A191486 -id:A191486 - cp's OEIS frontend

A046034 Numbers whose digits are primes.

2, 3, 5, 7, 22, 23, 25, 27, 32, 33, 35, 37, 52, 53, 55, 57, 72, 73, 75, 77, 222, 223, 225, 227, 232, 233, 235, 237, 252, 253, 255, 257, 272, 273, 275, 277, 322, 323, 325, 327, 332, 333, 335, 337, 352, 353, 355, 357, 372, 373, 375, 377, 522, 523, 525, 527, 532

Offset: 1

Eric W. Weisstein

If n is represented as a zerofree base-4 number (see A084544) according to n=d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n) = Sum_{j=0..m} c(d(j))*10^j, where c(k)=2,3,5,7 for k=1..4. - Hieronymus Fischer, May 30 2012

According to A153025, it seems that 5, 235 and 72335 are the only terms whose square is also a term, i.e., which are also in the sequence A275971 of square roots of the terms which are squares, listed in A191486. - M. F. Hasler, Sep 16 2016

			a(100)   = 2277,
a(10^3)  = 55327,
a(9881)  = 3233232,
a(10^4)  = 3235757,
a(10922) = 3333333,
a(10^5)  = 227233257.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
Eric Weisstein's World of Mathematics, Smarandache Sequences.
Index entries for 10-automatic sequences.

Cf. A046035, A118950, A019546 (primes), A203263, A035232, A039996, A085823, A052382, A084544, A084984, A017042, A001743, A001744, A014261, A014263, A153025, A191486, A193238, A202267, A202268, A211681, A365471 (complement).

a046034 n = a046034_list !! (n-1)
a046034_list = filter (all (`elem` "2357") . show ) [0..]
-- Reinhard Zumkeller, Jul 19 2011

[n: n in [2..532] | Set(Intseq(n)) subset [2, 3, 5, 7]];  // Bruno Berselli, Jul 19 2011

Table[FromDigits /@ Tuples[{2, 3, 5, 7}, n], {n, 3}] // Flatten (* Michael De Vlieger, Sep 19 2016 *)

is_A046034(n)=Set(isprime(digits(n)))==[1] \\ M. F. Hasler, Oct 12 2013

def A046034(n):
    m = (3*n+1).bit_length()-1>>1
    return int(''.join(('2357'[(3*n+1-(1<<(m<<1)))//(3<<((m-1-j)<<1))&3] for j in range(m)))) # Chai Wah Wu, Feb 08 2023

A055642(a(n)) = A193238(a(n)). - Reinhard Zumkeller, Jul 19 2011
From Hieronymus Fischer, Apr 20, May 30 and Jun 25 2012: (Start)
a(n) = Sum_{j=0..m-1} ((2*b(j)+1) mod 8 + floor(b(j)/4) - floor((b(j)-1)/4))*10^j, where m = floor(log_4(3*n+1)), b(j) = floor((3*n+1-4^m)/(3*4^j)).
a(n) = Sum_{j=0..m-1} A010877(A005408(b(j)) + A002265(b(j)) - A002265(b(j)-1))*10^j.
Special values:
a(1*(4^n-1)/3) = 2*(10^n-1)/9.
a(2*(4^n-1)/3) = 1*(10^n-1)/3.
a(3*(4^n-1)/3) = 5*(10^n-1)/9.
a(4*(4^n-1)/3) = 7*(10^n-1)/9.
Inequalities:
a(n) <= 2*(10^log_4(3*n+1)-1)/9, equality holds for n = (4^k-1)/3, k>0.
a(n) <= 2*A084544(n), equality holds iff all digits of A084544(n) are 1.
a(n) > A084544(n).
Lower and upper limits:
lim inf a(n)/10^log_4(n) = (7/90)*10^log_4(3) = 0.48232167706987..., for n -> oo.
lim sup a(n)/10^log_4(n) = (2/9)*10^log_4(3) = 1.378061934485343..., for n -> oo.
where 10^log_4(n) = n^1.66096404744...
G.f.: g(x) = (x^(1/3)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(4/3)*(2 + z(j) + 2*z(j)^2 + 2*z(j)^3 - 7*z(j)^4)/(1-z(j)^4), where z(j) = x^4^j.
Also g(x) = (x^(1/3)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(4/3)*(1-z(j))*(2 + 3*z(j) + 5*z(j)^2 + 7*z(j)^3)/(1-z(j)^4), where z(j)=x^4^j.
Also: g(x) = (1/(1-x))*(2*h_(4,0)(x) + h_(4,1)(x) + 2*h_(4,2)(x) + 2*h_(4,3)(x) - 7*h_(4,4)(x)), where h_(4,k)(x) = Sum_{j>=0} 10^j*x^((4^(j+1)-1)/3)*x^(k*4^j)/(1-x^4^(j+1)). (End)
Sum_{n>=1} 1/a(n) = 1.857333779940977502574887651449435985318556794733869779170825138954093657197... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024

More terms from Cino Hilliard, Aug 06 2006
Typo in second formula corrected by Hieronymus Fischer, May 12 2012
Two typos in example section corrected by Hieronymus Fischer, May 30 2012

A329147 Replace in n each nonzero digit d with prime(d).

Original entry on oeis.org

0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 20, 22, 23, 25, 27, 211, 213, 217, 219, 223, 30, 32, 33, 35, 37, 311, 313, 317, 319, 323, 50, 52, 53, 55, 57, 511, 513, 517, 519, 523, 70, 72, 73, 75, 77, 711, 713, 717, 719, 723, 110, 112, 113, 115, 117, 1111, 1113, 1117, 1119, 1123

Offset: 0

Bernard Schott, Nov 06 2019

Some properties:
No term has a digit 4, 6 or 8.
No term begins with 9, 10, 12, 15, 29, 39, 59, 79.
If a is regarded as a function a: n --> a(n) from N to N, then
1) a is neither increasing: a(9) > a(10) nor decreasing: a(3) < a(4),
2) a is not injective: a(92) = a(122) = 233,
3) a is not surjective: 4 and 15 are not terms. The integers that are not in this sequence are in A329149 and the integers that are obtained are in A329150, with increasing order.
Some primes remain primes: 2, 3, 5, 7, 19, 59, ...
Some primes become composites: 11, 13, 17, 23, 29, 31, ...
Some composites remain composites: 10, 14, 16, 18, 20, 21, 22,...
Some composites become primes: 4, 6, 8, 9, 12, 15, 24, 25, 26,...
When n > 4 ends respectively with 0, 1 or 3 then a(n) that ends with 0, 2, 5 is composite.
The sequence 9, 99, 999, ..., respectively 12, 1212, 121212, ... generates the same numbers 23, 2323, 232323, ... Analogously, 9, 92, 922, 922, ... and 12, 122, 1222, ... generate the same sequence 23, 233, 2333, 23333, .... For the numbers 91,9191,919191, ... the terms of the sequence are 232, 232232, 232232232, ... so palindromes. - Marius A. Burtea, Nov 07 2019
The numbers 113, 14113, 1441113, 144411113, ... determine the terms 225 = 15^2, 27225 = 165^2, 2772225 = 1665^2, ... (in A191486). The numbers 14, 14000, 14000000, ... determine the terms 27 = 3^3, 27000 = 30^3, 27000000 = 300^3, .... - Marius A. Burtea, Nov 12 2019

			As a(2) = prime(2) = 3, a(5) = prime(5) = 11 and a(8) = prime(8) = 19, a(258)= 31119.
As a(3) = prime(3) = 5, a(0) = 0 and a(7) = prime(7) = 17, hence a(307) = 5017.

Metin Sariyar, Table of n, a(n) for n = 0..10000

Cf. A000040, A329149, A329150.

Similar to A048380, A048385 and A322131.

v:=[0,2,3,5,7,11,13,17,19,23]; [0] cat [StringToInteger(&cat[IntegerToString(k): k in Reverse([v[m+1]: m in Intseq(n)])]): n in [1..60]]; // Marius A. Burtea, Nov 07 2019

a:= n-> (l-> parse(cat(seq(`if`(l[-i]=0, 0, ithprime(l[-i])),
             i=1..nops(l)))))(convert(n, base, 10)):
seq(a(n), n=0..80);  # Alois P. Heinz, Nov 07 2019

p[n_] := If[n > 0, Prime[n], 0]; a[n_] := FromDigits[Flatten @ IntegerDigits @ (p /@ IntegerDigits[n])]; Array[a, 60, 0] (* Amiram Eldar, Nov 06 2019 *)

a(n) = if (n, fromdigits(concat(apply(d -> if (d, digits(prime(d)), [0]), digits(n)))), 0) \\ Rémy Sigrist, Nov 07 2019

def A329147(n): return int("".join(map(str, ([0, 2, 3, 5, 7, 11, 13, 17, 19, 23][int(i)] for i in str(n)))))
print([A329147(n) for n in range(60)]) # Michael S. Branicky, Apr 10 2023

A030485 Squares composed of digits {2, 5, 7}.

Original entry on oeis.org

25, 225, 7225, 27225, 55225, 2772225, 227557225, 277722225, 27777222225, 72272257225, 2777772222225, 25772527522225, 277777722222225, 2775552752755225, 27522257555772225, 27777777222222225, 77525222275255225, 257727727257277225, 722555225555275225, 2275752775775227225

Offset: 1

Patrick De Geest, Dec 11 1999

We can easily prove that, except for the first term, all terms are of the form 100*m^2 + 100*m + 25 where mod(m, 10) is one of the numbers 1, 3, 6 or 8. Also we can show that all numbers of the form ((5 * 10^n - 5)/3)^2 where n is a natural number, are in the sequence. - Farideh Firoozbakht, Dec 09 2008

Zhao Hui Du, Table of n, a(n) for n = 1..45
Patrick De Geest, Squares containing at most three distinct digits, Index entries for related sequences

Subsequence of A191486. Also subsequence of A017330. Cf. A030487.

Flatten[Table[Select[FromDigits/@Tuples[{2, 5, 7}, n], IntegerQ[Sqrt[#]] &], {n, 17}]] (* The program takes a long time to run *) (* Harvey P. Dale, Jan 18 2015 *)
Select[(5Range[1, 9999, 2])^2, Complement[IntegerDigits[#], {2, 5, 7}] == {} &] (* Alonso del Arte, Feb 19 2020 *)

fromTernary(n, d)=sum(i=0,d-1,[2,5,7][(n\3^i)%3+1]*10^i)
v=List([25]);for(d=0,16,for(n=0,3^d-1,if(issquare(t=225+1000*fromTernary(n,d)), listput(v,t); print1(t", ")))); Vec(v) \\ Charles R Greathouse IV, Dec 22 2012

a(n) = A030487(n)^2. - M. F. Hasler, Dec 23 2012

Extended and corrected by author, May 08 2000

a(17)-a(19) from Farideh Firoozbakht, Dec 09 2008

A275971 Numbers n such that the decimal digits of n^2 are all prime.

Original entry on oeis.org

5, 15, 85, 165, 235, 485, 1665, 1885, 4835, 5765, 7585, 15085, 15885, 16665, 18365, 18915, 22885, 27115, 27885, 50235, 57665, 58115, 72335, 85635, 87885, 150915, 166665, 182415, 194235, 194365, 229635, 240365, 268835, 503515, 507665, 524915, 568835, 570415, 577515, 581165

Offset: 1

Zak Seidov, Aug 15 2016

Apparently 5, 235 and 72335 are the only terms using digits {2,3,5,7}.
a(n)/5 = {1, 3, 17, 33, 47, 97, 333, 377, 967, 1153, 1517, 3017, 3177, 3333, ...}; terms b(n) that have n 3's must be in the sequence since (5 b(n))^2 yields the decimal number 2 followed by (n-1) 7's then n 2's, and ending in 5 (i.e., 225, 27225, 2772225). Thus 5 b(n) = {15, 165, 1665, 16665, etc.} appears in this sequence. - Michael De Vlieger, Aug 15 2016
All terms are odd multiples of 5 (A017329), i.e., must end in 5, which is the only digit whose square ends in a prime digit. The sequence contains A030487 as an infinite proper subsequence which in turn contains all numbers of the form (5*10^n-5)/3 (these are the above 5 b(n)) as a proper subsequence. - M. F. Hasler, Sep 16 2016

			72335^2 = 5232352225 = A191486(23).

Michel Marcus, Table of n, a(n) for n = 1..1000

Cf. A191486, A017329, A030487.

w = Boole@! PrimeQ@ # & /@ RotateLeft@ Range[0, 9]; Sqrt@ Select[Range[10^6]^2, Total@ Pick[DigitCount@ #, w, 1] == 0 &] (* Michael De Vlieger, Aug 15 2016 *)

is(n)=#setintersect(Set(digits(n^2)), [0, 1, 4, 6, 8, 9])==0 \\ Charles R Greathouse IV, Sep 16 2016

def aupto(limit):
  alst = []
  for k in range(1, limit+1):
    if set(str(k*k)) <= set("2357"): alst.append(k)
  return alst
print(aupto(10**6)) # Michael S. Branicky, May 15 2021

a(n) = sqrt(A191486(n)).

More terms from Michel Marcus, Aug 17 2016

A361821 Perfect powers in A329150.

Original entry on oeis.org

25, 27, 32, 225, 2025, 2197, 2500, 3025, 3375, 7225, 11025, 13225, 21952, 22500, 27000, 27225, 55225, 70225, 112225, 133225, 172225, 195112, 202500, 207025, 235225, 250000, 255025, 302500, 319225, 511225, 555025, 570025, 722500, 1102500, 1113025, 1177225, 1311025

Offset: 1

Bernard Schott, Mar 25 2023

No term has a digit 4, 6 or 8.

Subsequences of squares are listed in Crossrefs.

			32 is a term since A329147(21) = 32 = 2^5.
2197 is a term since A329147(194) = 2197 = 13^3.
235225 is a term since A329147(123113) = 235225 = 485^2.

Intersection of A001597 and A329150.
Cf. A329147.
Subsequences of squares with specified digits: A058426 (0,2,5), A053919 (2,3,5), A030485 (2,5,7), A191486 (2,3,5,7).

p[n_] := If[n > 0, Prime[n], 0]; ppQ[n_] := GCD @@ FactorInteger[n][[;; , 2]] > 1; seq[ndigmax_] := Module[{t = Table[FromDigits[Flatten@ IntegerDigits@ (p /@ IntegerDigits[n])], {n, 0, 10^ndigmax - 1}]}, Union@ Select[t, 0 < # < 10^ndigmax && ppQ[#] &]]; seq[6] (* Amiram Eldar, Mar 26 2023 *)

f(n) = if (n, fromdigits(concat(apply(d -> if (d, digits(prime(d)), [0]), digits(n)))), 0); \\ A329147
lista(nn) = my(list = List(), m); for (n=0, nn, m = f(n); if ((m <= nn) && ispower(m), listput(list, m));); vecsort(Set(list)); \\ Michel Marcus, Mar 26 2023

A288854 The unique longest sequence of squares where each number (after the first) is obtained by prefixing a single digit to its predecessor.

Original entry on oeis.org

25, 625, 5625, 75625, 275625

Offset: 1

Bernard Schott, Jun 18 2017

This chain with five squares is the longest which exists in this context, there is no such sequence of length >= 6.
There are also only four chains of maximal length 4 with:
-> 25, 225, 1225, 81225. These four squares are the first terms of A061839.
-> 25, 225, 4225, 34225.
-> 25, 225, 7225, 27225. These four squares are the first terms of A191486.
-> 25, 625, 5625, 15625.
There are also only three chains of maximal length 3 with:
-> 3025, 93025, 893025.
-> 30625, 330625, 3330625.
-> 50625, 950625, 4950625.
See Crux Mathematicorum links.

			25 = 5^2; 625 = 25^2; 5625 = 75^2; 75625 = 275^2; 275625 = 525^2.

L. Csirmaz, Problem 526, solution, Crux Mathematicorum, page 280, Vol.7, Nov. 81.
Friend H. Kierstead, Jr., Problem 526, partial solution, Crux Mathematicorum, page 87, Vol.7, Mar. 81.

Cf. A061839, A191486.

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A046034 Numbers whose digits are primes.

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A329147 Replace in n each nonzero digit d with prime(d).

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A030485 Squares composed of digits {2, 5, 7}.

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A275971 Numbers n such that the decimal digits of n^2 are all prime.

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A361821 Perfect powers in A329150.

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A288854 The unique longest sequence of squares where each number (after the first) is obtained by prefixing a single digit to its predecessor.

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