A089026 a(n) = n if n is a prime, otherwise a(n) = 1.
1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1, 13, 1, 1, 1, 17, 1, 19, 1, 1, 1, 23, 1, 1, 1, 1, 1, 29, 1, 31, 1, 1, 1, 1, 1, 37, 1, 1, 1, 41, 1, 43, 1, 1, 1, 47, 1, 1, 1, 1, 1, 53, 1, 1, 1, 1, 1, 59, 1, 61, 1, 1, 1, 1, 1, 67, 1, 1, 1, 71, 1, 73, 1, 1, 1, 1, 1, 79, 1, 1, 1, 83, 1, 1, 1, 1, 1, 89, 1, 1, 1, 1, 1, 1, 1
Offset: 1
Keywords
Examples
From Larry Tesler (tesler(AT)pobox.com), Nov 08 2010: (Start) a(9) = (8*9*10)/(2^((5+2+1)-(3+1+0))*3^((3+1)-(2+0))*5^((2)-(1))*7^((1)-(1))) = 1 [composite]. a(10) = (8*9*10)/(2^((5+2+1)-(3+1+0))*3^((3+1)-(2+0))*5^((2)-(1))*7^((1)-(1))) = 1 [composite]. a(11) = (8*9*10*11*12)/(2^((6+3+1)-(3+1+0))*3^((4+1)-(2+0))*5^((2)-(1))*7^((1)-(1))) = 11 [prime]. (End)
References
- Paulo Ribenboim, The little book of big primes, Springer 1991, p. 106.
- L. Tesler, "Factorials and Primes", Math. Bulletin of the Bronx H.S. of Science (1961), 5-10. [From Larry Tesler (tesler(AT)pobox.com), Nov 08 2010]
Links
- G. C. Greubel, Table of n, a(n) for n = 1..5000
- The IMO Compendium, Problem 1, 9th Irish Mathematical Olympiad 1996.
- Index to sequences related to Olympiads.
Crossrefs
Programs
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MATLAB
a = [1:96]; a(isprime(a) == false) = 1; % Thomas Scheuerle, Oct 06 2022
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Magma
[IsPrime(n) select n else 1: n in [1..96]]; // Marius A. Burtea, Aug 02 2019
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Mathematica
digits=200; a=Table[If[PrimePi[n]-PrimePi[n-1]>0, n, 1], {n, 1, digits}]; Table[Numerator[(n/2)/(n-1)! ] + Floor[2/n] - 2*Floor[1/n], {n,1,200}] (* Alexander Adamchuk, May 20 2006 *) Range@ 120 /. k_ /; CompositeQ@ k -> 1 (* or *) Table[n Boole@ PrimeQ@ n, {n, 120}] /. 0 -> 1 (* or *) Table[If[PrimeQ@ n, n, 1], {n, 120}] (* Michael De Vlieger, Jul 02 2016 *)
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PARI
a(n) = n^isprime(n) \\ David A. Corneth, Oct 06 2022
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Python
from sympy import isprime def a(n): return n if isprime(n) else 1 print([a(n) for n in range(1, 97)]) # Michael S. Branicky, Oct 06 2022
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Sage
def A089026(n): if n == 4: return 1 f = factorial(n-1) return (f + 1) - n*(f//n) [A089026(n) for n in (1..96)] # Peter Luschny, Oct 16 2013
Formula
From Peter Luschny, Nov 29 2003: (Start)
a(n) = denominator(n! * Sum_{m=0..n} (-1)^m*m!*Stirling2(n+1, m+1)/(m+1)).
a(n) = denominator(n! * Sum_{m=0..n} (-1)^m*m!*Stirling2(n, m)/(m+1)). (End)
From Alexander Adamchuk, May 20 2006: (Start)
a(n) = numerator((n/2)/(n-1)!) + floor(2/n) - 2*floor(1/n).
a(n) = gcd(n,(n-1)!+1). - Jaume Oliver Lafont, Jul 17 2008, Jan 23 2009
a(1) = 1, a(2) = 2, then a(n) = 1 or a(n) = n = prime(m) = (Product q+k, k = 1 .. 2*floor(n/2+1)-q) / (Product prime(i)^(Sum (floor((n+1)/(prime(i)^w)) - floor(q/(prime(i)^w)) ), w = 1 .. floor(log[base prime(i)] n+1) ), i = 2 .. m-1) where q = prime(m-1). - Larry Tesler (tesler(AT)pobox.com), Nov 08 2010
a(n) = (n!*HarmonicNumber(n) mod n)+1, n != 4. - Gary Detlefs, Dec 03 2011
a(n) = denominator of (n!)/n^(3/2). - Arkadiusz Wesolowski, Dec 04 2011
a(n) = n^c(n), where c = A010051. - Wesley Ivan Hurt, Jun 16 2013
Conjecture: for n > 3, a(n) = gcd(n, A007406(n-1)). - Thomas Ordowski, Aug 02 2019
a(n) = 1 + c(n)*(n-1), where c = A010051. - Wesley Ivan Hurt, Jun 21 2025
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