A239715 Primes of the form m = 5^i + 5^j - 1, where i > j >= 0.

5, 29, 149, 15629, 15649, 15749, 16249, 18749, 391249, 393749, 1968749, 9765629, 9781249, 244140749, 244218749, 292968749, 30517968749, 152587890649, 152587891249, 152587893749, 152597656249, 152636718749, 3814697281249, 3814697656249, 19073486328749, 95367441406249

Offset: 1

Author

Hieronymus Fischer, Apr 14 2014

Keywords

• nonn

Comments

The base-5 representation of a term 5^i + 5^j - 1 has base-5 digital sum = 1 + 4*j == 1 (mod 4).

In base-5 representation the first terms are 10, 104, 1044, 1000004, 1000044, 1000444, 1004444, 1044444, 100004444, 100044444, 1000444444, 10000000004, 10000444444, ...

All terms after the first have the last digit 9, since 5^i == 5 (mod 10), and thus 5^i + 5^j == 0 (mod 10).

All terms which have i > j > 1 end with the last 2 digits …49, since 5^k == 25 (mod 100) for k > 1, and thus 5^i + 5^j == 50 (mod 100).

All terms which have i > j > 1 end with the last 3 digits ...249, ...649, or ...749, since 5^k == 125 (mod 1000) or 5^k == 625 (mod 1000) for k > 2, and thus 5^i + 5^j == 250 (mod 1000), or 5^i + 5^j == 650 (mod 1000), or 5^i + 5^j == 750 (mod 1000).

Numbers m = 5^i + 5^j - 1 with odd i and j are not terms. Example: 78249 = 5^7 + 5^3 - 1 = 3*26083.

Examples

			a(1) = 5, since 5 = 5^1 + 5^0 - 1 is prime.
a(3) = 149, since 149 = 5^3 + 5^2 - 1 is prime.
		

Links

• Hieronymus Fischer, Table of n, a(n) for n = 1..50

Programs

Smalltalk

A239715
  "Answers an array of the first n terms of A239715.
  Uses method primesWhichAreDistinctPowersOf: b withOffset: d from A239712.
  Usage: n A239715
  Answer: #(5 29 ... ) [a(1) ... a(n)]"
  ^self primesWhichAreDistinctPowersOf: 5 withOffset: -1