;

%p for n from 0 to q do

%p if type(sqrt(120*n+49), integer) then print(n);

%p fi; od; end:

%p A214429(1500); # _Peter Bala_, Dec 26 2024

%t CoefficientList[Series[x*(1+x+2*x^2+7*x^3+2*x^4+x^5+x^6)/((1-x)*(1- x^4)^2), {x,0,50}], x] (* _G. C. Greubel_, Aug 10 2018 *)

%t Select[(Range[0,500]^2-49)/120,IntegerQ] (* or *) LinearRecurrence[ {1,0,0,2,-2,0,0,-1,1},{0,1,2,4,11,15,18,23,37},80] (* _Harvey P. Dale_, Oct 23 2019 *)

%o (PARI) {a(n) = (((n*3 + 1) \ 4 * 10 + 5 + 2*(-1)^n)^2 - 49) / 120 }

%o (Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1+x+2*x^2+7*x^3+2*x^4+x^5+x^6)/((1-x)*(1-x^4)^2))); // _G. C. Greubel_, Aug 10 2018

%Y Cf. A093722, A204221, A379210.

%K nonn,easy

%O 0,3

%A _Michael Somos_, Jul 17 2012