A214429 Integers of the form (k^2 - 49) / 120.

0, 1, 2, 4, 11, 15, 18, 23, 37, 44, 49, 57, 78, 88, 95, 106, 134, 147, 156, 170, 205, 221, 232, 249, 291, 310, 323, 343, 392, 414, 429, 452, 508, 533, 550, 576, 639, 667, 686, 715, 785, 816, 837, 869, 946, 980, 1003, 1038, 1122, 1159, 1184, 1222, 1313, 1353

Offset: 0

Author

Michael Somos, Jul 17 2012

Keywords

• nonn
• easy

Comments

From Peter Bala, Dec 26 2024: (Start)

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).

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)

References

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

Links

• G. C. Greubel, Table of n, a(n) for n = 0..5000
• Eric Weisstein's World of Mathematics, Quintuple Product Identity
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,2,-2,0,0,-1,1).

Crossrefs

Cf. A093722, A204221, A379210.

Programs

Magma

m:=25; R:=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

Maple

A214429 := proc(q) local n;
for n from 0 to q do
 if type(sqrt(120*n+49), integer) then print(n);
fi; od; end:
A214429(1500); # Peter Bala, Dec 26 2024

Mathematica

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 *)
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 *)

PARI

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

Formula

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

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

From Peter Bala, Dec 26 2024: (Start)

a(n) is quasi-polynomial in n:

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

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

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)