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.

Previous Showing 21-28 of 28 results.

A355893 Let A090252(n) = Product_{i >= 1} prime(i)^e(i); then a(n) is the concatenation, in reverse order, of e_1, e_2, ..., ending at the exponent of the largest prime factor of A090252(n); a(1)=0 by convention.

Original entry on oeis.org

0, 1, 10, 100, 2, 1000, 20, 10000, 100000, 1000000, 3, 10000000, 100000000, 200, 1010, 1000000000, 10000000000, 100000000000, 1000000000000, 10000000000000, 100000000000000, 1000000000000000, 4, 10000000000000000
Offset: 1

Views

Author

N. J. A. Sloane, Aug 23 2022

Keywords

Comments

A090252 and A354169 are similar in many ways. This sequence and A355892 illustrate this.
This compressed format only make sense if all e_i are less than 10, that is, for n <= 24574.
It is believed that 6 does not appear in A090252, so 11 is missing from the present sequence.

Examples

			The initial terms of A090252 are:
1 -> 0
2 = 2^1 ->1
3 = 2^0 3^1 -> 10
5 = 2^0 3^0 5^1 -> 100
4 = 2^2 -> 2
7 = 2^0 3^0 5^0 7^1 -> 1000
9 = 2^0 3^2 -> 20
...
The terms, right-justified, for comparison with A355892, are:
.1 ...................................0
.2 ...................................1
.3 ..................................10
.4 .................................100
.5 ...................................2
.6 ................................1000
.7 ..................................20
.8 ...............................10000
.9 ..............................100000
10 .............................1000000
11 ...................................3
12 ............................10000000
13 ...........................100000000
14 .................................200
15 ................................1010
16 ..........................1000000000
17 .........................10000000000
18 ........................100000000000
19 .......................1000000000000
20 ......................10000000000000
21 .....................100000000000000
22 ....................1000000000000000
23 ...................................4
24 ...................10000000000000000
...
		

Crossrefs

See A354150 for indices of powers of 2 in A090252.

Programs

  • Mathematica
    nn = 24, s = Import["https://oeis.org/A090252/b090252.txt", "Data"][[1 ;; nn, -1]]; f[n_] := If[n == 1, 0, Function[g, FromDigits@ Reverse@ ReplacePart[Table[0, {PrimePi[g[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> Last@ # &, g]]@ FactorInteger@ n]; Array[f[s[[#]]] &, nn] (* Michael De Vlieger, Aug 24 2022 *)

Formula

a(n) = A054841(A090252(n)). - Stefano Spezia, Aug 24 2022

A167219 Numbers k such that there exists a positive integer B for which k = Sum_{i=0..m} (B^i)*a_i where the a_i are defined by k = Product_{i=0..m} prime(i+1)^a_i.

Original entry on oeis.org

3, 6, 10, 12, 24, 27, 36, 48, 96, 100, 144, 175, 192, 216, 273, 384, 486, 576, 768, 972, 1296, 1536, 1728, 2304, 3072, 3125, 6144, 9216, 12288, 13824, 17496, 19683, 20736, 24576, 36864, 46656, 49152, 62208, 69984, 98304, 110592, 147456, 196608, 331776, 393216, 589824
Offset: 1

Views

Author

Ctibor O. Zizka, Oct 30 2009

Keywords

Comments

Previous name: Numbers k such that there exists a solution to (p_m ^ a_m)*(p_m-1 ^ a_m-1)*...*(3^a_1)*(2^a_0) = (B^m)*a_m + (B^m-1)*a_m-1 + ... + (B^1)*a_1 + (B^0)*a_0 where k = (p_m ^ a_m)*(p_m-1 ^ a_m-1)*...*(3^a_1)*(2^a_0); a_m >= 1; a_(i= 0; p_0, p_1, ..., p_m are prime numbers; a_0, a_1, ..., a_m, B are integers.
B is the base in which we can express k as Sum_{i=0..m} B^i * a_i. B may also be seen as the variable in a polynomial, and k is then also an encoding of the polynomial (defined by the product of primes formula).
For k = (2^r)*3 we have B = (2^r)*3 - r.
A167221(n) is the smallest positive integer that yields a solution for k = a(n).
Negative B's can be obtained when the polynomial is an even function. This happens for instance when for k = 10, 100, 3125, ... - Michel Marcus, Aug 10 2022
From Peter Munn, Aug 13 2022: (Start)
Positive integers k such that k is a fixed point of a completely additive function f_B:N+ -> Z, B > 0, where f_B(prime(i+1)) = B^i for all i >= 0. Equivalently, since row B of A104244 is f_B, {a(n)} lists the columns of A104244 that contain their own column number.
If we require B to be negative instead, the sequence appears to start 10, 100, 3125, 1799875, 65610000, ... . Of these, 1799875 = 5^3 * 7 * 11^2 * 17 is the only k with only negative solutions (B = -11); the solutions for 65610000 are {4049, -4051}.
(End)
If p is the (k+1)-th prime and p is congruent to 1 modulo k, then p^p is a term with p^((p-1)/k) a solution for B. The list of such primes starts 3, 5, 7, 31, 97, 101, 331, ... . I suspect this list is infinite, meaning the greatest prime factor of the terms would be unbounded. - Peter Munn, Aug 15 2022

Examples

			For k = 10 = 2^1 * 3^0 * 5^1, k = B^0 * 1 + B^1 * 0 + B^2 * 1, so we have to solve the equation 10 = 1 + B^2 for a positive integer B, B = 3. But B=-3 works too. Thus 10 is a term.
For k = 12 = 2^2 * 3^1, k = B^0 * 2 + B^1 * 1, so we have to solve the equation 12 = 2 + B for a positive integer B. B = 10. Thus 12 is a term.
For k = 21 = 2^0 * 3^1 * 5^0 * 7^1, k = B^0 * 0 + B^1 * 1 + B^2 * 0 + B^3 * 1, so we have to solve the equation 21 = B + B^3 for an integer B. No such B exists, so 21 is not a term of this sequence.
From _Michel Marcus_, Aug 10 2022: (Start)
In other words:
  10 is a term because 10 = 5^1 * 3^0 * 2^1 and 101 in base 3 is 10.
  12 is a term because 12 = 3^1 * 2^2 and 12 in base 10 is 12. (End)
		

Crossrefs

A206284 describes the polynomial encoding used here.

Programs

  • PARI
    isok(k) = if (k>1, my(f=factor(k), v=primes(primepi(vecmax(f[,1])))); my(p=sum(i=1, #v, 'x^(i-1)*valuation(k,v[i]))); p -= k; my(c=-polcoef(p, 0)); my(q=(p+c)/x); my(d=divisors(c)); for (k=1, #d, if(subst(q, x, d[k]) == c/d[k], return(1)););); \\ Michel Marcus, Aug 08 2022
    
  • PARI
    \\ See PARI link \\ David A. Corneth, Aug 10 2022
    
  • Python
    from sympy import divisors, factorint, sieve
    def ok(n):
        if n < 2: return False
        f = factorint(n)
        a = [f[pi] if pi in f else 0 for pi in sieve.primerange(2, max(f)+1)]
        for B in range(1, n+1):
            polyB = sum(B**i*ai for i, ai in enumerate(a) if ai > 0)
            if polyB == n: return True
            elif polyB > n: return False
        return False
    print([k for k in range(10**4) if ok(k)]) # Michael S. Branicky, Aug 10 2022

Extensions

Edited by Jon E. Schoenfield, Mar 16 2022
Incorrect term 71 removed, new name and more terms from Michel Marcus, Aug 08 2022
a(41)-a(46) from Michael S. Branicky, Aug 10 2022

A287142 Least k such that the number of pairs of consecutive divisors of k equals n.

Original entry on oeis.org

1, 2, 6, 12, 72, 60, 180, 360, 420, 840, 1260, 3780, 2520, 5040, 13860, 36960, 41580, 27720, 55440, 83160, 166320, 277200, 491400, 471240, 360360, 1113840, 720720, 1081080, 3341520, 2162160, 2827440, 5405400, 4324320, 12972960, 6126120, 16576560, 28274400
Offset: 0

Views

Author

Michel Lagneau, May 20 2017

Keywords

Comments

a(n) is even for n > 0.
We observe numbers of the decimal form (abcabc) = 360360, 720720 and numbers of the decimal form (abcabc0) = 1081080, 2162160, 5405400, 4324320, 6126120.
Observation and questions: many terms are products of powers of a contiguous set of the smallest primes. Many early terms of a(n) are in A002182; e.g., a(35) - A002182(44). The smallest exception outside of the empty product a(0) = 1 is a(22) = 491400 = 2^3 * 3^3 * 5^2 * 7 * 13. In other words, many terms have A006530(a(n)) < A053669(a(n)); a(22) is the smallest exception. Other exceptions include {471240, 1113840, 3341520, 2827440, 16576560, 28274400, ...}. A000720(A053669(a(22))) - A000720(A006530(a(22))) = 1, but the first instance of 2 for this function is a(35) = 16576560. This is evident by mapping A054841 across a(n). Are there a finite number of terms of a(n) that are also in A002182? Are there a finite number of terms of a(n) that have A006530(a(n)) < A053669(a(n)); are they becoming less frequent as n increases? - Michael De Vlieger, May 20 2017
In other words, a(n) is the least integer with exactly n divisors that are oblong (A002378). - Bernard Schott, Jul 30 2022

Examples

			a(3) = 12 because the divisors of 12 are {1, 2, 3, 4, 6, 12} with three pairs of consecutive divisors (1, 2), (2, 3) and (3, 4).
		

Crossrefs

Essentially the same as A088726.

Programs

  • Maple
    with(numtheory):
    for n from 0 to 60 do:
    ii:=0:
      for k from 1 to 10^8 while(ii=0) do:
        d0:=divisors(k):n0:=nops(d0):c0:=0:
          for i from 1 to n0-1 do:
            if d0[i+1]=d0[i]+1
             then
              c0:=c0+1:
              else
             fi:
           od:
           if c0=n
           then
         ii:=1:printf(“%d %d \n”,n,k):
         else
         fi:
       od:
      od:
  • Mathematica
    Function[s, Function[t, ReplacePart[t, Map[#1 -> #2 & @@ # &, Transpose@{1 + Keys@ s, Values[s][[All, 1]]}]]]@ ConstantArray[0, Max@ Keys@ s]]@ KeySort@ PositionIndex@ Table[DivisorSum[n, 1 &, Divisible[n, # + 1] &], {n, 2 * 10^6}] (* Michael De Vlieger, May 20 2017, Version 10 *)
  • PARI
    isok(n,k) = {dk = divisors(k); ddk = vector(#dk-1, j, dk[j+1] - dk[j]); #select(x->x==1, ddk) == n;}
    a(n) = {my(k=1); while (!isok(n, k), k++); k;} \\ Michel Marcus, May 20 2017

Formula

a(n) = 2*A130317(n) for n >= 1. - Bernard Schott, Jul 30 2022

A050327 Number of factorizations into distinct squarefree factors indexed by prime signatures. A050326(A025487).

Original entry on oeis.org

1, 1, 0, 2, 0, 1, 0, 0, 5, 0, 1, 0, 4, 0, 0, 0, 1, 0, 0, 5, 0, 15, 0, 0, 0, 0, 2, 0, 16, 0, 0, 0, 0, 0, 0, 7, 0, 8, 0, 0, 1, 0, 23, 0, 0, 0, 1, 0, 5, 0, 0, 0, 0, 52, 14, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 68, 3, 0, 4, 0, 0, 40, 0, 0, 0, 0, 11, 0, 0, 0, 0, 0, 41
Offset: 1

Views

Author

Christian G. Bower, Oct 15 1999

Keywords

Comments

From Michael De Vlieger, Oct 06 2017: (Start)
Terms in A025487 that set records in this sequence: {1, 6, 30, 210, 420, 1260, 2310, 4620, 13860, 30030, 60060, 180180, 510510, 900900, 1021020, 3063060, 6126120, 9699690, ...}.
Conjecture: prime signatures corresponding to primorials A002110(i) with i > 1 set records in this sequence. (End)

Examples

			From _Michael De Vlieger_, Oct 06 2017: (Start)
First 20 values, with numbers in column "r" records, and the last column the concatenation of exponents of standard form prime decomposition of A025487(n):
  .
   n    a(n)   r   A025487(n) rev(A054841(A025487(n)))
  --------------------------------------------
   1      1    1          1   0
   2      1               2   1
   3      0               4   2
   4      2    2          6   11
   5      0               8   3
   6      1              12   21
   7      0              16   4
   8      0              24   31
   9      5    3         30   111
  10      0              32   5
  11      1              36   22
  12      0              48   41
  13      4              60   211
  14      0              64   6
  15      0              72   32
  16      0              96   51
  17      1             120   311
  18      0             128   7
  19      0             144   42
  20      5             180   221
  21      0             192   61
  22     15    4        210   1111
(End)
		

Crossrefs

Programs

  • Mathematica
    f[n_] := If[n <= 1, {{}}, Join @@ Table[Map[Prepend[#, d] &, Select[f[n/d], Min @@ # > d &]], {d, Select[Rest@ Divisors@ n, SquareFreeQ]}]]; Length[f@ #] & /@ Prepend[#, 1] &@ Sort@ Map[Times @@ Flatten@ MapIndexed[ConstantArray[Prime@ First@ #2, #1] &, #] &, Union@ Table[Sort[FactorInteger[n][[All, -1]], Greater], {n, 2, Product[Prime@ i, {i, 7}]}]] (* Michael De Vlieger, Oct 06 2017, after Gus Wiseman at A293243 *)

Formula

From Michael De Vlieger, Oct 06 2017: (Start)
a(n) = A050326(A025487(n)).
(End)

A167221 a(n) is the smallest positive number B that yields a solution for k = A167219(n).

Original entry on oeis.org

3, 5, 3, 10, 21, 9, 17, 44, 91, 7, 70, 5, 186, 71, 3, 377, 97, 285, 760, 194, 323, 1527, 574, 1148, 3062, 25, 6133, 4603, 12276, 4605, 2499, 2187, 5182, 24563, 18426, 7775, 49138, 12440, 9997, 98289, 36860, 73721, 196592, 82941, 393199, 294904, 786414, 49, 294907
Offset: 1

Views

Author

Ctibor O. Zizka, Oct 30 2009

Keywords

Comments

B is the base in which we can express k as Sum_{i=0..m} B^i * a_i. There is an isomorphism between (Z[B],+) and the positive rationals as the polynomials with integer coefficients considered as a group under addition are isomorphic to the positive rationals considered as a group under multiplication.

Examples

			For k = 21 = 2^0 * 3^1 * 5^0 * 7^1, k = B^0 * 0 + B^1 * 1 + B^2 * 0 + B^3 * 1, so we have to solve the equation 21 = B + B^3 for an integer B. No such B exists.
For k = 10 = 2^1 * 3^0 * 5^1, k = B^0 * 1 + B^1 * 0 + B^2 * 1, so we have to solve the equation 10 = 1 + B^2 for an integer B. B = +-3.
For k = 12 = 2^2 * 3^1, k = B^0 * 2 + B^1 * 1, so we have to solve the equation 12 = 2 + B for an integer B. B = 10. Are there any numbers other than k=12 for which B = 10 yields a solution?
		

Crossrefs

Programs

  • PARI
    lista(nn) = for (k=2, nn, my(f=factor(k), v=primes(primepi(vecmax(f[,1])))); my(p=sum(i=1, #v, 'x^(i-1)*valuation(k,v[i]))); p -= k; my(c=-polcoef(p, 0)); my(q=(p+c)/x); my(d=divisors(c)); for (k=1, #d, if(subst(q, x, d[k]) == c/d[k], print1(d[k], ", ")););); \\ Michel Marcus, Aug 09 2022

Extensions

Edited by Jon E. Schoenfield, Mar 16 2022
Corrected and extended by Michel Marcus, Aug 09 2022
a(41) and beyond from Michael S. Branicky, Aug 10 2022

A245500 Concatenation of multiplicities of prime divisors of highly composite numbers A002182(n).

Original entry on oeis.org

0, 1, 2, 11, 21, 31, 22, 41, 211, 311, 221, 411, 321, 421, 3111, 2211, 4111, 3211, 4211, 3311, 5211, 4311, 6211, 4221, 32111, 4411, 5221, 42111, 33111, 52111, 43111, 62111, 42211, 53111, 44111, 52211, 63111, 421111, 331111, 521111, 431111, 621111, 422111
Offset: 1

Views

Author

Michael De Vlieger, Jul 24 2014

Keywords

Comments

For prime decomposition of A002182(n) = 2^a * 3^b * 5^c * ..., a(n) = "abc..." converted to a decimal number.
In other words, each "place" read from left to right represents the n-th prime, starting with 2 at left and increasing to the right. A number in the "place" represents the multiplicity of the corresponding prime in A002182(n).
This notation is corrupt when any multiplicity exceeds 9. The smallest instance of this is at n = 221.
Similar to A054841 but multiplicities are in reverse order.
Given that the exponents e(i) (a,b,c... in the above) of the prime factorization are (weakly) decreasing, their concatenation remains unambiguous way beyond n = 221 (first instance where e(1) >= 10) and even beyond n = 8869 (first instance where e(2) >= 10). Only when e2 >= 10 + e(3) for the first time, in principle the first digit of e(2) could be mistaken for the last digit of e(1); yet it is unlikely if not impossible that e(2) < 10 and e(1) > 100. So the first ambiguous decomposition would require concat(e(1),e(2),e(3)) = concat(a',b',c') with a' > b' >= e(3), thus e(2) significantly larger than 10 + e(3) and e(1) much larger than 100. - M. F. Hasler, Jan 03 2020

Examples

			A002182(4)   = 12 = 2^2 * 3^1, thus a(4) = 21;
A002182(17)  = 2520 = 2^3 * 3^2 * 5^1 * 7^1, thus a(17) = 3211;
A002182(220) = 2^10 * 3^4 * 5^3 * 7^2 * 11 * ... * 53 (skipping no primes), thus a(220) cannot be represented using a single decimal place for the multiplicity 10.
		

Crossrefs

Programs

  • Mathematica
    encodePrimeSignature[n_Integer] :=
      Catch[FromDigits[Reverse[IntegerDigits[Apply[Plus,
           Which[n == 0, Throw["undefined"],
              n == 1, 0,
              Max[Last /@ FactorInteger @ n ] > 9, Throw["overflow"],
              True, Power[10, PrimePi[Abs[#]] - 1]] & /@
           Flatten[ConstantArray @@@ FactorInteger[n]] ]]]]];
    lst = FoldList[Max, 1, Table[DivisorSigma[0, n], {n, 2, 100000000}]];
    Map[encodePrimeSignature, Flatten[Position[lst, #, 1, 1] & /@ Union[lst]]]
  • PARI
    apply( A245500(n)=fromdigits(factor(A002182(n))[,2]~), [1..99]) \\ For n >= 8869, fromdigits must be replaced by (s->if(s,eval(concat([Str(e)|e<-s])))). - M. F. Hasler, Jan 03 2020

Extensions

Offset corrected to 1 by M. F. Hasler, Jan 03 2020

A299400 a(n) = concatenation of all (i, e_i) with e_i > 0, when n = Product_{i >= 1} prime(i)^e_i.

Original entry on oeis.org

0, 11, 21, 12, 31, 1121, 41, 13, 22, 1131, 51, 1221, 61, 1141, 2131, 14, 71, 1122, 81, 1231, 2141, 1151, 91, 1321, 32, 1161, 23, 1241, 101, 112131, 111, 15, 2151, 1171, 3141, 1222, 121, 1181, 2161, 1331, 131, 112141, 141, 1251, 2231, 1191, 151, 1421, 42, 1132, 2171
Offset: 1

Views

Author

M. F. Hasler, Mar 08 2018

Keywords

Comments

The conventional a(1) = 0 represents the empty concatenation.
Due to simple concatenation, this encoding of the positive integers becomes ambiguous from n = 613 = prime(112)^1 on, which has the same encoding a(n) = 1121 as 6 = prime(1)^1*prime(2)^1. To get a unique encoding, one could use, e.g., the digit 9 as delimiter to separate indices and exponents, written in base 9 as to use only digits 0..8, as soon as a term would be the duplicate of an earlier term (or for all n >= 613). Then one would have, e.g., a(613) = prime(134_9)^1 = 13491.
Sequence A067599 is based on the same idea, but uses the primes instead of their indices. In A037276 the prime factors are repeated, instead of giving the exponent. In A080670 exponents 1 are omitted. In A124010 only the prime signature is given. In A054841 the sum e_i*10^(i-1) is given, i.e., exponents are used as digits in base 10, while they are listed individually in the rows of A067255.

Examples

			2 = prime(1)^1 => a(2) = 11,
3 = prime(2)^1 => a(3) = 21,
4 = prime(1)^2 => a(4) = 12,
5 = prime(3)^1 => a(5) = 31,
6 = prime(1)^1*prime(2)^1 => a(1) = 1121,
7 = prime(3)^1 => a(7) = 41,
8 = prime(1)^3 => a(8) = 13, and so on.
		

Crossrefs

Cf. A067599 (decimal encoding of prime factorization).

Programs

  • Maple
    a:= n-> `if`(n=1, 0, parse(cat(seq([numtheory[pi]
           (i[1]), i[2]][], i=sort(ifactors(n)[2]))))):
    seq(a(n), n=1..60);  # Alois P. Heinz, Mar 16 2018
  • Mathematica
    Array[FromDigits@ Flatten@ Map[{PrimePi@ #1, #2} & @@ # &, FactorInteger@ #] &, 51] (* Michael De Vlieger, Mar 16 2018 *)
  • PARI
    A299400(n)=if(n=factor(n),eval(concat(apply(f->Str(primepi(f[1]),f[2]), Col(n)~))))

A279237 Let k_i be the multiplicity of prime(i) in the prime factorization of the n-th composite number C_n, and let k_i=0 if prime(i) is not a factor of C_n. Then a(n)=1*k_1+10*k_2+100*k_3+...+10^N*k_N, where N is the index of the largest prime factor in C_n.

Original entry on oeis.org

2, 11, 3, 20, 101, 12, 1001, 110, 4, 21, 102, 1010, 10001, 13, 200, 100001, 30, 1002, 111, 5, 10010, 1000001, 1100, 22, 10000001, 100010, 103, 1011, 10002, 120, 100000001, 14, 2000, 201, 1000010, 100002, 31, 10100, 1003, 10000010, 1000000001, 112, 10000000001, 1020, 6, 100100, 10011, 1000002
Offset: 1

Views

Author

Marcus Kylén, Dec 08 2016

Keywords

Examples

			The 1st composite number is 4 = 2^2, so a(1)=2.
The 2nd composite number is 6 = 3^1*2^1, so a(2)=11.
The 3rd composite number is 8 = 2^3, so a(3)=3.
The 4th composite number is 9 = 3^2*2^0, so a(4)=20.
The 5th composite number is 10 = 5^1*3^0*2^1, so a(5)=101.
The 6th composite number is 12 = 3^1*2^2, so a(6)=12.
The 7th composite number is 14 = 7^1*5^0*3^0*2^1, so a(7)=1001.
The 8th composite number is 15 = 5^1*3^1*2^0, so a(8)=110.
The 9th composite number is 16 = 2^4, so a(9)=4.
The 10th composite number is 18 = 3^2*2^1, so a(10)=21.
		

Crossrefs

Cf. A002808. Subset of A054841.

Programs

  • Mathematica
    Map[FromDigits@ Reverse@ Function[w, ReplacePart[#, Flatten@ Map[{PrimePi@ #1 -> #2} & @@ # &, w]] &@ ConstantArray[0, PrimePi@ Max@ w[[All, 1]]]]@ FactorInteger@ # &, Select[Range[4, 120], CompositeQ]] (* Michael De Vlieger, Dec 10 2016 *)

Formula

a(n) = 1*k_1+10*k_2+100*k_3+...+10^N*k_N, where k_i is the exponent of prime(i) in the factorization of the n-th composite number C_n, k_i=0 if prime(i) is not a factor in C_n. Also, N is the index of the largest prime factor of C_n, so that C_n = Product_{i=1..N} prime(i)^k_i.
Previous Showing 21-28 of 28 results.