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A001477
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The nonnegative integers.
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788
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77
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OFFSET
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0,3
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COMMENTS
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Although this is a list, and lists normally have offset 1, it seems better to make an exception in this case. - N. J. A. Sloane, Mar 13 2010
The subsequence 0,1,2,3,4 gives the known values of n such that 2^(2^n)+1 is a prime (see A019434, the Fermat primes). - N. J. A. Sloane, Jun 16 2010
Also: The identity map, defined on the set of nonnegative integers. The restriction to the positive integers yields the sequence A000027. - M. F. Hasler, Nov 20 2013
The number of partitions of 2n into exactly 2 parts. - Colin Barker, Mar 22 2015
The number of orbits of Aut(Z^7) as function of the infinity norm n of the representative lattice point of the orbit, when the cardinality of the orbit is equal to 8960 or 168.- Philippe A.J.G. Chevalier, Dec 29 2015
See A061579 for the transposed infinite square matrix, or triangle with rows reversed. - M. F. Hasler, Nov 09 2021
This is the unique sequence (a(n)) that satisfies the inequality a(n+1) > a(a(n)) for all n in N. This simple and surprising result comes from the 6th problem proposed by Bulgaria during the second day of the 19th IMO (1977) in Belgrade (see link and reference). - Bernard Schott, Jan 25 2023
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REFERENCES
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Maurice Protat, Des Olympiades à l'Agrégation, suite vérifiant f(n+1) > f(f(n)), Problème 7, pp. 31-32, Ellipses, Paris 1997.
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LINKS
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The IMO Compendium, Problem 6, 19th IMO 1977.
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FORMULA
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a(n) = n.
a(0) = 0, a(n) = a(n-1) + 1.
G.f.: x/(1-x)^2.
When seen as array: T(k, n) = n + (k+n)*(k+n+1)/2. Main diagonal is 2*n*(n+1) (A046092), antidiagonal sums are n*(n+1)*(n+2)/2 (A027480). - Ralf Stephan, Oct 17 2004
a(n+1) = det(C(i+1,j), 1 <= i, j <= n), where C(n,k) are binomial coefficients. - Mircea Merca, Apr 06 2013
a(0)=0, a(n>0) = 2*z(-1)^[( |z|/z + 3 )/2] + ( |z|/z - 1 )/2 for z = A130472(n>0); a 1 to 1 correspondence between integers and naturals. - Adriano Caroli, Mar 29 2015
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EXAMPLE
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Triangular view:
0
1 2
3 4 5
6 7 8 9
10 11 12 13 14
15 16 17 18 19 20
21 22 23 24 25 26 27
28 29 30 31 32 33 34 35
36 37 38 39 40 41 42 43 44
45 46 47 48 49 50 51 52 53 54
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MAPLE
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[ seq(n, n=0..100) ];
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MATHEMATICA
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CoefficientList[ Series[x/(x - 1)^2, {x, 0, 76}], x] (* Robert G. Wilson v, May 23 2013 *)
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PROG
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(Magma) [ n : n in [0..100]];
(PARI) A001477(n)=n /* first term is a(0) */
(Haskell)
a001477 = id
(Python)
def a(n): return n
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CROSSREFS
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When written as an array, the rows/columns are A000217, A000124, A152948, A152950, A145018, A167499, A166136, A167487... and A000096, A034856, A055998, A046691, A052905, A055999... (with appropriate offsets); cf. analogous lists for A000027 in A185787.
Cf. A061579 (transposed matrix / reversed triangle).
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KEYWORD
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AUTHOR
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STATUS
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approved
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