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A278491
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After a(0)=0, numbers n such that (A002828(1+n) = 1) and (A002828(4+n) = 4).
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5
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0, 3, 24, 35, 99, 120, 195, 323, 440, 483, 675, 728, 899, 1155, 1368, 1443, 1763, 1848, 2115, 2499, 2808, 2915, 3363, 3480, 3843, 4355, 4760, 4899, 5475, 5624, 6083, 6723, 7224, 7395, 8099, 8280, 8835, 9603, 10200, 10403, 11235, 11448, 12099, 12995, 13688, 13923, 14883, 15128, 15875, 16899, 17688, 17955, 19043, 19320, 20163
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OFFSET
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0,2
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COMMENTS
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The definition implies that after 0 these are also all numbers n such that (A002828(1+n) = 1), (A002828(2+n) = 2), (A002828(3+n) = 3) and (A002828(4+n) = 4).
Because A002828 obtains value 1 only at squares, every term must be one less than a square.
In the terms of tree defined by edge relation A255131(child) = parent, ("the least squares beanstalk"), these numbers are the nodes with four children (maximum possible).
Either of the above facts implies that this is a subsequence of A276573.
Indexing starts from zero, because a(0)=0 is a special case in this sequence, as it is only number which is its own child in the least squares beanstalk tree.
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LINKS
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FORMULA
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a(0) = 0, and for n >= 1, a(n) = A273324(n)^2 - 1.
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PROG
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(PARI)
\\ (For a more intelligent way to generate the terms, check Altug Alkan's PARI-code for A273324).
istwo(n:int)=my(f); if(n<3, return(n>=0); ); f=factor(n>>valuation(n, 2)); for(i=1, #f[, 1], if(bitand(f[i, 2], 1)==1&&bitand(f[i, 1], 3)==3, return(0))); 1
isthree(n:int)=my(tmp=valuation(n, 2)); bitand(tmp, 1)||bitand(n>>tmp, 7)!=7
A002828(n)=if(issquare(n), !!n, if(istwo(n), 2, 4-isthree(n))) \\ From _Charles R Greathouse_ IV, Jul 19 2011
i=0; n=0; while(i <= 10000, if(isA278491(n), write("b278491.txt", i, " ", n); i++); n++ );
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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