A107740 - OEIS

A107740

Number of numbers m such that prime(n) = m + (digit sum of m).

1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 2, 2, 1, 1, 0, 1, 1, 0, 1

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OFFSET

1,26

COMMENTS

a(A049084(A006378(n))) = 0; a(A049084(A048521(n))) > 0. [Corrected by Reinhard Zumkeller, Sep 27 2014]

a(n) <= 2 for n <= 10^5. Conjecture: sequence is bounded.

I would rather conjecture the opposite. Of course a(n) >= m implies n >= A006064(m), having more than A230857(m) digits, i.e., 14, 25 and 1111111111125 digits of n, for a(n) = 3, 4, 5. - M. F. Hasler, Nov 09 2018

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

FORMULA

a(n) = A230093(prime(n)), i.e.: A107740 = A230093 o A000040. - M. F. Hasler, Nov 08 2018

EXAMPLE

A000040(26) = 101 = 91 + (9 + 1) = 100 + (1 + 0 + 0): a(26) = # {91, 100} = 2.

MATHEMATICA

Table[p=Prime[n]; c=0; i=1; While[i<p, If[i+Total[IntegerDigits[i]]==p, c=c+1]; i++]; c, {n, 105}] (* Jayanta Basu, May 03 2013 *)

PROG

(Haskell)

a107740 n = length [() | let p = a000040 n,

m <- [max 0 (p - 9 * a055642 p) .. p - 1],