A135769 Pentagonal numbers > 0 which are not the difference of two other pentagonal numbers > 0.

1, 5, 12, 51, 92, 117, 176, 330, 477, 852, 1080, 4347, 9801, 13776, 24512, 27270, 39285, 69876, 88452, 124272, 137562, 220992, 268182, 315792, 354051, 403782, 523626, 612162, 629532, 699392, 796797, 919242, 1119312, 1158522, 1190376

Offset: 1

Author

R. J. Mathar and M. F. Hasler, Feb 07 2008

Keywords

• nonn

Comments

A subsequence of A136113, obtained by omitting A136113(A135771(k)), k=1,2,3,... ; i.e. those which are not the difference of two larger pentagonal numbers, but the difference of a larger and a smaller pentagonal number.

The definition ("...two other...") excludes the case P(n) = P(m)-P(n), cf. comment by R. J. Mathar in A000326.

Examples

			See A135768 for a list of P(n) which are in A136113 but not in A135769.
		

Links

• Donovan Johnson, Table of n, a(n) for n = 1..200

Crossrefs

Cf. A000326, A136112-A136118, a(n) = A000326(A135768(n)), A135771 = A136112 \ A135768.

Programs

PARI

P(n)=n*(3*n-1)/2
isPent(t)=P(sqrtint((t*2)\3)+1)==t
for( i=1,999,for( j=1,(P(i)-1)\3, isPent(P(i)+P(j))&j!=i&next(2)); print1(P(i)","))

Formula

P(n)=n*(3*n-1)/2 <=> n*(n-1/3) = (2/3)*P(n), thus m = P(n) <=> m = P([sqrt(2m/3)]+1)

and m = P(n) <=> 24m+1 = (6n-1)^2, useful for investigating the possibility of writing P(n)=P(n')+P(n"): this is possible whenever (6n-1)^2=(6n'-1)^2+(6n"-1)^2-1.