The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A306701 Numbers k such that the sum of the digits and the sum of the square of the digits of k are prime factors of k. 1
 110, 111, 120, 133, 210, 803, 1010, 1011, 1020, 1100, 1101, 1110, 1200, 1330, 2010, 2023, 2100, 8030, 10010, 10011, 10020, 10100, 10101, 10110, 10200, 11000, 11001, 11010, 11100, 12000, 12110, 13300, 20010, 20100, 20230, 21000, 80300, 100010, 100011, 100020 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Subsequence of A062713. The corresponding pairs of primes factors are (2, 2), (3, 3), (3, 5), (7, 19), (3, 5), (11, 73), (2, 2), (3, 3), (3, 5), (2, 2), (3, 3), (3, 3), (3, 5), (7, 19), (3, 5), (7, 17), (3, 5), (11, 73), (2, 2), ... with many repetitions of some pairs. {a(n)} = E1 union E2 where E1 = {110, 111, 120, 133, 210, ...} is the subsequence of primitive values, and E2 = {1100, 1110, 1200, ...} is the subsequence of the non-primitive values. The subsequence E1 = F1 union F2 where F1 is the finite subsequence having distinct decimal digits. F1 starts with 120, 210, 803, 615784, 1742386, 2341678, 3059824, 3075914, 5910374, 6157840, 6285901, 8234716, 8734126, 8951024, 9150428, 12946078, ... with the corresponding pairs of prime factors (3, 5), (3, 5), (11, 73), (31, 191), (31, 179), (31, 179), (31, 199), (29, 181), (29, 181), (31, 191), (31, 211), (31, 179), (31, 179), (29, 191), (29, 191), (37, 251), ... We observe repetitions of some pairs : (3, 5), (31, 191), (31, 179), (29, 181), (29, 171), ... F2, the set of numbers having at least two identical digits, is infinite because it contains all the numbers of the form 10^k + 10^j + 1 with k > j > 0. - Giovanni Resta, May 08 2019 If k is in the sequence then so is 10*k. - David A. Corneth, May 08 2019 LINKS Georg Fischer, Table of n, a(n) for n = 1..300 EXAMPLE 120 is an element because 1 + 2 + 0 = 3 and 1^2 + 2^2 + 0^2 = 5 are prime factors of 120; 615784 is an element because 6 + 1 + 5 + 7 + 8 + 4 = 31 and 6^2 + 1^2 + 5^2 + 7^2 + 8^2 + 4^2 = 191 are prime factors of 615784 = 2^3 * 13 * 31 * 191. MAPLE with(numtheory):nn:=20000: for n from 1 to nn do: x:=convert(n, base, 10):n0:=nops(x): s1:=sum(‘x[i]’, ‘i’=1..n0):s2:=sum(‘x[i]^2’, ‘i’=1..n0):   if isprime(s1) and isprime(s2) and irem(n, s1)=0 and irem(n, s2)=0    then     printf(`%d, `, n):     else    fi: od: MATHEMATICA sd2Q[n_]:=Module[{idn=Total[IntegerDigits[n]], idn2=Total[ IntegerDigits[ n]^2], f= FactorInteger[ n][[All, 1]]}, AllTrue[{idn, idn2}, PrimeQ] && MemberQ[ f, idn]&&MemberQ[f, idn2]]; Select[Range[101000], sd2Q] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 15 2020 *) PROG (MATLAB) m=1; for u=1:200000     digit=dec2base(u, 10)-'0'; digit1=digit.^2;  s1=sum(digit); s2=sum(digit1);       if and(isprime(s1)==1, isprime(s2)==1)          if and(mod(u, s1)==0, mod(u, s2)==0)             sol(m)=u;             m=m+1;          end       end end sol % Marius A. Burtea, May 08 2019 (PARI) isok(n) = my(d=digits(n), sd = vecsum(d), sd2 = sum(k=1, #d, d[k]^2)); isprime(sd) && isprime(sd2) && !(n%sd) && !(n%sd2); \\ Michel Marcus, May 11 2019 CROSSREFS Cf. A003132, A007953, A062713 Sequence in context: A267138 A039724 A008944 * A106004 A113556 A286943 Adjacent sequences:  A306698 A306699 A306700 * A306702 A306703 A306704 KEYWORD nonn,base AUTHOR Michel Lagneau, May 08 2019 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified June 17 22:14 EDT 2021. Contains 345086 sequences. (Running on oeis4.)