

A340907


Numbers m without zero digits such that pod(q) = pod(k) = pod(m) where q = k + pod(k) and k = m + pod(m) where pod is the product of digits, A007954.


3



262713, 267338, 283628, 342713, 351678, 432713, 451676, 516469, 516657, 516675, 622713, 634838, 651674, 716655, 728364, 851673, 857297, 916465, 1262713, 1267338, 1283628, 1342713, 1351678, 1432713, 1451676, 1516469, 1516657, 1516675, 1622713, 1634838, 1651674
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

The idea of this sequence comes from a remark of Amiram Eldar in Discussion section of A327750 (m + pod(m) = k with pod(k) = pod(m)) in September 2019.
Question: is it possible to get a longer string of integers with this rule?
The product of digits of a(n) is a multiple of 6. In terms below 10^10 however all products of digits of a(n) are a multiple of 36. Is that product a multiple of 36 for all a(n)?
The least term k such that k + 6 is here is k = 56516718. Are there consecutive terms that differ by less than 6? (End)


REFERENCES

Roman Fedorov, Alexei Belov, Alexander Kovaldzhi, and Ivan Yashchenko, MoscowMathematical Olympiads, 20002005, Level A, Problem 2, 2003; MSRI, 2011, p. 15 and 97.


LINKS



EXAMPLE

262713 + pod(262713) = 262713 + 504 = 263217, whose product of digits is also 504, and 263217 + 504 = 263721 whose product of digits is again 504; hence, m=262713, k=263217, q=263721 and pod(m)=pod(k)=pod(q)=504, so 262713 is a term.


MATHEMATICA

pod[n_] := Times @@ IntegerDigits[n]; seqQ[n_] := Module[{p = pod[n], k, q}, k = n + p; q = k + pod[k]; p > 0 && Equal @@ {p, pod[k], pod[q]}]; Select[Range[2*10^6], seqQ] (* Amiram Eldar, Jan 26 2021 *)


PROG

(PARI) isok(m) = my(pm=vecprod(digits(m)), k=m+pm, pk=vecprod(digits(k)), q=k+pk, pq=vecprod(digits(q))); pm && (pm==pk) && (pk==pq); \\ Michel Marcus, Jan 26 2021


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



