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Search: id:A154028
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| A154028 |
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List of pairs of numbers:{n*(n+1)/2,n!} such that F(n!)=n*(n+1)/2. |
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1
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| 0, 1, 1, 1, 3, 2, 6, 6, 10, 24, 15, 120, 21, 720, 28, 5040, 36, 40320, 45, 362880, 55, 3628800, 66, 39916800, 78, 479001600, 91, 6227020800, 105, 87178291200, 120, 1307674368000, 136, 20922789888000, 153, 355687428096000, 171
(list; graph; listen)
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OFFSET
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0,5
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COMMENT
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If you have two recursions ( addition and multiplication):
a(0)=0;a(n)=n+a(n-1);a(n)=n*(n+1)/2;
and
b(0)=1;b(n)=n*a(n-1):a(n)=n!;
then you can form a function F such that:
F(n!)=n*(n+1)/2.
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FORMULA
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{n*(n+1)/2,n!}
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MATHEMATICA
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Flatten[Table[{n*(n + 1)/2, n!}, {n, 0, 20}]]
(*addition*)
a[0] = 0; a[n_] := a[n] = n + a[n - 1];
Table[a[n] - n*(n + 1)/2, {n, 0, 20}] (*multiplication*)
b[0] = 1; b[n_] := b[n] = n*b[n - 1];
Table[b[n] - n!, {n, 0, 20}]
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CROSSREFS
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Sequence in context: A060408 A098071 A023360 this_sequence A157793 A096375 A062200
Adjacent sequences: A154025 A154026 A154027 this_sequence A154029 A154030 A154031
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KEYWORD
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nonn,tabf
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AUTHOR
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Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 04 2009
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