A014979 Numbers that are both triangular and pentagonal.

0, 1, 210, 40755, 7906276, 1533776805, 297544793910, 57722156241751, 11197800766105800, 2172315626468283465, 421418033734080886426, 81752926228785223683195, 15859646270350599313653420

Offset: 1

Author

Glenn Johnston (glennj(AT)sonic.net)

Keywords

• nonn
• easy

Examples

			G.f. = x^2 + 210*x^3 + 40755*x^4 + 7906276*x^5 + 1533776805*x^6 + ...
a(4) = 40755 which is 285*(285-1)/2 = 165*(3*165-1)/2.
		

References

• J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 210, p. 61, Ellipses, Paris 2008.
• L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 22.

Links

• C. Gill, solution to question no. 8, Mathematical Miscellany, 1 (1836), pp. 220-225, at p. 223.
• J. C. Su, On some properties of two simultaneous polygonal sequences, JIS 10 (2007) 07.10.4, example 4.2.
• Eric Weisstein's World of Mathematics, Pentagonal Triangular Number.
• Index entries for linear recurrences with constant coefficients, signature (195,-195,1).

Crossrefs

Cf. A046174, A046175, A076139, A108281.

Programs

Mathematica

a[ n_] := ChebyshevU[ 2 n - 3, 7] / 14 + ChebyshevT[ 2 n - 3, 7] / 84 - 1/12; (* Michael Somos, Feb 24 2015 *)
LinearRecurrence[{195,-195,1},{0,1,210},20] (* Harvey P. Dale, May 19 2017 *)

PARI

{a(n) = polchebyshev( 2*n - 3, 2, 7) / 14 + polchebyshev( 2*n - 3, 1, 7) / 84 - 1 / 12}; /* Michael Somos, Jun 16 2011 */

Formula

a(n) = 194 * a(n-1) - a(n-2) + 16.

G.f.: x^2 * (1 + 15*x) / ((1 - x) * (1 - 194*x + x^2)).

a(n)=((((1+sqrt(3))^(4*n-1)-(1-sqrt(3))^(4*n-1))/(2^(2*n+1)*sqrt(3)))^2)/2-1/8. - John Sillcox (johnsillcox(AT)hotmail.com), Sep 01 2003

a(n+1) = 97*a(n)+8+7*(192*a(n)^2+32*a(n)+1)^(1/2) - Richard Choulet, Sep 19 2007

a(n) = A076139(2*n - 3) = A108281(2 - n). for all n in Z. - Michael Somos, Jun 16 2011

Extensions

Corrected and extended by Warut Roonguthai

Edited by N. J. A. Sloane, Jul 24 2006