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%I #19 Nov 26 2022 18:12:10
%S 1,3,5,7,8,9,10,11,13,19,20,21,23,25,29,30,31,33,34,35,36,37,43,44,45,
%T 49,50,55,56,58,59,61,62,63,66,68,70,71,72,74,75,77,79,80,81,83,85,91,
%U 93,94,103,104,106,108,115,117,118,119,124,125,127,128,131,138,139,143,144,153,154,155,157
%N Numbers k such that A000217(k)^A000217(k+1) mod A000217(k+2) is a triangular number.
%C It appears that in most of these cases, A000217(k)^A000217(k+1) mod A000217(k+2) is either 1 or A000217(k).
%H Robert Israel, <a href="/A335575/b335575.txt">Table of n, a(n) for n = 1..10000</a>
%e a(3) = 5 is a member because A000217(5..7) are 15, 21, 28, and 15^21 == 15 (mod 28) where 15 is a triangular number.
%p tri:= n -> n*(n+1)/2:
%p istri:= n -> issqr(1+8*n):
%p select( n -> istri(tri(n) &^ tri(n+1) mod tri(n+2)), [$1..1000]);
%t Position[Partition[Accumulate[Range[200]],3,1],_?(OddQ[Sqrt[1+8*PowerMod[ #[[1]], #[[2]],#[[3]]]]]&),1,Heads->False]//Flatten (* _Harvey P. Dale_, Nov 26 2022 *)
%o (PARI) tri(n) = n*(n+1)/2; \\ A000217
%o isok(n) = ispolygonal(lift(Mod(tri(n), tri(n+2))^tri(n+1)), 3); \\ _Michel Marcus_, Jan 26 2021
%Y Cf. A000217, A340877.
%K nonn
%O 1,2
%A _J. M. Bergot_ and _Robert Israel_, Jan 26 2021
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