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Search: id:A136117
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| A136117 |
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Pentagonal numbers (A000326) which are the sum of 2 other positive pentagonal numbers. |
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+0
6
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| 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
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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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 (jvospost3(AT)gmail.com), Dec 18 2007
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FORMULA
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a(n)=A000326(A136116(n))=A000326(m)+A136114(m) where m is the index of the n-th nonzero term in A136114 or A136115.
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EXAMPLE
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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).
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PROGRAM
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(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
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CROSSREFS
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Cf. A000326, A136112-A136118, A007527.
Adjacent sequences: A136114 A136115 A136116 this_sequence A136118 A136119 A136120
Sequence in context: A118216 A114838 A036191 this_sequence A007621 A051971 A075004
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KEYWORD
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nonn
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AUTHOR
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M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Dec 15 2007; corrected Dec 25 2007
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