A054842 -id:A054842 - cp's OEIS frontend

A246468 Given a number of k digits x = d_(k)10^(k-1) + d_(k-1)10^(k-2) + … + d_(2)10 + d_(1), consider y = p_(1)^d_(1)p_(2)^d_(2)…p_(k)^d_(k), where p_(i) is the i-th prime. Sequence lists the numbers x such that y / x is an integer.

1, 2, 4, 8, 12, 16, 24, 36, 48, 54, 81, 96, 128, 135, 144, 162, 225, 288, 375, 486, 576, 625, 648, 675, 768, 972, 1296, 1575, 1875, 2187, 2268, 2625, 2646, 2688, 3087, 3136, 3375, 3528, 3675, 3888, 3969, 4116, 4374, 4802, 5145, 5292, 5488, 5625, 6048, 6174, 7056

Offset: 1

Paolo P. Lava, Aug 27 2014

a(n) = x such that A054842(x)/x is an integer.

			x = 48 -> y = 2^8*3^4 = 20736 and 20736 / 48 = 432.
x = 972 -> y = 2^2*3^7*5^9 = 17085937500 and 17085937500 / 972 = 17578125.

Paolo P. Lava, Table of n, a(n) for n = 1..1000

Cf. A054842, A189398, A246469.

with(numtheory):P:=proc(q) local a,b,k,n;
for n from 1 to q do a:=n; b:=1;
for k from 1 to ilog10(n)+1 do b:=b*ithprime(k)^(a mod 10); a:=trunc(a/10);
od; if type(b/n,integer) then print(n); fi; od; end: P(10^9);

A246469 Given a number of k digits x = d_(k)10^(k-1) + d_(k-1)10^(k-2) + … + d_(2)10 + d_(1), consider y = p_(1)^d_(k)p_(2)^d_(k-1)…p_(k)^d_(1), where p_(i) is the i-th prime. Sequence lists the numbers x such that y / x is an integer.

Original entry on oeis.org

1, 2, 4, 8, 18, 27, 36, 48, 54, 64, 72, 96, 125, 135, 162, 225, 375, 432, 486, 625, 648, 675, 864, 972, 1225, 1250, 1323, 1350, 1575, 1701, 1715, 1875, 2250, 2646, 2835, 2916, 3375, 3528, 3645, 3675, 3750, 3969, 4116, 4375, 4536, 4725, 4860, 5145, 5488, 5832

Offset: 1

Paolo P. Lava, Aug 27 2014

a(n) = x such that A189398(x) / x is an integer.

			x = 48 -> y = 2^4*3^8 = 104976 and 104976 / 48 = 2187.
x = 972 -> y = 2^9*3^7*5^2 = 27993600 and 27993600 / 972 = 28800.

Paolo P. Lava, Table of n, a(n) for n = 1..1000

Cf. A054842, A189398, A246468.

with(numtheory):P:=proc(q) local a,b,k,n;
for n from 1 to q do a:=n; b:=1;
for k from 1 to ilog10(n)+1 do b:=b*ithprime(ilog10(n)+2-k)^(a mod 10); a:=trunc(a/10);
od; if type(b/n,integer) then print(n); fi; od; end: P(10^9);

A262478 a(n) = Sum_{i >= 0} d_i(n) * p_(i + 1) where d_i(n) = i-th digit of n in base 3, and p_i = i-th prime.

Original entry on oeis.org

0, 2, 4, 3, 5, 7, 6, 8, 10, 5, 7, 9, 8, 10, 12, 11, 13, 15, 10, 12, 14, 13, 15, 17, 16, 18, 20, 7, 9, 11, 10, 12, 14, 13, 15, 17, 12, 14, 16, 15, 17, 19, 18, 20, 22, 17, 19, 21, 20, 22, 24, 23, 25, 27, 14, 16, 18, 17, 19, 21, 20, 22, 24, 19, 21, 23, 22, 24, 26, 25, 27, 29, 24, 26, 28, 27

Offset: 0

James Burling, Sep 23 2015

d_i(n) can be found using either of the following formulas:
* d_i(n) = floor(n / 3^i) mod 3;
* d_i(n) = floor(n / 3^i) - 3 * floor(n / 3^(i + 1)).

			The base 3 representation of n = 5 is 12 so a(5) = 2 * 2 + 1 * 3 = 7.
The base 3 representation of n = 12 is 110 so a(12) = 0 * 2 + 1 * 3 + 1 * 5 = 8.

James Burling, Table of n, a(n) for n = 0..10000

Similar method, different base for n: A089625 (base 2).

Similar method, uses product for sum index for multiplication: A019565 (base 2), A101278 (base 3), A054842 (base 10).

Table[Sum[IntegerDigits[n, 3][[-i]] Prime@ i, {i, IntegerLength[n, 3]}], {n, 0, 81}] (* Michael De Vlieger, Sep 24 2015 *)

a(n) = my(d = Vecrev(digits(n, 3))); sum(k=1, #d, d[k]*prime(k)); \\ Michel Marcus, Sep 24 2015

a(n) = Sum_{i >= 0} p_(i + 1) * (floor(n / 3^i) - 3 * floor(n / 3^(i + 1))).

A263042 a(n) = Sum_{i >= 1} d_i(n) * prime(i) where d_i(n) is the i-th digit of n in base 10, and prime(i) is the i-th prime.

Original entry on oeis.org

0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36

Offset: 0

James Burling, Oct 08 2015

Digits are counted from the right, so d_1(n) is the ones digit, d_2(n) is the tens digit, etc.
d_i(n) can be found using either of the following formulas:
* d_i(n) = floor(n / 10^(i-1)) mod 10;
* d_i(n) = floor(n / 10^(i-1)) - 10 * floor(n / 10^i).
From Derek Orr, Dec 24 2015: (Start)
For n < 1000, this sequence may be written as a series of 10 X 10 subtables:
Subtable 1:
   0,  2,  4,  6,  8, 10, 12, 14, 16, 18
   3,  5,  7,  9, 11, 13, 15, 17, 19, 21
   6,  8, 10, 12, 14, 16, 18, 20, 22, 24
   9, 11, 13, 15, 17, 19, 21, 23, 25, 27
  12, 14, 16, 18, 20, 22, 24, 26, 28, 30
  15, 17, 19, 21, 23, 25, 27, 29, 31, 33
  18, 20, 22, 24, 26, 28, 30, 32, 34, 36
  21, 23, 25, 27, 29, 31, 33, 35, 37, 39
  24, 26, 28, 30, 32, 34, 36, 38, 40, 42
  27, 29, 31, 33, 35, 37, 39, 41, 43, 45
Subtable 2:
   5,  7,  9, 11, 13, 15, 17, 19, 21, 23
   8, 10, 12, 14, 16, 18, 20, 22, 24, 26
  11, 13, 15, 17, 19, 21, 23, 25, 27, 29
  14, 16, 18, 20, 22, 24, 26, 28, 30, 32
  17, 19, 21, 23, 25, 27, 29, 31, 33, 35
  20, 22, 24, 26, 28, 30, 32, 34, 36, 38
  23, 25, 27, 29, 31, 33, 35, 37, 39, 41
  26, 28, 30, 32, 34, 36, 38, 40, 42, 44
  29, 31, 33, 35, 37, 39, 41, 43, 45, 47
  32, 34, 36, 38, 40, 42, 44, 46, 48, 50
Subtable 3:
  10, 12, 14, 16, 18, 20, 22, 24, 26, 28
  13, 15, 17, 19, 21, 23, 25, 27, 29, 31
  16, 18, 20, 22, 24, 26, 28, 30, 32, 34
  19, 21, 23, 25, 27, 29, 31, 33, 35, 37
  22, 24, 26, 28, 30, 32, 34, 36, 38, 40
  25, 27, 29, 31, 33, 35, 37, 39, 41, 43
  28, 30, 32, 34, 36, 38, 40, 42, 44, 46
  31, 33, 35, 37, 39, 41, 43, 45, 47, 49
  34, 36, 38, 40, 42, 44, 46, 48, 50, 52
  37, 39, 41, 43, 45, 47, 49, 51, 53, 55
  ...
Each subtable is 10 X 10. Let T_n(j,k) = the element in the j-th row of the k-th column of subtable n. T_n(1,1) = 5*(n-1). T_n(j,1) = 5*(n-1)+3*(j-1). T_n(1,k) = 5*(n-1)+2*(k-1). Altogether, T_n(j,k) = 5*(n-1)+3*(j-1)+2*(k-1) = 5*n+3*j+2*k-10.
(End)

			For n = 12, the digits are 2 and 1 and the corresponding primes are 2 and 3, so a(12) = (first digit * first prime) + (second digit * second prime) = 2 * 2 + 1 * 3 = 4 + 3 = 7.

James Burling, Table of n, a(n) for n = 0..10000

Similar method, different base for n: A089625 (base 2), A262478 (base 3).

Similar method, uses product instead of sum: A019565 (base 2), A101278 (base 3), A054842 (base 10).

Table[Sum_{m=0}^{infinity} (Floor[n/10^(m)] - 10*Floor[n/10^(m+1)])*Prime(m+1), {n,0,500}] (* G. C. Greubel, Oct 08 2015 *)

a(n) = if (n==0, d = [0], d=Vecrev(digits(n))); sum(i=1,#d, d[i]*prime(i)); \\ Michel Marcus, Oct 10 2015

vector(200,n,n--;sum(i=1,#digits(n),Vecrev(digits(n))[i]*prime(i))) \\ Derek Orr, Dec 24 2015

a(n) = Sum_{i >= 0} prime(i + 1) * (floor(n / 10^i) - 10 * floor(n / 10^(i + 1))).

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

Ctibor O. Zizka, Oct 30 2009

nonn

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.

			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?

Cf. A054841, A054842, A167219.

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

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

cp's OEIS Frontend

A246468 Given a number of k digits x = d_(k)*10^(k-1) + d_(k-1)*10^(k-2) + … + d_(2)*10 + d_(1), consider y = p_(1)^d_(1)*p_(2)^d_(2)*…*p_(k)^d_(k), where p_(i) is the i-th prime. Sequence lists the numbers x such that y / x is an integer.

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A246469 Given a number of k digits x = d_(k)*10^(k-1) + d_(k-1)*10^(k-2) + … + d_(2)*10 + d_(1), consider y = p_(1)^d_(k)*p_(2)^d_(k-1)*…*p_(k)^d_(1), where p_(i) is the i-th prime. Sequence lists the numbers x such that y / x is an integer.

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A262478 a(n) = Sum_{i >= 0} d_i(n) * p_(i + 1) where d_i(n) = i-th digit of n in base 3, and p_i = i-th prime.

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Formula

A263042 a(n) = Sum_{i >= 1} d_i(n) * prime(i) where d_i(n) is the i-th digit of n in base 10, and prime(i) is the i-th prime.

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

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A246468 Given a number of k digits x = d_(k)10^(k-1) + d_(k-1)10^(k-2) + … + d_(2)10 + d_(1), consider y = p_(1)^d_(1)p_(2)^d_(2)…p_(k)^d_(k), where p_(i) is the i-th prime. Sequence lists the numbers x such that y / x is an integer.

A246469 Given a number of k digits x = d_(k)10^(k-1) + d_(k-1)10^(k-2) + … + d_(2)10 + d_(1), consider y = p_(1)^d_(k)p_(2)^d_(k-1)…p_(k)^d_(1), where p_(i) is the i-th prime. Sequence lists the numbers x such that y / x is an integer.