

A136117


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


6



70, 92, 852, 925, 1247, 1426, 1926, 2625, 3577, 5192, 6305, 6501, 7107, 7740, 7957, 8177, 8626, 9560, 10292, 12927, 13207, 14652, 15555, 16172, 18095, 20475, 20827, 21901, 22265, 22632, 23002, 23751, 24130, 28497, 29330, 31032, 33227, 33675
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OFFSET

1,1


COMMENTS

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


LINKS

Table of n, a(n) for n=1..38.


FORMULA

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


EXAMPLE

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).


PROG

(PARI) P(n)=n*(3*n1)>>1 /* a.k.a. A000326 */
isPent(t)=P(sqrtint(t<<1\3)+1)==t
for(i=1, 299, for(j=1, (i+1)\sqrt(2), isPent(P(i)P(j))&print1(P(i)", ")next(2)))
/* The following is much faster, at the cost of implementing sum2sqr(), cf. A133388*/
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}
A136117vect(n, i)=vector(n, j, i=A136117next(i)) /* 2nd arg =0 by default but allows one to start elsewhere */
A136117(n, i)={until(!n, i=A136117next(i)); i} \\  M. F. Hasler, Dec 25 2007


CROSSREFS

Cf. A000326, A136112A136118, A007527.
Sequence in context: A114838 A036191 A165632 * A224553 A156718 A007621
Adjacent sequences: A136114 A136115 A136116 * A136118 A136119 A136120


KEYWORD

nonn


AUTHOR

M. F. Hasler, Dec 15 2007; corrected Dec 25 2007


STATUS

approved



