A164322 Numbers m such that m = prime(P) + sigma(P), where P is product of digits of m.

14, 67673, 25688777

Offset

1,1

Comments

The product of the digits of next term (if it exists) is greater than 2*10^8.

The sequence is finite with all terms < 10^70. Indeed, since P is a product of primes 2, 3, 5, 7, we have sigma(P) < (2/1)*(3/2)*(5/4)*(7/6)*P = (35/8)*P. For an L-digit m, we have P <= 9^L, and using Dusart's upper bound for prime(9^L), we get prime(P) + sigma(P) < prime(9^L) + (35/8)*9^L < 10^(L-1) <= m whenever L > 70. - Max Alekseyev, Jun 12 2026

No other terms below 10^16. The sequence is likely complete. - Max Alekseyev, Jun 13 2026

Links

Table of n, a(n) for n=1..3.

Example

25688777 = prime(2*5*6*8*8*7*7*7) + sigma(2*5*6*8*8*7*7*7), so 25688777 is in the sequence.

Mathematica

Do[If[n=Prime[m]+DivisorSigma[1, m]; m==Apply[Times, IntegerDigits[n]], Print[n]], {m, 200000000}]

Crossrefs

Cf. A000040, A000203, A164323, A164324.

Sequence in context: A344668 A013754 A073940 * A372084 A159430 A395234

Adjacent sequences: A164319 A164320 A164321 * A164323 A164324 A164325

Keyword

base,bref,fini,more,nonn,changed

Author

Farideh Firoozbakht, Aug 13 2009

Status

approved