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a(n) = prime(n+1) mod (10^k) where k is the least positive integer such that floor(prime(n)/(10^k)) = floor(prime(n+1)/(10^k)) and prime(n) denotes the n-th prime number.
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%I #12 Jun 02 2020 14:09:54

%S 3,5,7,11,3,7,9,23,9,31,7,41,3,7,53,9,61,7,71,3,9,83,9,97,101,3,7,9,

%T 13,27,31,7,9,49,51,7,63,7,73,9,81,91,3,7,9,211,23,7,9,33,9,41,51,7,

%U 63,9,71,7,81,3,93,307,11,3,7,31,7,47,9,53,9,67,73,9

%N a(n) = prime(n+1) mod (10^k) where k is the least positive integer such that floor(prime(n)/(10^k)) = floor(prime(n+1)/(10^k)) and prime(n) denotes the n-th prime number.

%C In other words, a(n) is the smallest suffix to be overlaid on the decimal representation of the n-th prime number to obtain the next prime number.

%C This sequence has similarities with A274206; here we consider consecutive prime numbers, there consecutive nonnegative integers.

%C There are no two consecutive equal terms.

%H Rémy Sigrist, <a href="/A335301/b335301.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) <= prime(n+1) with equality iff prime(n+1) is the least prime number with its number of digits and leading digit.

%e For n = 42:

%e - prime(42) = 181 and prime(43) = 191,

%e - floor(181/(10^1)) = 18 <> 19 = floor(191/(10^1)),

%e - floor(181/(10^2)) = 1 = floor(191/(10^2)),

%e - so a(42) = 191 mod (10^2) = 91.

%o (PARI) { base=10; p=2; forprime (q=p+1, 379, for (k=0, oo, m=base^k; if (q\m == p\m, print1 (q%m", "); p=q; break))) }

%Y Cf. A274206, A335302 (binary variant).

%K nonn,base

%O 1,1

%A _Rémy Sigrist_, May 31 2020