%I #10 Jan 25 2021 14:41:16

%S 70,92,852,925,1247,1426,1926,2625,3577,5192,6305,6501,7107,7740,7957,

%T 8177,8626,9560,10292,12927,13207,14652,15555,16172,18095,20475,20827,

%U 21901,22265,22632,23002,23751,24130,28497,29330,31032,33227,33675

%N Pentagonal numbers (A000326) which are the sum of 2 other positive pentagonal numbers.

%C It is conjectured that every integer and hence every pentagonal number, greater than 33066, hence greater than A000326(149) = 33227, can be represented as the sum of three pentagonal numbers. - _Jonathan Vos Post_, Dec 18 2007

%F a(n) = A000326(A136116(n)) = A000326(m)+A136114(m) where m is the index of the n-th nonzero term in A136114 or A136115.

%e a(1)=70=P(7) is the least pentagonal number which can be written as sum of two other pentagonal numbers, P(7)=P(5)+P(5).

%o (PARI) P(n)=n*(3*n-1)>>1 /* a.k.a. A000326 */

%o isPent(t)=P(sqrtint(t<<1\3)+1)==t

%o for(i=1,299,for(j=1,(i+1)\sqrt(2),isPent(P(i)-P(j)) && print1(P(i)",") || next(2)))

%o /* The following is much faster, at the cost of implementing sum2sqr(), cf. A133388*/

%o A136117next(i)=i=sqrtint(i\3*2)*6+5; until(0, for(j=2,#t=sum2sqr((i+=6)^2+1), t[j]%6==[5,5] && break(2)));i^2\24

%o A136117vect(n,i)=vector(n,j,i=A136117next(i)) /* 2nd arg =0 by default but allows one to start elsewhere */

%o A136117(n,i)=until(!n--,i=A136117next(i));i \\ _M. F. Hasler_, Dec 25 2007

%Y Cf. A000326, A136112-A136118, A007527.

%K nonn

%O 1,1

%A _M. F. Hasler_, Dec 15 2007; corrected Dec 25 2007