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a(n) is the first pentagonal number that is nontrivially the sum of two pentagonal numbers of the type P(p) + P(p+n) (we always have P(k) = P(0) + P(k)).
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%I #9 Dec 24 2015 10:25:43

%S 70,1926,6305,92,22632,34580,49051,66045,85562,1426,925,159251,188860,

%T 220992,255647,292825,852,2625,7107,466767,516560,568876,623715,

%U 681077,5192,803370,7957,935755,1005732,1078232,22265,8626,1310870,3577

%N a(n) is the first pentagonal number that is nontrivially the sum of two pentagonal numbers of the type P(p) + P(p+n) (we always have P(k) = P(0) + P(k)).

%C The sequence is globally increasing with, at first sight, "unpredictible holes"; the regular values satisfy a(n) = (2523/2)*n^2 + (1189/2)*n + 70, which is approximately (35.5176013*n + 8.3690808)^2. In fact, we have to solve (6*k-1)^2 = 2*(6*p + 3*n - 1)^2 + 18*n^2 - 1 (Eq. 1) with given n that is X^2 = 2Y^2 + 18*n^2 - 1 where X = 6*k - 1 and Y = 6*p + 3*n - 1. p = 20*n + 5 and k = 29*n + 7 always give a solution of Eq. 1 but it is not certain that it is the best, i.e., the first.

%e a(0) = 70 because the first interesting relation is P(70) = P(35) + P(35), i.e., 70 = 2*35.

%e a(1) = 1926 because the first nonobvious relation is P(36) = 1926 = P(25) + P(26).

%p for n from 1 to n ; a:=proc(k) if type (sqrt(18*k^2-6*k+1-9*n^2)/6-(3*n-1)/6,integer)=true then k*(3*k-1)/2 else fi end : seq (a(k),k=n..100000) od;

%K nonn

%O 1,1

%A _Richard Choulet_, Dec 20 2007