A097228 Numbers n such that the product of digits of n equals the concatenation of pi(d)'s where d runs through the digits of n.

27, 38, 127, 138, 289, 298, 1127, 1138, 1289, 1298, 11127, 11138, 11289, 11298, 111127, 111138, 111289, 111298, 1111127, 1111138, 1111289, 1111298, 11111127, 11111138, 11111289, 11111298, 111111127, 111111138, 111111289, 111111298

Offset: 1

Author

Farideh Firoozbakht, Aug 15 2004

Keywords

• base
• nonn

Comments

This sequence is infinite because if n is in the sequence then the number with an arbitrary number of 1's in front of n is also in the sequence. Are 27, 38, 289 and 298 the only nontrivial terms (i.e., terms whose first digit is not 1) in this sequence? The next term is greater than 3*10^8.

There are no more nontrivial terms; i.e., all terms in the sequence are 27, 38, 289, 298 prepended with zero or more 1's. To see this, note that a nontrivial term must have at most 21 digits since 9^22 < 10^21, i.e., has 21 digits. Searching through all numbers in A009994 of at most 21 digits that do not start with 1 shows that there are no more nontrivial terms. - Chai Wah Wu, Aug 10 2017

Examples

			298 is in the sequence because 2*9*8 = 144 = concatenate(pi(2), pi(9), pi(8)) = concatenate(1, 4, 4).
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 11, 0, 0, 0, -10).

Crossrefs

Cf. A097227.

Programs

Mathematica

h[a_]:=(v1={};Do[l=Length[a];v1=Join[v1, IntegerDigits[a[[n]]]], {n, l}]; FromDigits[v1]);v={};Do[h1=IntegerDigits[n];l=Length[h1]; p=Product[h1[[k]], {k, l}];s=Sum[h1[[k]], {k, l}];If[p>0&& p==h[PrimePi[h1]], v=Append[v, n];Print[v]], {n, 300000000}]
LinearRecurrence[{0,0,0,11,0,0,0,-10},{27,38,127,138,289,298,1127,1138,1289,1298},30] (* Harvey P. Dale, Jan 01 2019 *)

Python

from _future_ import division
A097228_list = [27,38] + [1000*(10**k-1)//9+d for k in range(20) for d in [127,138,289,298]] # Chai Wah Wu, Aug 10 2017

Formula

From Chai Wah Wu, Aug 10 2017: (Start)

a(n) = 11*a(n-4) - 10*a(n-8) for n > 10.

G.f.: x*(-1600*x^9 - 1620*x^8 - 380*x^7 - 270*x^6 - 120*x^5 - 8*x^4 + 138*x^3 + 127*x^2 + 38*x + 27)/(10*x^8 - 11*x^4 + 1). (End)