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Search: id:A135768
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| A135768 |
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Indices of pentagonal numbers > 0 which are not the difference of 2 other pentagonal numbers > 0. |
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+0
5
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| 1, 2, 3, 5, 6, 8, 9, 11, 15, 18, 24, 27, 54, 81, 96, 128, 135, 162, 216, 243, 288, 303, 384, 423, 459, 486, 519, 591, 639, 648, 683, 729, 783, 864, 879, 891
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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A subsequence of A136112, obtained by omitting A136112(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.
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FORMULA
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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 write P(n)=P(n')+P(n"): this is possible whenever (6n-1)^2=(6n'-1)^2+(6n"-1)^2.
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EXAMPLE
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Indices of the following numbers are not here but in A136112:
P_23 = P_24 -P_7
P_51 = P_66 -P_42
P_71 = P_74 -P_21
P_72 = P_80 - P_35
P_99 = P_104 - P_32
P_123 = P_144 - P_75
P_239 = P_249 - P_70
P_263 = P_274 - P_77
P_311 = P_324 - P_91
P_359 = P_374 - P_105
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PROGRAM
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(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))&next(2)); print1(i", "))
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CROSSREFS
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Cf. A000326, A136112-A136118, A135769(n) = A000326(a(n)), A135771 =.A136112 \ A135768.
Sequence in context: A027563 A000534 A136112 this_sequence A127936 A096276 A075725
Adjacent sequences: A135765 A135766 A135767 this_sequence A135769 A135770 A135771
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KEYWORD
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more,nonn
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AUTHOR
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R. J. Mathar (mathar(AT)strw.leidenuniv.nl) and M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Feb 07 2008
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