A377851 Smallest multiplier which can complete the square for n-polygonal numbers, together with a constant offset.

8, 1, 24, 8, 40, 3, 56, 16, 72, 5, 88, 24, 104, 7, 120, 32, 136, 9, 152, 40, 168, 11, 184, 48, 200, 13, 216, 56, 232, 15, 248, 64, 264, 17, 280, 72, 296, 19, 312, 80, 328, 21, 344, 88, 360, 23, 376, 96, 392, 25, 408, 104, 424, 27, 440, 112, 456, 29, 472

Offset

3,1

Comments

This smallest multiplier is also the only multiplier that is relatively prime to the offset.

The n-polygonal numbers, indexed by x, are P(n,x) = (n-2)*(x-1)*x/2 + x = A139601(n-3,x).

S(x) = P(n,x)*a(n) + A181318(n-4) completes the square in that quadratic, ensuring S(x) is a square for all x.

Links

Paolo Xausa, Table of n, a(n) for n = 3..10000

Wikipedia, Completing the square.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Formula

a(n) = 8*(n-2)/gcd(n,4)^2. - Andrew Howroyd, Nov 10 2024

From Stefano Spezia, Nov 13 2024: (Start)

G.f.: x^3*(8 + x + 24*x^2 + 8*x^3 + 24*x^4 + x^5 + 8*x^6)/(1 - x^4)^2.

E.g.f.: (4 + 32*x + 6*cos(x) + 2*(16*x - 5)*cosh(x) + 3*x*sin(x) + (5*x - 64)*sinh(x))/4. (End)

Example

For n=7, the heptagonal numbers are h(x) = x*(5*x-3)/2 and with multiplier a(7) = 40 and offset A181318(7-4) = 9 become 40*h(x)+9 = (10*x - 3)^2.

Mathematica

Table[8*(n - 2)/GCD[n, 4]^2, {n, 3, 100}] (* Paolo Xausa, Dec 07 2024 *)

Prog

(PARI) a(n) = 8*(n-2)/gcd(n, 4)^2 \\ Andrew Howroyd, Nov 10 2024

Crossrefs

Cf. A181318 (offsets).

Sequence in context: A317640 A125235 A183892 * A019432 A211796 A138505

Adjacent sequences: A377847 A377848 A377849 * A377852 A377853 A377854

Keyword

nonn,easy

Author

Jonathan Dushoff, Nov 09 2024

Status

approved