A136117 - OEIS

A136117

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

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 n-th 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*n-1)>>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, A136112-A136118, A007527.

Sequence in context: A036191 A165632 A295807 * 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