A339805 - OEIS

A339805

Numbers k such that the sum of decimal digits of k is the sum of primes dividing k+1 (with repetition).

5, 17, 47, 97, 98, 159, 279, 359, 485, 489, 749, 879, 1679, 1979, 2399, 2499, 3968, 5669, 6749, 7199, 7799, 8099, 8639, 9719, 12799, 19199, 25599, 31999, 37499, 39599, 44799, 68599, 78399, 78749, 79379, 94499, 134999, 143999, 146999, 161999, 172799, 175999, 194399, 199679, 209999, 218699, 259999

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OFFSET

1,1

COMMENTS

Numbers k such that A007953(k) = A001414(k+1).

If m is not divisible by 10, A007953(10^k*m-1) = A007953(m) - 1 + 9*k while A001414(10^k*m) = A001414(m) + 7*k. Thus if in addition A001414(m) - A007953(m) is odd and positive, 10^k*m-1 is in the sequence where k = (A001414(m) - A007953(m)+1)/2.

Are there infinitely many terms that do not end in 9?

LINKS

Robert Israel, Table of n, a(n) for n = 1..156

EXAMPLE

a(4) = 97 is in the sequence because the sum of digits of 97 is 9+7 = 16 and the sum of primes dividing 98=2*7*7 is 2+7+7 = 16.

MAPLE

sod:= n -> convert(convert(n, base, 10), `+`):

spf:= proc(n) local t; add(t[1]*t[2], t=ifactors(n)[2]) end proc:

select(t -> sod(t) = spf(t+1), [$1..10^6]);

PROG

(PARI) isok(m) = my(f=factor(m+1)); sumdigits(m) == f[, 1]~*f[, 2]; \\ Michel Marcus, Dec 18 2020

CROSSREFS