OFFSET
1,2
COMMENTS
From David A. Corneth, May 25 2024: (Start)
a(n) cannot be equal to 2, 4, 6, 10, 12, 14, 18 (mod 20).
We can ease verification of terms in this sequence by rewriting A062273(k). A062273(k) can be written as blocks of 10 digits followed by some remaining digits like this:
A062273(k) = (B * Sum_{i = 0..(floor(k/10)-1)} (10^10)^i)*10^(k mod 10) + floor(B / (10 ^ (9 - (k-1) mod 10))) = (B * Sum_{i = 0..(floor(k/10)-1)} (10^10)^i)*10^(k mod 10) + floor(B / (10 ^ (9 - (k-1) mod 10))) where B is a block of 10 digits starting with (((k-1)*k)/2 + 1) mod 10 and each successive digit is one more than the previous except after 9 there is 0.
Using modular arithmetic and a simplification for geometric sums we can compute A062273(k) mod k and see if the result is 0.
Also terms are 11670334251, 12057978588, 15925928169. Any missing term must have gcd(t, floor(10^10/9)) > 1. (End)
LINKS
David A. Corneth, PARI program
EXAMPLE
From David A. Corneth, May 25 2024: (Start)
48 is in the sequence as 48 divides A062273(48). We may write A062273(48) as (9012345678 * Sum_{i = 0..(floor(48)-1)} (10^10)^i)*10^8 + 90123456
= 9012345678 * ((10^10)^4 - 1) / (10^10 - 1) * 10^8 + 90123456
= 1001371742 * ((10^10)^4 - 1) / ((10^10 - 1)/9) * 10^8 + 90123456 and using modular arithmetic (note that (10^10 - 1)/9 is coprime to 48 so we can compute the inverse of the modular class of (10^10 - 1)/9 mod 48. (End)
PROG
(PARI) \\ See PARI link
CROSSREFS
KEYWORD
base,nonn,more
AUTHOR
Amarnath Murthy, Nov 12 2002
EXTENSIONS
More terms from Mark Hudson (mrmarkhudson(AT)hotmail.com), Feb 11 2003
Offset corrected and a(12)-a(16) added by Amiram Eldar, May 21 2024
a(17)-a(20) from Michael S. Branicky, May 22 2024
a(21)-a(23) from David A. Corneth and Michael S. Branicky, May 24 2024
a(24)-a(31) from David A. Corneth, May 25 2024
STATUS
approved