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A252769
Numbers n such that the sum of the pentagonal numbers P(n), P(n+1), P(n+2) and P(n+3) is equal to the heptagonal number H(m) for some m.
2
94, 5908, 366282, 22703656, 1407260470, 87227445564, 5406694364578, 335127823158352, 20772518341453326, 1287561009346947940, 79808010061169319034, 4946809062783150832248, 306622353882494182280422, 19005639131651856150553996, 1178043003808532587152067410
OFFSET
1,1
COMMENTS
Also positive integers x in the solutions to 12*x^2-5*y^2+32*x+3*y+36 = 0, the corresponding values of y being A252770.
FORMULA
a(n) = 63*a(n-1)-63*a(n-2)+a(n-3).
G.f.: 2*x*(7*x-47) / ((x-1)*(x^2-62*x+1)).
a(n) = 2*(-2/3+1/240*(31+8*sqrt(15))^(-n)*(80-27*sqrt(15)+(31+8*sqrt(15))^(2*n)*(80+27*sqrt(15)))). - Colin Barker, Mar 03 2016
EXAMPLE
94 is in the sequence because P(94)+P(95)+P(96)+P(97) = 13207+13490+13776+14065 = 54538 = H(148).
MATHEMATICA
LinearRecurrence[{63, -63, 1}, {94, 5908, 366282}, 30] (* Harvey P. Dale, Mar 04 2015 *)
PROG
(PARI) Vec(2*x*(7*x-47)/((x-1)*(x^2-62*x+1)) + O(x^100))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Dec 21 2014
STATUS
approved