A249164 Numbers n such that the triangular number T(n) is equal to the sum of the pentagonal numbers P(m) and P(m+1) for some m.
1, 3, 113, 331, 11121, 32483, 1089793, 3183051, 106788641, 311906563, 10464197073, 30563660171, 1025384524561, 2994926790243, 100477219209953, 293472261783691, 9845742098050881, 28757286728011523, 964782248389776433, 2817920627083345611, 94538814600100039601
Offset: 1
Examples
113 is in the sequence because T(113) = 6441 = 3151+3290 = P(46)+P(47).
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).
Programs
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PARI
Vec(-x*(x+1)^2*(11*x^2+1)/((x-1)*(x^2-10*x+1)*(x^2+10*x+1)) + O(x^100))
Formula
a(n) = a(n-1)+98*a(n-2)-98*a(n-3)-a(n-4)+a(n-5).
G.f.: -x*(x+1)^2*(11*x^2+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)).
Comments