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A226540
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Maximum of the proper divisors of the triangular numbers.
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1
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1, 3, 5, 5, 7, 14, 18, 15, 11, 33, 39, 13, 35, 60, 68, 51, 57, 95, 105, 77, 23, 138, 150, 65, 117, 189, 203, 145, 155, 248, 264, 187, 119, 315, 333, 37, 247, 390, 410, 287, 301, 473, 495, 345, 47, 564, 588, 245, 425, 663, 689, 477, 495, 770, 798, 551, 59, 885
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OFFSET
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2,2
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COMMENTS
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LINKS
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FORMULA
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a(4n) = 4n^2 + n, 4n+1 <= a(4n+1) <= (8n^2 + 6n + 1)/3, 4n+3 <= a(4n+2) <= (8n^2 + 10n + 3)/3, a(4n+3) = 4n^2 + 7n + 3. - Charles R Greathouse IV, Jun 10 2013
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EXAMPLE
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For n = 28 we have n*(n+1)/2 = 406 and its proper divisors are 1, 2, 7, 14, 29, 58, 203. Hence a(28) = 203.
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MAPLE
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with(numtheory); A226540:=proc(q) local a, n;
for n from 2 to q do a:=sort([op(divisors(n*(n+1)/2))]);
print(a[nops(a)-1]); od; end: A226540(10^6);
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MATHEMATICA
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Table[Divisors[(n(n+1))/2][[-2]], {n, 2, 60}] (* Harvey P. Dale, Apr 09 2021 *)
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PROG
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(PARI) a(n)=if(n==2, return(1)); my(p=factor(n/gcd(n, 2))[1, 1], q=factor((n+1)/gcd(n+1, 2))[1, 1]); binomial(n+1, 2)/min(p, q) \\ Charles R Greathouse IV, Jun 10 2013
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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