cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A077728 a(n) = A077727(n)/n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 109, 0, 103, 1009, 1009, 103, 1009, 1009, 101, 1009, 10099, 1009, 1009, 1036, 1012, 1027, 1018, 1007, 1027, 1027, 10333, 10189, 10153, 100333, 10117, 10171, 10083, 1027, 10126, 10051, 1007497, 100729, 100714, 100693, 100909
Offset: 1

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Author

Amarnath Murthy, Nov 19 2002

Keywords

Crossrefs

Cf. A077727.

Extensions

Corrected and extended by Ray Chandler, Sep 05 2003

A225903 The smallest number beginning with n whose distinct prime factors are the first n primes.

Original entry on oeis.org

16, 24, 30, 420, 50820, 60060, 7147140, 87297210, 9369900540, 103515091680, 11030826957150, 126152548291770, 13387011595197240, 143910374648370330, 15372244564712285250, 162945792385950223650, 17304843151387913751630, 1876614101750511535732320
Offset: 1

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Author

Keywords

Comments

a(3)=30 is the only term with fewer than 1000 digits whose superscripts are all 1.
Though counterexamples are possible, it appears that the sequence is strictly increasing (confirmed for n < 350, and counterexamples are increasingly unlikely statistically thereafter).

Examples

			For a(6), the number 60060 = 2^2 * 3 * 5 * 7 * 11 * 13. The only number smaller whose factors contains the first 6 primes is 30030, which does not begin with 6.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Block[{p = Prime[n], ba = Product[Prime@k, {k, n}], d = IntegerDigits@ n, mu = 1}, While[d != Take[IntegerDigits[mu*ba], Length@d] || Max[ First /@ FactorInteger[mu]] > p, mu++]; mu*ba]; Array[a, 20] (* Giovanni Resta, May 27 2013 *)
  • R
    library(gmp); primes<-function(n) { x=as.bigz(rep(2,n)); for(i in 2:n) x[i]=nextprime(x[i-1]); as.vector(x[1:n]) }
    newmin<-function(b,d) { if(d>length(b)) return();
        while(1) { b[d]=b[d]+1; if((x=prod(pr^b))>v) return()
            if(substr(x,1,ndig(i))==as.character(i)) { v<<-x; return() }
            if(b[d]==2) {b[d]=1; newmin(b,d+1); b[d]=2 }
            newmin(b,d+1)
        }
    }
    y=as.bigz(rep(0,50))
    for(i in 1:50) {
        pr=primes(i); b=rep(1,i)
        while(substr((v=prod(pr^b)),1,ndig(i))!=as.character(i)) b[1]=b[1]+1;
        while(b[1]>1) { b[1]=b[1]-1; newmin(b,2) }
        if(y[i]>v) y[i]=v;
    }
Showing 1-2 of 2 results.