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.

Showing 1-10 of 16 results. Next

A138790 Numbers k such that A138793(k) is prime.

Original entry on oeis.org

61, 946
Offset: 1

Views

Author

Artur Jasinski, Mar 30 2008, Mar 31 2008

Keywords

Comments

There are no more primes for k <= 5000.
a(3) > 20000. - Robert Price, Mar 24 2015

Examples

			a(1) = 61 because the number 160695...654321 is prime.
		

Crossrefs

Programs

  • Mathematica
    b = {}; a = {}; Do[w = RealDigits[n]; w = First[w]; Do[AppendTo[a, w[[k]]], {k, 1, Length[w]}]; p = FromDigits[Reverse[a]]; If[PrimeQ[p], Print[n]; AppendTo[b, p]], {n, 1, 2000}]; b (* Artur Jasinski, Mar 30 2008 *)
    Select[Range[1, 1000], PrimeQ[lst = {}; Do[lst = Join[lst, IntegerDigits[n]], {n, 1, #}]; FromDigits[Reverse[lst]]] &] (* Robert Price, Mar 24 2015 *)

A138962 a(1) = 1, a(n) = the smallest prime divisor of A138793(n).

Original entry on oeis.org

1, 3, 3, 29, 3, 3, 19, 3, 3, 457, 3, 3, 16087, 3, 3, 35963, 3, 3, 167, 3, 3, 7, 3, 3, 13, 3, 3, 953, 3, 3, 7, 3, 3, 548636579, 3, 3, 19, 3, 3, 71, 3, 3, 13, 3, 3, 89, 3, 3, 114689, 3, 3, 17, 3, 3, 12037, 3, 3, 7, 3, 3
Offset: 1

Views

Author

Artur Jasinski, Apr 04 2008

Keywords

Comments

a(61) > 10^11. - Robert Price, Mar 22 2015

Crossrefs

Programs

  • Mathematica
    b = {}; a = {}; Do[w = RealDigits[n]; w = First[w]; Do[AppendTo[a, w[[k]]], {k, 1, Length[w]}]; p = FromDigits[Reverse[a]]; AppendTo[b, First[First[FactorInteger[p]]]], {n, 1, 31}]; b (* Artur Jasinski, Apr 04 2008 *)
    lst = {}; Table[First[First[FactorInteger[FromDigits[Reverse[lst = Join[lst,IntegerDigits[n]]]]]]], {n, 1, 60}] (* Robert Price, Mar 22 2015 *)
  • PARI
    f(n) = my(D = Vec(concat(apply(s->Str(s), [1..n])))); eval(concat(vector(#D, k, D[#D-k+1]))); \\ A138793
    a(n) = my(k=f(n)); forprime(p=2, 10^6, if(k%p == 0, return(p))); if(n == 1, 1, vecmin(factor(k)[,1])); \\ Daniel Suteu, May 27 2022

Formula

a(n) = A020639(A138793(n)). - Daniel Suteu, May 27 2022

Extensions

a(32)-a(60) from Robert Price, Mar 22 2015

A138794 a(n) = A138793(n+1)-A138793(n).

Original entry on oeis.org

20, 300, 4000, 50000, 600000, 7000000, 80000000, 900000000, 1000000000, 1100000000000, 210000000000000, 31000000000000000, 4100000000000000000, 510000000000000000000, 61000000000000000000000
Offset: 1

Views

Author

Artur Jasinski, Mar 30 2008

Keywords

Comments

First differences of A138793

Crossrefs

Programs

  • Mathematica
    b = {}; a = {}; Do[w = RealDigits[n]; w = First[w]; Do[AppendTo[a, w[[k]]], {k, 1, Length[w]}]; p = FromDigits[Reverse[a]]; AppendTo[b, p], {n, 1, 31}]; c = {}; Do[AppendTo[c, b[[n + 1]] - b[[n]]], {n, 1, Length[b] - 1}]; c (*Artur Jasinski*)

Formula

a(n) = A138793(n+1)-A138793(n)

A138963 a(1) = 1, a(n) = the largest prime divisor of A138793(n).

Original entry on oeis.org

1, 7, 107, 149, 953, 218107, 402859, 4877, 379721, 4349353, 169373, 182473, 1940144339383, 2184641, 437064932281, 5136696159619, 67580875919190833, 1156764458711, 464994193118899, 4617931439293, 1277512103328491957510030561, 8177269604099
Offset: 1

Views

Author

Artur Jasinski, Apr 04 2008

Keywords

Comments

For the smallest prime divisors of A138793 see A138962.

Crossrefs

Programs

  • Mathematica
    b = {}; a = {}; Do[w = RealDigits[n]; w = First[w]; Do[AppendTo[a, w[[k]]], {k, 1, Length[w]}]; p = FromDigits[Reverse[a]]; AppendTo[b, First[Last[FactorInteger[p]]]], {n, 1, 31}]; b (* Artur Jasinski, Apr 04 2008 *)
    lst = {}; Table[First[Last[FactorInteger[FromDigits[Reverse[lst = Join[lst,IntegerDigits[n]]]]]]], {n, 1, 10}] (* Robert Price, Mar 22 2015 *)
  • PARI
    f(n) = my(D = Vec(concat(apply(s->Str(s), [1..n])))); eval(concat(vector(#D, k, D[#D-k+1]))); \\ A138793
    a(n) = if(n == 1, 1, vecmax(factor(f(n))[,1])); \\ Daniel Suteu, May 26 2022

Formula

a(n) = A006530(A138793(n)). - Daniel Suteu, May 26 2022

A138795 a(n) = (A138793(n+1)-A138793(n))/10^n.

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 8, 9, 1, 110, 2100, 31000, 410000, 5100000, 61000000, 710000000, 8100000000, 91000000000, 20000000000, 1200000000000, 22000000000000, 320000000000000, 4200000000000000, 52000000000000000, 620000000000000000
Offset: 1

Views

Author

Artur Jasinski, Mar 30 2008

Keywords

Comments

First differences of A138793 divided by 10^n

Crossrefs

Programs

  • Mathematica
    b = {}; a = {}; Do[w = RealDigits[n]; w = First[w]; Do[AppendTo[a, w[[k]]], {k, 1, Length[w]}]; p = FromDigits[Reverse[a]]; AppendTo[b, p], {n, 1, 61}]; c = {}; Do[AppendTo[c, (b[[n + 1]] - b[[n]])/(10^n)], {n, 1, Length[b] - 1}]; c (*Artur Jasinski*)

Formula

a(n) = A138793(n+1)-A138793(n)

A000422 Concatenation of numbers from n down to 1.

Original entry on oeis.org

1, 21, 321, 4321, 54321, 654321, 7654321, 87654321, 987654321, 10987654321, 1110987654321, 121110987654321, 13121110987654321, 1413121110987654321, 151413121110987654321, 16151413121110987654321, 1716151413121110987654321, 181716151413121110987654321
Offset: 1

Views

Author

R. Muller

Keywords

Comments

The first prime term in this sequence is a(82) (see A176024). - Artur Jasinski, Mar 30 2008
For n < 10^4, a(n)/A000217(n) is an integer for n = 1, 2, and 18. The integers are 1, 7 (prime), and 1062667552123515268933651, respectively. - Derek Orr, Sep 04 2014

References

  • F. Smarandache, "Properties of the Numbers", University of Craiova Archives, 1975; Arizona State University Special Collections, Tempe, AZ

Crossrefs

Programs

  • Maple
    a[1]:= 1:
    for n from 2 to 100 do
    a[n]:= n*10^(1+ilog10(a[n-1])) + a[n-1]
    od:
    seq(a[n],n=1..100); # Robert Israel, Sep 05 2014
    # second Maple program:
    a:= proc(n) a(n):= `if`(n=1, 1, parse(cat(n, a(n-1)))) end:
    seq(a(n), n=1..22);  # Alois P. Heinz, Jan 12 2021
  • Mathematica
    b = {}; a = {}; Do[w = RealDigits[n]; w = First[w]; Do[PrependTo[a, w[[Length[w] - k + 1]]], {k, 1, Length[w]}]; p = FromDigits[a]; AppendTo[b, p], {n, 1, 30}]; b (* Artur Jasinski, Mar 30 2008 *)
    Table[FromDigits[Flatten[IntegerDigits/@Range[n,1,-1]]],{n,20}] (* Harvey P. Dale, Jul 06 2019 *)
  • PARI
    a(n)=my(t=n);forstep(k=n-1,1,-1,t=t*10^#Str(k)+k);t \\ Charles R Greathouse IV, Jul 15 2011
    
  • PARI
    A000422(n,p=1,L=1)=sum(k=1,n,k*p*=L+(k==L&&!L*=10)) \\ M. F. Hasler, Nov 02 2016
    
  • Python
    def a(n): return int("".join(map(str, range(n, 0, -1))))
    print([a(n) for n in range(1, 19)]) # Michael S. Branicky, Dec 08 2021

Formula

a(n+1) = (n+1)*10^len(a(n)) + a(n), where len(k) = number of digits in k.
a(n) = Sum_{k=1..n} k*10^(A058183(k) - (1+floor(log10(k)))). - Alexander Goebel, Mar 07 2020
From Serge Batalov, Dec 08 2021: (Start)
a(n) = ((n*9-1)*10^n+1)/9^2 for n < 10,
a(n) = ((n*99-1)*10^(2*n-19)-89)/99^2*10^10 + (8*10^10+1)/9^2 for 10 <= n < 100,
a(n) = ((n*999-1)*10^(3*n-299)-989)/999^2*10^191 + c2 for 10^2 <= n < 10^3,
a(n) = ((n*9999-1)*10^(4*n-3999)-9989)/9999^2*10^2892 + c3 for 10^3 <= n < 10^4,
a(n) = ((n*99999-1)*10^(5*n-49999)-99989)/99999^2*10^38893 + c4 for 10^4 <= n < 10^5,
a(n) = ((n*999999-1)*10^(6*n-599999)-999989)/999999^2*10^488894 + c5 for 10^5 <= n < 10^6,
where
c2 = (98*10^191 + 879*10^10 + 121)/99^2 = a(99),
c3 = (998*10^2701 - 989)/999^2*10^191 + c2 = a(999),
c4 = (9998*10^36001 - 9989)/9999^2*10^2892 + c3 = a(9999),
c5 = (99998*10^450001 - 99989)/99999^2*10^38893 + c4 = a(99999).
(End)

Extensions

Edited by N. J. A. Sloane, Dec 03 2021

A037123 a(n) = a(n-1) + sum of digits of n.

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 46, 48, 51, 55, 60, 66, 73, 81, 90, 100, 102, 105, 109, 114, 120, 127, 135, 144, 154, 165, 168, 172, 177, 183, 190, 198, 207, 217, 228, 240, 244, 249, 255, 262, 270, 279, 289, 300, 312, 325, 330, 336, 343, 351, 360, 370, 381
Offset: 0

Views

Author

Vasiliy Danilov (danilovv(AT)usa.net), Jun 15 1998

Keywords

Comments

Sum of digits of A007908(n). - Franz Vrabec, Oct 22 2007
Also digital sum of A138793(n) for n > 0. - Bruno Berselli, May 27 2011
Sum of the digital sum of i for i from 0 to n. - N. J. A. Sloane, Nov 13 2013

References

  • N. Agronomof, Sobre una función numérica, Revista Mat. Hispano-Americana 1 (1926), 267-269.
  • Maurice d'Ocagne, Sur certaines sommations arithmétiques, J. Sciencias Mathematicas e Astronomicas 7 (1886), 117-128.

Crossrefs

Cf. also A074784, A231688, A231689.
Partial sums of A007953.

Programs

  • Magma
    [ n eq 0 select 0 else &+[&+Intseq(k): k in [0..n]]: n in [0..56] ];  // Bruno Berselli, May 27 2011
  • Maple
    # From N. J. A. Sloane, Nov 13 2013:
    digsum:=proc(n,B) local a; a := convert(n, base, B):
    add(a[i], i=1..nops(a)): end;
    f:=proc(n,k,B) global digsum; local i;
    add( digsum(i,B)^k,i=0..n); end;
    lprint([seq(digsum(n,10),n=0..100)]); # A007953
    lprint([seq(f(n,1,10),n=0..100)]); #A037123
    lprint([seq(f(n,2,10),n=0..100)]); #A074784
    lprint([seq(f(n,3,10),n=0..100)]); #A231688
    lprint([seq(f(n,4,10),n=0..100)]); #A231689
  • Mathematica
    Table[Plus@@Flatten[IntegerDigits[Range[n]]], {n, 0, 200}] (* Enrique Pérez Herrero, Oct 12 2015 *)
    a[0] = 0; a[n_] := a[n - 1] + Plus @@ IntegerDigits@ n; Array[a, 70, 0] (* Robert G. Wilson v, Jul 06 2018 *)
  • PARI
    a(n)=n*(n+1)/2-9*sum(k=1,n,sum(i=1,ceil(log(k)/log(10)),floor(k/10^i)))
    
  • PARI
    a(n)={n++;my(t,i,s);c=n;while(c!=0,i++;c\=10);for(j=1,i,d=(n\10^(i-j))%10;t+=(10^(i-j)*(s*d+binomial(d,2)+d*9*(i-j)/2));s+=d);t} \\ David A. Corneth, Aug 16 2013
    
  • Perl
    for $i (0..100){ @j = split "", $i; for (@j){ $sum += $; } print "$sum,"; } __END_ # gamo(AT)telecable.es
    

Formula

a(n) = Sum_{k=0..n} s(k) = Sum_{k=0..n} A007953(k), where s(k) denote the sum of the digits of k in decimal representation. Asymptotic expression: a(n-1) = Sum_{k=0..n-1} s(k) = 4.5*n*log_10(n) + O(n). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Sep 07 2002
a(n) = n*(n+1)/2 - 9*Sum_{k=1..n} Sum_{i=1..ceiling(log_10(k))} floor(k/10^i). - Benoit Cloitre, Aug 28 2003
From Hieronymus Fischer, Jul 11 2007: (Start)
G.f.: Sum_{k>=1} ((x^k - x^(k+10^k) - 9x^(10^k))/(1-x^(10^k)))/(1-x)^2.
a(n) = (1/2)*((n+1)*(n - 18*Sum_{k>=1} floor(n/10^k)) + 9*Sum_{k>=1} (1 + floor(n/10^k))*floor(n/10^k)*10^k).
a(n) = (1/2)*((n+1)*(2*A007953(n)-n) + 9*Sum_{k>=1} (1+floor(n/10^k))*floor(n/10^k)*10^k). (End)
a(n) = A007953(A053064(n)). - Reinhard Zumkeller, Oct 10 2008
From Wojciech Raszka, Jun 14 2019: (Start)
a(10^k - 1) = 10*a(10^(k - 1) - 1) + 45*10^(k - 1) for k > 0.
a(n) = a(n mod m) + MSD*a(m - 1) + (MSD*(MSD - 1)/2)*m + MSD*((n mod m) + 1), where m = 10^(A055642(n) - 1), MSD = A000030(n). (End)

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Sep 07 2002

A104759 Concatenation of digits of natural numbers from n down to 1.

Original entry on oeis.org

1, 21, 321, 4321, 54321, 654321, 7654321, 87654321, 987654321, 1987654321, 1987654321, 101987654321, 1101987654321, 11101987654321, 211101987654321, 1211101987654321, 31211101987654321, 131211101987654321, 4131211101987654321, 14131211101987654321, 514131211101987654321
Offset: 1

Views

Author

Alexandre Wajnberg & Juliette Bruyndonckx, Apr 23 2005

Keywords

Examples

			a(11) = a(10) because no number may begin with 0.
a(9)= [123456789]101112131415...=987654321
a(10)=[1234567891]01112131415...=1987654321
a(11)=[12345678910]1112131415...=01987654321=1987654321
a(12)=[123456789101]112131415...=101987654321
a(13)=[1234567891011]12131415...=1101987654321
a(14)=[12345678910111]2131415...=11101987654321
a(15)=[123456789101112]131415...=211101987654321
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{t = Reverse@ Flatten@ IntegerDigits@ Range@ n, k}, Reap@ For[k = 1, k <= Length@ t, k++, Sow[FromDigits@ Take[t, -k]]] // Flatten // Rest]; f@ 14 (* Michael De Vlieger, Mar 23 2015 *)
    lst = {}; Do[lst = Join[lst, IntegerDigits[n]], {n, 1, 100}]; Table[FromDigits[Reverse[lst[[Range[1, n]]]]], {n, 1, Length[lst]}] (* Robert Price, Mar 24 2015 *)

Formula

a(n) = A138793(n) mod 10^(n-1). - R. J. Mathar, Sep 17 2011

A116504 Number of distinct prime divisors of the concatenation of n,...,1.

Original entry on oeis.org

0, 2, 2, 2, 3, 2, 2, 3, 3, 3, 2, 3, 3, 4, 5, 4, 6, 8, 4, 5, 4, 5, 4, 5, 6, 7, 5, 5, 7, 8, 3, 6, 5, 7, 8, 6, 4, 3, 6, 5, 8, 6, 3, 7, 6, 5, 7, 7, 3, 6, 3, 7, 9, 9, 3, 4, 4, 6, 3, 3, 5, 8, 5, 6, 7, 7, 4, 8, 8, 4, 8, 4, 7, 8, 10, 3, 7, 6, 4, 7, 7, 1, 3, 8, 3, 8, 5, 4, 5, 7, 11, 9, 6
Offset: 1

Views

Author

Parthasarathy Nambi, Mar 20 2006

Keywords

Examples

			87654321 = 3*3*1997*4877, distinct prime divisors are 3, 1997 and 4877, hence a(8) = 3.
		

Crossrefs

Programs

  • Mathematica
    b = {}; a = {}; Do[w = RealDigits[n]; w = First[w]; Do[PrependTo[a, w[[Length[w] - k + 1]]], {k, 1, Length[w]}]; p = FromDigits[a]; m = FactorInteger[p]; AppendTo[b, Length[m]], {n, 1, 30}]; b (* Artur Jasinski, Mar 30 2008 *)
    Table[PrimeNu[FromDigits[Flatten[IntegerDigits/@Range[n,1,-1]]]],{n,95}] (* Harvey P. Dale, Oct 03 2015 *)
  • PARI
    {a="";for(n=1,58,a=concat(n,a);print1(omega(eval(a)),","))}

Extensions

Edited and extended by Klaus Brockhaus, Mar 29 2006
Terms a(59)-a(93) from Sean A. Irvine, Nov 04 2009
a(90) corrected by Sean A. Irvine, Nov 02 2024

A116505 Number of distinct prime divisors of the concatenation of 1..n.

Original entry on oeis.org

0, 2, 2, 2, 3, 3, 2, 4, 3, 3, 6, 4, 3, 3, 3, 3, 4, 5, 6, 6, 8, 6, 4, 5, 4, 6, 5, 5, 4, 7, 3, 5, 6, 2, 7, 5, 4, 4, 6, 8, 5, 7, 4, 4, 9, 7, 5, 7, 6, 9, 3, 3, 4, 9, 5, 4, 6, 4, 4, 6, 3, 7, 4, 9, 6, 8, 3, 7, 7, 6, 5, 5, 3, 9, 5, 4, 5, 6, 6, 7, 4, 7, 6, 3, 5, 7, 6, 5, 9, 8, 6, 6, 7, 5, 6, 5, 2, 9, 5, 9
Offset: 1

Views

Author

Parthasarathy Nambi, Mar 20 2006

Keywords

Comments

Dario Alpern's factorization program was used for n > 43.

Examples

			123456 = 2*2*2*2*2*2*3*643, with distinct prime divisors 2, 3 and 643. Hence, a(6) = 3.
		

Crossrefs

Programs

  • Mathematica
    Table[PrimeNu[FromDigits[Flatten[IntegerDigits[Range[n]]]]], {n, 30}] (* Jan Mangaldan, Jul 07 2020 *)
  • PARI
    {a="";for(n=1,43,a=concat(a,n);print1(omega(eval(a)),", "))}

Extensions

Edited and extended by Klaus Brockhaus, Mar 29 2006
Terms 59-100 from Sean A. Irvine, Nov 04 2009
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